Prove the identity: $\dfrac{1}{2} \left(\cos t + \sqrt{3} \sin t\right) = \cos \left(\dfrac{\pi}{3} - t\right)$
$\begin{aligned}
LHS & = \dfrac{1}{2} \left(\cos t + \sqrt{3} \sin t\right) \\\\
& = \dfrac{1}{2} \cos t + \dfrac{\sqrt{3}}{2} \sin t \\\\
& = \cos \dfrac{\pi}{3} \cos t + \sin \dfrac{\pi}{3} \sin t \\\\
& = \cos \left(\dfrac{\pi}{3} - t\right) = RHS
\end{aligned}$
Hence proved.
Prove the identity: $\dfrac{\sin x + \cos x}{\cos^3 x} = \tan^3 x + \tan^2 x + \tan x + 1$
$\begin{aligned}
RHS & = \tan^3 x + \tan^2 x + \tan x + 1 \\\\
& = \tan^2 x \left(\tan x + 1\right) + 1 \left(\tan x + 1\right) \\\\
& = \left(1 + \tan^2 x\right) \left(1 + \tan x\right) \\\\
& = \sec^2 x \left(1 + \tan x\right) \\\\
& = \dfrac{1}{\cos^2 x} \left(1 + \dfrac{\sin x}{\cos x}\right) \\\\
& = \dfrac{\sin x + \cos x}{\cos^3 x} = LHS
\end{aligned}$
Hence proved.
Prove the identity: $\left(\dfrac{1 + \sin \alpha}{1 + \cos \alpha}\right) \left(\dfrac{1 + \sec \alpha}{1 + \text{cosec} \alpha}\right) = \tan \alpha$
$\begin{aligned}
LHS & = \left(\dfrac{1 + \sin \alpha}{1 + \cos \alpha}\right) \left(\dfrac{1 + \sec \alpha}{1 + \text{cosec} \alpha}\right) \\\\
& = \left(\dfrac{1 + \sin \alpha}{1 + \cos \alpha}\right) \left(\dfrac{1 + \dfrac{1}{\cos \alpha}}{1 + \dfrac{1}{\sin \alpha}}\right) \\\\
& = \left(\dfrac{1 + \sin \alpha}{1 + \cos \alpha}\right) \left[\dfrac{\left(1 + \cos \alpha\right) /\ \cos \alpha}{\left(1 + \sin \alpha\right) /\ \sin \alpha}\right] \\\\
& = \dfrac{\sin \alpha}{\cos \alpha} \\\\
& = \tan \alpha = RHS
\end{aligned}$
Hence proved.
Solve the following system of equations: $\begin{cases} 5 \log_2 x = \log_2 y^3 - \log_{\sqrt{2}} 2 \\ \log_2 y = 8 - \log_{\sqrt{2}} x \end{cases}$
Given system of equations:
$5 \log_2 x = \log_2 y^3 - \log_{\sqrt{2}} 2$ $\;\;\; \cdots \; (1)$
$\log_2 y = 8 - \log_{\sqrt{2}} x$ $\;\;\; \cdots \; (2)$
We have from equation $(1)$, $\;$ $5 \log_2 x = 3 \log_2 y - \log_{\sqrt{2}} \left(\sqrt{2}\right)^2$
i.e. $\;$ $5 \log_2 x - 3 \log_2 y = - 2$ $\;\;\; \cdots \; (3)$
We have from equation $(2)$, $\;$ $\log_2 y = 8 - \dfrac{\log_2 x}{\log_2 \sqrt{2}}$
i.e. $\;$ $\log_2 y = 8 - \dfrac{\log_2 x}{\log_2 2^{\frac{1}{2}}}$
i.e. $\;$ $\log_2 y = 8 - \dfrac{\log_2 x}{\dfrac{1}{2}}$
i.e. $\;$ $2 \log_2 x + \log_2 y = 8$ $\;\;\; \cdots \; (4)$
Multiplying equation $(4)$ by $3$ gives
$6 \log_2 x + 3 \log_2 y = 24$ $\;\;\; \cdots \; (5)$
Adding equations $(3)$ and $(5)$ gives
$11 \log_2 x = 22$
i.e. $\;$ $\log_2 x = 2$
i.e. $\;$ $x = 2^2 = 4$
Substituting the value of $x$ in equation $(4)$ gives
$2 \log_2 4 + \log_2 y = 8$
i.e. $\;$ $\log_2 y = 8 - 2 \log_2 2^2$
i.e. $\;$ $\log_2 y = 8 - 2 \times 2 = 4$
i.e. $\;$ $y = 2^4 = 16$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(4, \; 16\right) \right\}$
Solve the following system of equations: $\begin{cases} x + y = 4 + \sqrt{y^2 + 2} \\ \log x - 2 \log 2 = \log \left(1 + \dfrac{1}{2} y\right) \end{cases}$
Given system of equations:
$x + y = 4 + \sqrt{y^2 + 2}$ $\;\;\; \cdots \; (1)$
$\log x - 2 \log 2 = \log \left(1 + \dfrac{1}{2} y\right)$ $\;\;\; \cdots \; (2)$
We have from equation $(2)$,
$\log x - \log 4 = \log \left(1 + \dfrac{y}{2}\right)$
i.e. $\;$ $\log \left(\dfrac{x}{4}\right) = \log \left(1 + \dfrac{y}{2}\right)$ $\;\;\; \cdots \; (3)$
Taking antilog on both sides of equation $(3)$ gives
$\dfrac{x}{4} = 1 + \dfrac{y}{2}$
i.e. $\;$ $x - 4 = 2y$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$ equation $(1)$ becomes
$2y + y = \sqrt{y^2 +2}$
i.e. $\;$ $3y = \sqrt{y^2 + 2}$
i.e. $\;$ $9y^2 = y^2 + 2$
i.e. $\;$ $8y^2 = 2$
i.e. $\;$ $y^2 = \dfrac{1}{4}$ $\implies$ $y = \pm \dfrac{1}{2}$
When $y = \dfrac{-1}{2}$, we have from equation $(4)$
$x = 4 - 2 \times \dfrac{1}{2} = 3$
Substituting $\left(x, y\right) = \left(3, \dfrac{-1}{2}\right)$ in equation $(1)$ gives
$3 - \dfrac{1}{2} = 4 + \sqrt{\dfrac{1}{4} + 2}$
i.e. $\;$ $\dfrac{5}{2} = 4 + \dfrac{3}{2}$
i.e. $\;$ $\dfrac{5}{2} = \dfrac{11}{2}$ $\;$ which is not true.
$\implies$ $\left(3, \dfrac{-1}{2}\right)$ is not a valid solution.
When $y = \dfrac{1}{2}$, we have from equation $(4)$
$x = 4 + 2 \times \dfrac{1}{2} = 5$
Substituting $\left(x, y\right) = \left(5, \dfrac{1}{2}\right)$ in equation $(1)$ gives
$5 + \dfrac{1}{2} = 4 + \sqrt{\dfrac{1}{4} + 2}$
i.e. $\;$ $\dfrac{11}{2} = 4 + \dfrac{3}{2}$
i.e. $\;$ $\dfrac{11}{2} = \dfrac{11}{2}$ $\;$ which is true.
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(5, \; \dfrac{1}{2}\right) \right\}$
Solve the following system of equations: $\begin{cases} \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{2}{15} \\ \log_3 x + \log_3 y = 1 + \log_3 5 \end{cases}$
Given system of equations:
$\dfrac{1}{x} - \dfrac{1}{y} = \dfrac{2}{15}$ $\;\;\; \cdots \; (1)$
$\log_3 x + \log_3 y = 1 + \log_3 5$ $\;\;\; \cdots \; (2)$
We have from equation $(2)$,
$\log_3 x + \log_3 y - \log_3 5 = 1$
i.e. $\;$ $\log_3 \left(\dfrac{x \; y}{5}\right) = 1$
i.e. $\;$ $\dfrac{x \; y}{5} = 3^1 = 3$
$\implies$ $x = \dfrac{15}{y}$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(1)$ becomes
$\dfrac{y}{15} - \dfrac{1}{y} = \dfrac{2}{15}$
i.e. $\;$ $y^2 - 2y - 15 = 0$
i.e. $\;$ $\left(y - 5\right) \left(y + 3\right) = 0$
i.e. $\;$ $y = 5$ $\;$ or $\;$ $y = -3$
When $y = -3$, the term $\log_3 y$ in equation $(2)$ becomes $\log_3 \left(-3\right)$.
But logarithm of a negative number is not defined.
$\therefore \;$ $y = -3$ is not a valid solution.
When $y = 5$, we have from equation $(3)$, $\;\;$ $x = \dfrac{15}{5} = 3$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(3, \; 5\right) \right\}$
Solve the following system of equations: $\begin{cases} \log_2 x + 2 \log_2 y = 3 \\ x^2 + y^4 = 16 \end{cases}$
Given system of equations:
$\log_2 x + 2 \log_2 y = 3$ $\;\;\; \cdots \; (1)$
$x^2 + y^4 = 16$ $\;\;\; \cdots \; (2)$
Equation $(1)$ $\implies$ $\log_2 x + \log_2 y^2 = 3$
i.e. $\;$ $\log_2 \left(x y^2\right) = 3$
i.e. $\;$ $x y^2 = 2^3 = 8$
i.e. $\;$ $y^2 = \dfrac{8}{x}$ $\implies$ $y^4 = \dfrac{64}{x^2}$ $\;\;\; \cdots \; (3)$
Substituting the value of $y^4$ from equation $(3)$ in equation $(2)$ gives
$x^2 + \dfrac{64}{x^2} = 16$
i.e. $\;$ $x^4 - 16 x^2 + 64 = 0$
i.e. $\;$ $\left(x^2 - 8\right)^2 = 0$
i.e. $\;$ $x^2 - 8 = 0$
i.e. $\;$ $x^2 = 8$ $\implies$ $x = \pm 2 \sqrt{2}$
When $x = -2 \sqrt{2}$, the term $\log_2 x$ in equation $(1)$ becomes $\log_2 \left(-2\sqrt{2}\right)$.
But logarithm of a negative number is not defined.
$\implies$ $x - 2 \sqrt{2}$ is not a valid solution.
When $x = + 2 \sqrt{2}$, we have from from equation $(1)$
$\log_2 2 \sqrt{2} + 2 \log_2 y = 3$
i.e. $\;$ $\log_2 2 + \log_2 2^{\frac{1}{2}} + 2 \log_2 y = 3$
i.e. $\;$ $1 + \dfrac{1}{2} + 2 \log_2 y = 3$
i.e. $\;$ $2 \log_2 y = \dfrac{3}{2}$
i.e. $\;$ $\log_2 y = \dfrac{3}{4}$
$\implies$ $y = 2^{\frac{3}{4}} = \left(2^3\right)^{\frac{1}{4}} = \sqrt[4]{8}$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2 \sqrt{2}, \; \sqrt[4]{8}\right) \right\}$
Solve the following system of equations: $\begin{cases} \log_4 x - \log_x y = \dfrac{7}{6} \\ xy = 16 \end{cases}$
Given system of equations:
$\log_4 x - \log_x y = \dfrac{7}{6}$ $\;\;\; \cdots \; (1)$
$xy = 16$ $\;\;\; \cdots \; (2)$
Equation $(1)$ $\implies$ $6 \log_4 x - 6 \log_x y = 7$ $\;\;\; \cdots \; (3)$
Equation $(2)$ $\implies$ $y = \dfrac{16}{x}$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$, equation $(3)$ becomes
$6 \log_4 x - 6 \log_x \left(\dfrac{16}{x}\right) = 7$
i.e. $\;$ $6 \log_4 x - 6 \log_x 16 + 6 \log_x x = 7$
i.e. $\;$ $6 \log_4 x - 6 \left(\dfrac{\log_4 16}{\log_4 x}\right) + 6 \times 1 = 7$
i.e. $\;$ $6 \log_4 x - 6 \left(\dfrac{2}{\log_4 x}\right) = 1$
i.e. $\;$ $6 \left(\log_4 x\right)^2 - \log_4 x - 12 = 0$
i.e. $\;$ $6 \left(\log_4 x\right)^2 - 9 \log_4 x + 8 \log_4 x - 12 = 0$
i.e. $\;$ $3 \log_4 x \left(2 \log_4 x - 3\right) + 4 \left(2 \log_4 x - 3\right) = 0$
i.e. $\;$ $\left(3 \log_4 x + 4\right) \left(2 \log_4 x - 3\right) = 0$
i.e. $\;$ $\log_4 x = \dfrac{-4}{3}$ $\;$ or $\;$ $\log_4 x = \dfrac{3}{2}$
i.e. $\;$ $x = 4^{\frac{-4}{3}} = \left(4^{-4}\right)^{\frac{1}{3}} = \dfrac{1}{\sqrt[3]{256}}$ $\;$ is an irrational number
or $\;$ $x = 4^{\frac{3}{2}} = \left(4^3\right)^{\frac{1}{2}} = \sqrt{64} = 8$
When $\;$ $x = \dfrac{1}{\sqrt[3]{256}}$, $\;$ the base of the term $\;$ $\log_x y$ $\;$ in equation $(1)$ becomes irrational.
But, base of a logarithm cannot be irrational.
$\therefore \;$ $x = \dfrac{1}{\sqrt[3]{256}}$ $\;$ is not a valid solution.
When $\;$ $x = 8$, $\;$ we have from equation $(4)$, $\;$ $y = \dfrac{16}{8} = 2$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(8, \; 2\right) \right\}$
Solve the following system of equations: $\begin{cases} x^{\log y} = 2 \\ xy = 20 \end{cases}$
Given system of equations:
$x^{\log y} = 2$ $\;\;\; \cdots \; (1)$
$xy = 20$ $\;\;\; \cdots \; (2)$
Taking log on both sides of equation $(1)$ gives
$\log \left(x^{\log y}\right) = \log 2$
i.e. $\;$ $\log y \times \log x = \log 2$ $\;\;\; \cdots \; (3)$
Equation $(2)$ $\implies$ $x = \dfrac{20}{y}$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$, equation $(3)$ becomes
$\log y \times \log \left(\dfrac{20}{y}\right) = \log 2$
i.e. $\;$ $\log y \left[\log 20 - \log y\right] = \log 2$
i.e. $\;$ $\left(\log y\right)^2 - \log 20 \; \log y + \log 2 = 0$
i.e. $\;$ $\left(\log y\right)^2 - \left[\log \left(10 \times 2\right)\right] \log y + \log 2 = 0$
i.e. $\;$ $\left(\log y\right)^2 - \left[\log 10 + \log 2\right] \log y + \log 2 = 0$
i.e. $\;$ $\left(\log y\right)^2 - \left[1 + \log 2\right] \log y + \log 2 = 0$
i.e. $\;$ $\left(\log y\right)^2 - \log y - \log 2 \; \log y + \log 2 = 0$
i.e. $\;$ $\log y \left(\log y - 1\right) - \log 2 \left(\log y - 1\right) = 0$
i.e. $\;$ $\left(\log y -1\right) \left(\log y - \log 2\right) = 0$
i.e. $\;$ $\log y = 1$ $\;$ or $\;$ $\log y = \log 2$
i.e. $\;$ $y = 10^1 = 10$ $\;$ or $\;$ $y = 2$
When $\;$ $y = 10$, $\;$ we have from equation $(4)$, $\;$ $x = \dfrac{20}{10} = 2$
When $\;$ $y = 2$, $\;$ we have from equation $(4)$, $\;$ $x = \dfrac{20}{2} = 10$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2, \; 10\right), \; \left(10, \; 2\right) \right\}$
Solve the following system of equations: $\begin{cases} y - \log_3 x = 1 \\ x^y = 3^{12} \end{cases}$
Given system of equations:
$y - \log_3 x = 1$ $\;\;\; \cdots \; (1)$
$x^y = 3^{12}$ $\;\;\; \cdots \; (2)$
Equation $(1)$ $\implies$ $\log_3 x = y - 1$
i.e. $\;$ $x = 3^{y - 1}$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(2)$ becomes
$\left(3^{y - 1}\right)^y = 3^{12}$
i.e. $\;$ $y \left(y - 1\right) = 12$
i.e. $\;$ $y^2 - y - 12 = 0$
i.e. $\;$ $\left(y - 4\right) \left(y + 3\right) = 0$
i.e. $\;$ $y = 4$ $\;$ or $\;$ $y = -3$
When $\;$ $y = 4$, $\;$ we have from equation $(3)$
$x = 3^{4 - 1} = 3^3 = 27$
When $\;$ $y = -3$, $\;$ we have from equation $(3)$
$x = 3^{-3 - 1} = 3^{-4} = \dfrac{1}{81}$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(27, \; 4\right), \; \left(\dfrac{1}{81}, \; -3\right) \right\}$
Solve the following system of equations: $\begin{cases} 3^x \times 2^y = 972 \\ \log_{\sqrt{3}} \left(x - y\right) = 2 \end{cases}$
Given system of equations:
$3^x \times 2^y = 972$ $\;\;\; \cdots \; (1)$
$\log_{\sqrt{3}} \left(x - y\right) = 2$ $\;\;\; \cdots \; (2)$
Equation $(2)$ $\implies$ $x - y = \sqrt{3}^2 = 3$
i.e. $\;$ $x = y + 3$ $\;\;\; \cdots \; (3)$
Substituting the value of $x$ from equation $(3)$ in equation $(1)$ gives
$3^{y + 3} \times 2^y = 972$
i.e. $\;$ $3^y \times 3^3 \times 2^y = 972$
i.e. $\;$ $6^y = \dfrac{972}{27} = 36 = 6^2$
$\implies$ $y = 2$
Substituting the value of $y$ in equation $(3)$ gives $\;\;$ $x = 2 + 3 = 5$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(5, \; 2\right) \right\}$
Solve the following system of equations: $\begin{cases} 4^{\frac{x}{y}} - 3 \times 4^{\frac{5y-x}{y}} = 16 \\ \sqrt{x} - \sqrt{2y} = \sqrt{12} - \sqrt{8} \end{cases}$
Given system of equations:
$4^{\frac{x}{y}} - 3 \times 4^{\frac{5y-x}{y}} = 16$ $\;\;\; \cdots \; (1)$
$\sqrt{x} - \sqrt{2y} = \sqrt{12} - \sqrt{8}$ $\;\;\; \cdots \; (2)$
We have from equation $(1)$
$4^{\frac{x}{y}} - 3 \times 4^5 \times 4^{\frac{-x}{y}} = 16$
i.e. $\;$ $4^{\frac{x}{y}} - \dfrac{3072}{4^{\frac{x}{y}}} = 16$
i.e. $\;$ $\left(4^{\frac{x}{y}}\right)^2 - 16 \times 4^{\frac{x}{y}} - 3072 = 0$
i.e. $\;$ $\left(4^{\frac{x}{y}}\right)^2 - 64 \times 4^{\frac{x}{y}} + 48 \times 4^{\frac{x}{y}} - 3072 = 0$
i.e. $\;$ $4^{\frac{x}{y}} \left(4^{\frac{x}{y}} - 64\right) + 48 \left(4^{\frac{x}{y}} - 64\right) = 0$
i.e. $\;$ $\left(4^{\frac{x}{y}} - 64\right) \left(4^{\frac{x}{y}} + 48\right) = 0$
i.e. $\;$ $4^{\frac{x}{y}} = 64$ $\;$ or $\;$ $4^{\frac{x}{y}} = -48$
i.e. $\;$ $\dfrac{x}{y} = \log_4 64$ $\;$ or $\;$ $\dfrac{x}{y} = \log_4 \left(-48\right)$
But logarithm of a negative number is not defined.
$\therefore \;$ $\dfrac{x}{y} = \log_4 \left(-48\right)$ $\;$ is not a valid solution.
Now, $\;$ $\dfrac{x}{y} = \log_4 64$ $\implies$ $\dfrac{x}{y} = \log_4 4^3$
i.e. $\;$ $\dfrac{x}{y} = 3 \log_4 4 = 3$
$\implies$ $x = 3y$ $\;\;\; \cdots \; (3)$
Substituting the value of $x$ in equation $(2)$ gives
$\sqrt{3y} - \sqrt{2y} = \sqrt{12} - \sqrt{8}$
i.e. $\;$ $\sqrt{y} \left(\sqrt{3} - \sqrt{2}\right) = 2 \sqrt{3} - 2 \sqrt{2}$
i.e. $\;$ $\sqrt{y} \left(\sqrt{3} - \sqrt{2}\right) = 2 \left(\sqrt{3} - \sqrt{2}\right)$
i.e. $\;$ $\sqrt{y} = 2$ $\implies$ $y = 4$
Substituting the value of $y$ in equation $(3)$ gives $\;$ $x = 3 \times 4 = 12$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(12, \; 4\right) \right\}$
Solve the following system of equations: $\begin{cases} 7^x - 16y = 0 \\ 4^x - 49y = 0 \end{cases}$
Given system of equations:
$7^x - 16y = 0$ $\;\;\; \cdots \; (1)$
$4^x - 49y = 0$ $\;\;\; \cdots \; (2)$
Multiplying equation $(1)$ with $49$ gives
$49 \times 7^x - 784 y = 0$
i.e. $\;$ $7^2 \times7^x - 784 y = 0$
i.e. $\;$ $7^{x + 2} = 784 y$ $\;\;\; \cdots \; (1a)$
Multiplying equation $(2)$ with $16$ gives
$16 \times 4^x - 784 y = 0$
i.e. $\;$ $4^2 \times4^x - 784 y = 0$
i.e. $\;$ $4^{x + 2} = 784 y$ $\;\;\; \cdots \; (2a)$
We have from equations $(1)$ and $(2)$
$7^{x+2} = 4^{x+2}$ $\;\;\; \cdots \; (3)$
i.e. $\;$ $\dfrac{7^{x+2}}{4^{x+2}} = 1$
i.e. $\;$ $\left(\dfrac{7}{4}\right)^{x+2} = 1$
i.e. $\;$ $x + 2 = \log_{\frac{7}{4}} 1$
i.e. $\;$ $x + 2 = 0$ $\implies$ $x = - 2$
Substituting the value of $x$ in equation $(1a)$ gives
$7^{-2 + 2} = 784 y$
i.e. $\;$ $7^0 = 784 y$
i.e. $\;$ $784 y = 1$ $\implies$ $y = \dfrac{1}{784}$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(-2, \; \dfrac{1}{784}\right) \right\}$
Solve the following system of equations: $\begin{cases} 3 \times 2^x + 2 \times 3^y = 2.75 \\ 2^x - 3^y = -0.75 \end{cases}$
Given system of equations:
$3 \times 2^x + 2 \times 3^y = 2.75$ $\;\;\; \cdots \; (1)$
$2^x - 3^y = -0.75$ $\;\;\; \cdots \; (2)$
Multiply equation $(2)$ with $2$ and adding to equation $(1)$ gives
$5 \times 2^x = 1.25$
i.e. $\;$ $2^x = 0.25 = \dfrac{1}{4} = \left(\dfrac{1}{2}\right)^2 = 2^{-2}$
i.e. $\;$ $x = \log_2 2^{-2} = -2 \log_2 2 = -2$
Substituting the value of $x$ in equation $(2)$ gives
$2^{-2} - 3^y = -0.75$
i.e. $\;$ $3^y = 0.25 + 0.75 = 1$
i.e. $\;$ $y = \log_3 1 = 0$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(-2, 0\right) \right\}$
Solve the following system of equations: $\begin{cases} \log_4 x - \log_2 y = 0 \\ x^2 - 5y^2 + 4 = 0 \end{cases}$
Given system of equations:
$\log_4 x - \log_2 y = 0$ $\;\;\; \cdots \; (1)$
$x^2 - 5y^2 + 4 = 0$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be written as $\;\;$ $\dfrac{\log_2 x}{\log_2 4} = \log_2 y$
i.e. $\;$ $\dfrac{\log_2 x}{\log_2 2^2} = \log_2 y$
i.e. $\;$ $\dfrac{\log_2 x}{2 \log_2 2} = \log_2 y$
i.e. $\;$ $\dfrac{\log_2 x}{2} = \log_2 y$
i.e. $\;$ $\log_2 x = 2 \log_2 y$
i.e. $\;$ $\log_2 x = \log_2 y^2$ $\;\;\; \cdots \; (1a)$
Taking antilog to the base $2$ on both sides of equation $(1a)$ gives
$x = y^2$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(2)$ becomes
$x^2 - 5x^2 + 4 = 0$
i.e. $\;$ $\left(x - 4\right) \left(x - 1\right) = 0$
i.e. $\;$ $x = 4$ $\;$ or $\;$ $x = 1$
Substituting the values of $x$ in equation $(3)$ give
when $\;$ $x = 4$ $\implies$ $y^2 = 4$ $\implies$ $y = \pm 2$
when $\;$ $x = 1$ $\implies$ $y^2 = 1$ $\implies$ $y = \pm 1$
When $\;$ $y = -2$ $\;$ or $\;$ $y = -1$, $\;$ the term $\;$ $\log_2 y$ $\;$ in equation $(1)$ becomes $\;$ $\log_2 \left(-2\right)$ $\;$ and $\;$ $\log_2 \left(-1\right)$ $\;$ respectively.
But logarithm of a negative number is not defined.
$\implies$ $y = -2$ $\;$ and $\;$ $y = -1$ $\;$ are not valid solutions to the given system of equations.
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(4, 2\right), \; \left(1, 1\right) \right\}$
Solve the following system of equations: $\begin{cases} 2^x + 2^y = 12 \\ x + y = 5 \end{cases}$
Given system of equations:
$2^x + 2^y = 12$ $\;\;\; \cdots \; (1)$
$x + y = 5$ $\;\;\; \cdots \; (2)$
Let $\;$ $2^x = p$ $\;\;\; \cdots \; (3a)$
$\implies$ $x = \log_2 p$ $\;\;\; \cdots \; (3b)$
and $\;$ $2^y = q$ $\;\;\; \cdots \; (4a)$
$\implies$ $y = \log_2 q$ $\;\;\; \cdots \; (4b)$
In view of equations $(3a)$ and $(4a)$ equation $(1)$ becomes
$p + q = 12$ $\;\;\; \cdots \; (5)$
In view of equations $(3b)$ and $(4b)$ equation $(2)$ becomes
$\log_2 p + \log_ q = 5$
i.e. $\;$ $\log_2 \left(pq\right) = 5$
i.e. $\;$ $pq = 2^5 = 32$ $\;\;\; \cdots \; (6)$
Now, $\;$ $\left(p - q\right)^2 = \left(p + q\right)^2 - 4pq$
i.e. $\;$ $\left(p - q\right)^2 = 144 - 128 = 16$ $\;\;\;$ [by equations $(5)$ and $(6)$]
i.e. $\;$ $p - q = 4$ $\;\;\; \cdots \; (7a)$ $\;$ or $\;$ $p - q = -4$ $\;\;\; \cdots \; (7b)$
Solving equations $(5)$ and $(7a)$ simultaneously gives
$2p = 16$ $\implies$ $p = 8$ $\;\;\; \cdots \; (8a)$
Substituting the value of $p$ from equation $(8a)$ in equation $(5)$ gives
$q = 4$ $\;\;\; \cdots \; (8b)$
Solving equations $(5)$ and $(7b)$ simultaneously gives
$2p = 8$ $\implies$ $p = 4$ $\;\;\; \cdots \; (9a)$
Substituting the value of $p$ from equation $(9a)$ in equation $(5)$ gives
$q = 8$ $\;\;\; \cdots \; (9b)$
In view of equations $(8a)$ and $(8b)$, equations $(3b)$ and $(4b)$ become
$x = \log_2 8 = \log_2 2^3 = 3 \log_2 2 = 3$
$y = \log_2 4 = \log_2 2^2 = 2 \log_2 2 = 2$
In view of equations $(9a)$ and $(9b)$, equations $(3b)$ and $(4b)$ become
$x = \log_2 4 = \log_2 2^2 = 2 \log_2 2 = 2$
$y = \log_2 8 = \log_2 2^3 = 3 \log_2 2 = 3$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2, 3\right), \; \left(3, 2\right) \right\}$
Solve the following system of equations: $\begin{cases} 3^{2x} - 2^y = 77 \\ 3^x - 2^{\frac{y}{2}} = 7 \end{cases}$
Given system of equations:
$3^{2x} - 2^y = 77$ $\;\;\; \cdots \; (1)$
$3^x - 2^{\frac{y}{2}} = 7$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be written as
$\left(3^x\right)^2 - \left(2^{\frac{y}{2}}\right)^2 = 77$
i.e. $\;$ $\left(3^x + 2 ^{\frac{y}{2}}\right) \left(3^x - 2^{\frac{y}{2}}\right) = 77$
i.e. $\;$ $\left(3^x + 2^{\frac{y}{2}}\right) \times 7 = 77$ $\;\;\;$ [in view of equation $(2)$]
i.e. $\;$ $3^x + 2^{\frac{y}{2}} = 11$ $\;\;\; \cdots \; (3)$
Adding equations $(2)$ and $(3)$ gives
$2 \times 3^x = 18$
i.e. $\;$ $3^x = 9 = 3^2$
$\implies$ $x = 2$
Substituting the value of $x$ in equation $(3)$ gives
$3^2 + 2^{\frac{y}{2}} = 11$
i.e. $\;$ $2^{\frac{y}{2}} = 2 = 2^1$
$\implies$ $\dfrac{y}{2} = 1$ $\implies$ $y = 2$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2, 2\right) \right\}$
Solve the following system of equations: $\begin{cases} 4^{\left(x-y\right)^2 - 1} = 1 \\ 5^{x + y} = 125 \end{cases}$
Given system of equations:
$4^{\left(x-y\right)^2 - 1} = 1$ $\;\;\; \cdots \; (1)$
$5^{x + y} = 125$ $\;\;\; \cdots \; (2)$
Equation $(1)$ $\implies$ $\left(x - y\right)^2 - 1 = \log_4 1$
i.e. $\;$ $\left(x - y\right)^2 - 1 = 0$
i.e. $\;$ $\left(x - y\right)^2 = 1$
i.e. $\;$ $x - y = 1$ $\;\;\; \cdots \; (3a)$ $\;$ or $\;$ $x - y = -1$ $\;\;\; \cdots \; (3b)$
Equation $(2)$ $\implies$ $x + y = \log_5 125$
i.e. $\;$ $x + y = \log_5 5^3$
i.e. $\;$ $x + y = 3 \log_5 5$
i.e. $\;$ $x + y = 3$ $\;\;\; \cdots \; (4)$
Solving equations $(3a)$ and $(4)$ simultaneously [adding equations $(3a)$ and $(4)$] gives
$2x = 4$ $\implies$ $x = 2$
Substituting the value of $x$ in equation $(4)$ gives
$y = 3 - 2 = 1$
Solving equations $(3b)$ and $(4)$ simultaneously [adding equations $(3b)$ and $(4)$] gives
$2x = 2$ $\implies$ $x = 1$
Substituting the value of $x$ in equation $(4)$ gives
$y = 3 - 1 = 2$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(1, 2\right), \; \left(2, 1\right) \right\}$
Solve the following system of equations: $\begin{cases} 4^{x+y} = 128 \\ 5^{3x-2y-3} = 1 \end{cases}$
Given system of equations:
$4^{x+y} = 128$ $\;\;\; \cdots \; (1)$
$5^{3x-2y-3} = 1$ $\;\;\; \cdots \; (2)$
Equation $(1)$ $\implies$ $x + y = \log_4 128$
i.e. $\;$ $x + y = \log_4 2^7$
i.e. $\;$ $x + y = 7 \log_4 2$
i.e. $\;$ $x + y = 7 \log_4 4^{\frac{1}{2}}$
i.e. $\;$ $x + y = \dfrac{7}{2} \log_4 4$
i.e. $\;$ $x + y = \dfrac{7}{2}$
i.e. $\;$ $2x + 2y = 7$ $\;\;\; \cdots \; (3)$
Equation $(2)$ $\implies$ $3x - 2y - 3 = \log_5 1$
i.e. $\;$ $3x - 2y - 3 = 0$
i.e. $\;$ $3x - 2y = 3$ $\;\;\; \cdots \; (4)$
Solving equations $(3)$ and $(4)$ simultaneously [adding equations $(3)$ and $(4)$] gives
$5x = 10$ $\implies$ $x = 2$
Substituting the value of $x$ in equation $(3)$ gives
$y = \dfrac{7 - 4}{2} = \dfrac{3}{2}$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2, \dfrac{3}{2}\right) \right\}$
Solve the following system of equations: $\begin{cases} \log \sqrt{5 - x} + \log 2 = \log \left(x + 3\right) \\ x^2 + 7x - 8 = 0 \end{cases}$
Given system of equations:
$\log \sqrt{5 - x} + \log 2 = \log \left(x + 3\right)$ $\;\;\; \cdots \; (1)$
$x^2 + 7x - 8 = 0$ $\;\;\; \cdots \; (2)$
Solving equation $(2)$ gives
$\left(x + 8\right) \left(x - 1\right) = 0$
i.e. $\;$ $x = -8$ $\;$ or $\;$ $x = 1$
When $\;$ $x = -8$, $\;$ the term $\;$ $\log \left(x + 3\right)$ $\;$ in equation $(1)$ becomes $\;$ $\log \left(-8 + 3\right) = \log \left(-5\right)$
But logarithm of a negative number is not defined.
$\therefore \;$ $x = -8$ $\;$ is not a valid solution to the given pair of equations.
Substituting $\;$ $x = 1$ $\;$ in equation $(1)$ gives
$\log \sqrt{5 - 1} + \log 2 = \log \left(1 + 3\right)$
i.e. $\;$ $\log \sqrt{4} + \log 2 = \log 4$
i.e. $\;$ $\log 2 + \log 2 = \log 4$
i.e. $\;$ $2 \log 2 = \log 4$
i.e. $\;$ $\log 2^2 = \log 4$ $\implies$ $\log 4 = \log 4$
$\therefore \;$ $x = 1$ $\;$ satisfies both equations $(1)$ and $(2)$.
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{1 \right\}$
Solve the equation: $\;$ $2 \log_2 \left(\dfrac{x - 7}{x - 1}\right) + \log_2 \left(\dfrac{x - 1}{x + 1}\right) = 1$
Given equation: $\;\;$ $2 \log_2 \left(\dfrac{x - 7}{x - 1}\right) + \log_2 \left(\dfrac{x - 1}{x + 1}\right) = 1$
i.e. $\;$ $\log_2 \left(\dfrac{x - 7}{x - 1}\right)^2 + \log_2 \left(\dfrac{x - 1}{x + 1}\right) = 1$
i.e. $\;$ $\log_2 \left[\left(\dfrac{x - 7}{x - 1}\right)^2 \times \left(\dfrac{x - 1}{x + 1}\right)\right] = 1$
i.e. $\;$ $\dfrac{\left(x - 7\right)^2}{\left(x + 1\right) \left(x - 1\right)} = 2^1 = 2$
i.e. $\;$ $x^2 - 14x + 49 = 2x^2 - 2$
i.e. $\;$ $x^2 + 14x - 51 = 0$
i.e. $\;$ $\left(x + 17\right) \left(x - 3\right) = 0$
i.e. $\;$ $x + 17 = 0$ $\;$ or $\;$ $x - 3 = 0$
i.e. $\;$ $x = -17$ $\;$ or $\;$ $x = 3$
When $\;$ $x = 3$, $\;$ the term $\;$ $\log_2 \left(\dfrac{x - 7}{x - 1}\right)$ $\;$ in the given problem becomes
$\log_2 \left(\dfrac{3 - 7}{3 - 1}\right) = \log \left(-2\right)$
But, logarithm of a negative number is not defined.
When $\;$ $x = -17$, $\;$ the given problem becomes
$2 \log_2 \left(\dfrac{-17 - 7}{-17 - 1}\right) + \log_2 \left(\dfrac{-17 - 1}{-17 + 1}\right) = 1$
i.e. $\;$ $\log_2 \left(\dfrac{-24}{-18}\right)^2 + \log_2 \left(\dfrac{-18}{-16}\right) = 1$
i.e. $\;$ $\log_2 \left(\dfrac{4}{3}\right)^2 + \log_2 \left(\dfrac{9}{8}\right) = 1$
i.e. $\;$ $\log_2 \left(\dfrac{16}{9} \times \dfrac{9}{8}\right) = 1$
i.e. $\;$ $\log_2 2 = 1$
i.e. $\;$ $1 = 1$ $\;\;$ which is true.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-17 \right\}$
Solve the equation: $\;$ $5^{\log x} = 50 - x^{\log 5}$
Given equation: $\;\;$ $5^{\log x} = 50 - x^{\log 5}$ $\;\;\; \cdots \; (1)$
Let $\;$ $5^{\log x} = p$
Then, by definition $\;$ $\log_5 p = \log x$ $\;\;\; \cdots \; (2)$
$\implies$ $x = 10^{\log_5 p}$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(1)$ becomes
$p = 50 - \left(10^{\log_5 p}\right)^{\log 5}$
i.e. $\;$ $p = 50 - \left[10^{\left(\frac{\log p}{\log 5}\right)}\right]^{\log 5}$
i.e. $\;$ $p = 50 - 10^{\log p}$
i.e. $\;$ $p = 50 - p$
i.e. $\;$ $2p = 50$ $\implies$ $p = 25$
Substituting the value of $p$ in equation $(2)$ gives
$\log_5 25 = \log x$
i.e. $\;$ $\log x = \log_5 5^2 = 2 \log_5 5 = 2$
$\implies$ $x = 10^2 = 100$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{100 \right\}$
Solve the equation: $\;$ $\log^2 \left(100x\right) - \log^2 \left(10 x\right) + \log^2 x = 6$
Given equation: $\;\;$ $\log^2 \left(100x\right) - \log^2 \left(10 x\right) + \log^2 x = 6$
i.e. $\;$ $\left(\log 100 + \log x\right)^2 - \left(\log 10 + \log x\right)^2 + \log^2 x = 6$
i.e. $\;$ $\left(2 + \log x\right)^2 - \left(1 + \log x\right)^2 + \log^2 x = 6$
i.e. $\;$ $\log^2 x + 4 \log x + 4 - \left(\log^2 x + 2 \log x + 1\right) + \log^2 x = 6$
i.e. $\;$ $\log^2 x + 2 \log x - 3 = 0$
i.e. $\;$ $\log^2 x + 3 \log x - \log x - 3 = 0$
i.e. $\;$ $\log x \left(\log x + 3\right) - 1 \left(\log x + 3\right) = 0$
i.e. $\;$ $\left(\log x + 3\right) \left(\log x - 1\right) = 0$
i.e. $\;$ $\log x + 3 = 0$ $\;$ or $\;$ $\log x - 1 = 0$
i.e. $\;$ $\log x = -3$ $\;$ or $\;$ $\log x = 1$
i.e. $\;$ $x = 10^{-3}$ $\;$ or $\;$ $x = 10^1 = 10$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^{-3}, \; 10 \right\}$
Solve the equation: $\;$ $\sqrt{\log_9 \left(9x^8\right) \cdot \log_3 \left(3x\right)} = \log_3 x^3$
Given equation: $\;\;$ $\sqrt{\log_9 \left(9x^8\right) \cdot \log_3 \left(3x\right)} = \log_3 x^3$
i.e. $\;$ $\sqrt{\left(\log_9 9 + \log_9 x^8\right) \cdot \left(\log_3 3 + \log_3 x\right)} = 3 \log_3 x$
i.e. $\;$ $\sqrt{\left(1 + 8 \times \dfrac{\log_3 x}{\log_3 9}\right) \cdot \left(1 + \log_3 x\right)} = 3 \log_3 x$
i.e. $\;$ $\sqrt{\left(1 + 8 \times \dfrac{\log_3 x}{\log_3 3^2}\right) \cdot \left(1 + \log_3 x\right)} = 3 \log_3 x$
i.e. $\;$ $\sqrt{\left(1 + 8 \times \dfrac{\log_3 x}{2 \log_3 3}\right) \cdot \left(1 + \log_3 x\right)} = 3 \log_3 x$
i.e. $\;$ $\sqrt{\left(1 + 4 \log_3 x\right) \cdot \left(1 + \log_3 x\right)} = 3 \log_3 x$
i.e. $\;$ $1 + 5 \log_3 x + 4 \left(\log_3 x\right)^2 = 9 \left(\log_3 x\right)^2$
i.e. $\;$ $5 \left(\log_3 x\right)^2 - 5 \log_3 x - 1 = 0$
i.e. $\;$ $\log_3 x = \dfrac{5 \pm \sqrt{25 + 20}}{10} = \dfrac{5 \pm \sqrt{45}}{10} = \dfrac{5 \pm 3 \sqrt{5}}{10}$
$\implies$ $x = 3^{\frac{5 + 3 \sqrt{5}}{10}}$ $\;$ or $\;$ $x = 3^{\frac{5 - 3 \sqrt{5}}{10}}$
i.e. $\;$ $x = 3.619$ $\;$ or $\;$ $x = 0.8289$
When $\;$ $x = 3^{\frac{5 - 3 \sqrt{5}}{10}} = 0.8289$, $\;$ the given equation becomes
$\sqrt{\log_9 \left(9 \times 0.8289^8\right) \cdot \log_3 \left(3 \times 0.8289\right)} = \log_3 \left(0.8289\right)^3$
i.e. $\;$ $\sqrt{\log_9 \left(2.006\right) \cdot \log_3 \left(2.4867\right)} = \log_3 \left(0.5695\right)$
i.e. $\;$ $\sqrt{0.3168 \times 0.8292} = -0.5125$
i.e. $\;$ $0.5125 = -0.5125$ $\;\;$ which is not possible.
$\implies$ $x = 3^{\frac{5 - 3 \sqrt{5}}{10}}$ $\;$ is not a valid solution.
When $\;$ $x = 3^{\frac{5 + 3 \sqrt{5}}{10}} = 3.619$, $\;$ the given equation becomes
$\sqrt{\log_9 \left(9 \times 3.619^8\right) \cdot \log_3 \left(3 \times 3.619\right)} = \log_3 \left(3.619\right)^3$
i.e. $\;$ $\sqrt{\log_9 \left(265820.24\right) \cdot \log_3 \left(10.857\right)} = \log_3 \left(47.3986\right)$
i.e. $\;$ $\sqrt{5.6830 \times 2.1707} = 3.5122$
i.e. $\;$ $3.5122 = 3.5122$
$\implies$ $x = 3^{\frac{5 + 3 \sqrt{5}}{10}}$ $\;$ is a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{3^{\frac{5 + 3 \sqrt{5}}{10}} \right\}$
Solve the equation: $\;$ $\log_4 \left(\log_2 x\right) + \log_2 \left(\log_4 x\right) = 2$
Given equation: $\;\;$ $\log_4 \left(\log_2 x\right) + \log_2 \left(\log_4 x\right) = 2$ $\;\;\; \cdots \; (1)$
Now, $\;$ $\log_4 x = \dfrac{\log_2 x}{\log_2 4} = \dfrac{\log_2 x}{\log_2 2^2} = \dfrac{\log_2 x}{2 \log_2 2}$
i.e. $\;$ $\log_4 x = \dfrac{\log_2 x}{2}$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$\log_4 \left(\log_2 x\right) + \dfrac{1}{2} \log_2 \left(\log_2 x\right) = 2$ $\;\;\; \cdots \; (3)$
Let $\;$ $\log_2 x = p$ $\;\;\; \cdots \; (4)$
Then equation $(3)$ becomes
$\log_4 p + \dfrac{1}{2} \log_2 p = 2$
i.e. $\;$ $2 \log_4 p + \log_2 p = 4$
i.e. $\;$ $2 \times \dfrac{\log_2 p}{\log_2 4} + \log_2 p = 4$
i.e. $\;$ $2 \times \dfrac{\log_2 p}{\log_2 2^2} + \log_2 p = 4$
i.e. $\;$ $2 \times \dfrac{\log_2 p}{2 \log_2 2} + \log_2 p = 4$
i.e. $\;$ $\log_2 p + \log_2 p = 4$
i.e. $\;$ $2 \log_2 p = 4$
i.e. $\;$ $\log_2 p = 2$
i.e. $\;$ $p = 2^2 = 4$
Substituting the value of $p$ in equation $(4)$ gives
$\log_2 x = 4$ $\implies$ $x = 2^4 = 16$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{16 \right\}$
Solve the equation: $\;$ $x^{3 \log x - \frac{1}{\log x}} = \sqrt[3]{10}$
Given equation: $\;\;$ $x^{3 \log x - \frac{1}{\log x}} = \sqrt[3]{10}$
i.e. $\;$ $\left(x^{3 \log x - \frac{1}{\log x}}\right)^3 = \left(10^{\frac{1}{3}}\right)^3$
i.e. $\;$ $x^{9 \log x - \frac{3}{\log x}} = 10$ $\;\;\; \cdots \; (1)$
Taking logarithim to base $10$ on both sides of equation $(1)$ gives
i.e. $\;$ $\log \left(x^{9 \log x - \frac{3}{\log x}}\right) = \log 10$
i.e. $\;$ $\left(9 \log x - \frac{3}{\log x}\right) \log x = 1$
i.e. $\;$ $9 \left(\log x\right)^2 - 3 = 1$
i.e. $\;$ $9 \left(\log x\right)^2 = 4$
i.e. $\;$ $3 \log x = \pm 2$
i.e. $\;$ $\log x = \dfrac{2}{3}$, $\;$ or $\;$ $\log x = \dfrac{-2}{3}$
$\implies$ $x = 10^{\frac{2}{3}}$ $\;$ or $\;$ $x = 10^{\frac{-2}{3}}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^{\frac{-2}{3}}, \; 10^{\frac{2}{3}} \right\}$
Solve the equation: $\;$ $\log \left(x^3 + 27\right) - 0.5 \log \left(x^2 + 6x + 9\right) = 3 \log \sqrt[3]{7}$
Given equation: $\;\;$ $\log \left(x^3 + 27\right) - 0.5 \log \left(x^2 + 6x + 9\right) = 3 \log \sqrt[3]{7}$
i.e. $\;$ $\log \left[\left(x + 3\right) \left(x^2 -3x + 9\right)\right] - 0.5 \log \left(x + 3\right)^2 = \log \left(\sqrt[3]{7}\right)^3$
i.e. $\;$ $\log \left[\left(x + 3\right) \left(x^2 - 3x + 9\right)\right] - 0.5 \times 2 \log \left(x + 3\right) = \log 7$
i.e. $\;$ $\log \left[\left(x + 3\right) \left(x^2 - 3x + 9\right)\right] - \log \left(x + 3\right) = \log 7$
i.e. $\;$ $\log \left[\dfrac{\left(x + 3\right) \left(x^2 - 3x + 9\right)}{x + 3}\right] = \log 7$
i.e. $\;$ $\log \left(x^2 - 3x + 9\right) = \log 7$ $\;\;\; \cdots \; (1)$
Taking antilog to base $10$ on both sides of equation $(1)$ gives
$x^2 - 3x + 9 = 7$
i.e. $\;$ $x^2 - 3x + 2 = 0$
i.e. $\;$ $\left(x - 2\right) \left(x - 1\right) = 0$
i.e. $\;$ $x = 2$ $\;$ or $\;$ $x = 1$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{1, \; 2 \right\}$
Solve the equation: $\;$ $\log_2 \left(2x^2\right) \times \log_2 \left(16 x\right) = \dfrac{9}{2} \log^2_2 x$
Given equation: $\;\;$ $\log_2 \left(2x^2\right) \times \log_2 \left(16 x\right) = \dfrac{9}{2} \log^2_2 x$
i.e. $\;$ $\left(\log_2 2 + \log_2 x^2\right) \times \left(\log_2 16 + \log_2 x\right) = \dfrac{9}{2} \left(\log_2 x\right)^2$
i.e. $\;$ $\left(1 + 2 \log_2 x\right) \times \left(\log_2 2^4 + \log_2 x\right) = \dfrac{9}{2} \left(\log_2 x\right)^2$
i.e. $\;$ $\left(1 + 2 \log_2 x\right) \times \left(4 \log_2 2 + \log_2 x\right) = \dfrac{9}{2} \left(\log_2 x\right)^2$
i.e. $\;$ $\left(1 + 2 \log_2 x\right) \times \left(4 + \log_2 x\right) = \dfrac{9}{2} \left(\log_2 x\right)^2$ $\;\;\; \cdots \; (1)$
Let $\;$ $\log_2 x = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$\left(1 + 2p\right) \left(4 + p\right) = \dfrac{9}{2} p^2$
i.e. $\;$ $4 + 9p + 2p^2 = \dfrac{9}{2} p^2$
i.e. $\;$ $8 + 18 p + 4p^2 = 9 p^2$
i.e. $\;$ $5p^2 - 18p - 8 = 0$
i.e. $\;$ $\left(5p + 2\right) \left(p - 4\right) = 0$
i.e. $\;$ $5p + 2 = 0$ $\;$ or $\;$ $p - 4 = 0$
i.e. $\;$ $p = \dfrac{-2}{5}$, $\;$ or $\;$ $p = 4$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = \dfrac{-2}{5}$, $\;$ we have from equation $(2)$
$\log_2 x = \dfrac{-2}{5}$ $\implies$ $x = 2^{\frac{-2}{5}}$
when $\;$ $p = 4$, $\;$ we have from equation $(2)$
$\log_2 x = 4$ $\implies$ $x = 2^4 = 16$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2^{\frac{-2}{5}}, \; 16 \right\}$
Solve the equation: $\;$ $\log_2 \left(4^{x+1} + 4\right) \times \log_2 \left(4^x + 1\right) = \log_{\frac{1}{\sqrt{2}}} \sqrt{\dfrac{1}{8}}$
Given equation: $\;\;$ $\log_2 \left(4^{x+1} + 4\right) \times \log_2 \left(4^x + 1\right) = \log_{\frac{1}{\sqrt{2}}} \sqrt{\dfrac{1}{8}}$
i.e. $\;$ $\log_2 \left(4^x \times 4 + 4\right) \times \log_2 \left(4^x + 1\right) = \dfrac{\log_2 \sqrt{\dfrac{1}{8}}}{\log_2 \dfrac{1}{\sqrt{2}}}$
i.e. $\;$ $\log_2 \left[4 \left(4^x + 1\right)\right] \times \log_2 \left(4^x + 1\right) = \dfrac{\log_2 2^{\frac{-3}{2}}}{\log_2 2^{\frac{-1}{2}}}$
i.e. $\;$ $\left[\log_2 4 + \log_2 \left(4^x + 1\right)\right] \times \log_2 \left(4^x + 1\right) = \dfrac{\dfrac{-3}{2} \log_2 2}{\dfrac{-1}{2} \log_2 2}$
i.e. $\;$ $\log_2 2^2 \times \log_2 \left(4^x + 1\right) + \left[\log_2 \left(4^x + 1\right)\right]^2 = 3$
i.e. $\;$ $\left[\log_2 \left(4^x + 1\right)\right]^2 + 2 \log_2 \left(4^x + 1\right) - 3 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $\log_2 \left(4^x + 1\right) = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$p^2 + 2p - 3 = 0$
i.e. $\;$ $\left(p + 3\right) \left(p - 1\right) = 0$
i.e. $\;$ $p = -3$ $\;$ or $\;$ $p = 1$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = -3$,
$\log_2 \left(4^x + 1\right) = -3$
i.e. $\;$ $4^x + 1 = 2^{-3} = \dfrac{1}{8}$
i.e. $\;$ $4^x = \dfrac{-7}{8}$ $\implies$ $x = \log_4 \left(\dfrac{-7}{8}\right)$
But, logarithim of a negative number is not defined.
$\therefore \;$ $p = -3$ $\;$ is not a valid solution.
when $\;$ $p = 1$,
$\log_2 \left(4^x + 1\right) = 1$
i.e. $\;$ $4^x + 1 = 2^1 = 2$
i.e. $\;$ $4^x = 1$
i.e. $\;$ $x = \log_4 1$ $\implies$ $x = 0$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0 \right\}$
Solve the equation: $\;$ $\log 2 + \log \left(4^{x-2} + 9\right) = 1 + \log \left(2^{x-2} + 1\right)$
Given equation: $\;\;$ $\log 2 + \log \left(4^{x-2} + 9\right) = 1 + \log \left(2^{x-2} + 1\right)$
i.e. $\;$ $\log \left(4^{x-2} + 9\right) - \log \left(2^{x-2} + 1\right) = 1 - \log 2$
i.e. $\;$ $\log \left[\dfrac{4^{x-2} + 9}{2^{x-2} + 1}\right] = \log 10 - \log2$
i.e. $\;$ $\log \left[\dfrac{\left(2^2\right)^{x-2} + 9}{2^{x-2} + 1}\right] = \log \left(\dfrac{10}{2}\right)$
i.e. $\;$ $\log \left[\dfrac{2^{2x-4} + 9}{2^{x-2} + 1}\right] = \log 5$ $\;\;\; \cdots \; (1)$
Taking antilog to base $10$ on both sides of equation $(1)$ gives
$\dfrac{2^{2x-4} + 9}{2^{x-2} + 1} = 5$
i.e. $\;$ $2^{2x-4} + 9 = 5 \times 2^{x-2} + 5$
i.e. $\;$ $\dfrac{2^{2x}}{2^4} - 5 \times \dfrac{2^x}{2^2} + 4 = 0$
i.e. $\;$ $2^{2x} - 5 \times 2^x \times 2^2 + 4 \times 2^4 = 0$
i.e. $\;$ $\left(2^x\right)^2 - 20 \times 20^x + 64 = 0$
i.e. $\;$ $\left(2^x\right)^2 - 16 \times 2^x - 4 \times 2^x + 64 = 0$
i.e. $\;$ $2^x \left(2^x - 16\right) - 4 \left(2^x - 16\right) = 0$
i.e. $\;$ $\left(2^x - 4\right) \left(2^x - 16\right) = 0$
i.e. $\;$ $2^x - 4 = 0$ $\;$ or $\;$ $2^x - 16 = 0$
i.e. $\;$ $2^x = 4 = 2^2$ $\;$ or $\;$ $2^x = 16 = 2^4$
$\implies$ $x = 2$ $\;$ or $\;$ $x = 4$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2, \; 4 \right\}$
Solve the equation: $\;$ $\log_3 \left[\sqrt{130 - 7^{\log_x \left(6 - x\right)}}\right] = 2$
Given equation: $\;\;$ $\log_3 \left[\sqrt{130 - 7^{\log_x \left(6 - x\right)}}\right] = 2$
i.e. $\;$ $\dfrac{1}{2} \log_3 \left[130 - 7^{\log_x \left(6 - x\right)}\right] = 2$
i.e. $\;$ $\log_3 \left[130 - 7^{\log_x \left(6 - x\right)}\right] = 4$
i.e. $\;$ $130 - 7^{\log_x \left(6 - x\right)} = 3^4 = 81$
i.e. $\;$ $7^{\log_x \left(6 - x\right)} = 49 = 7^2$
i.e. $\;$ $\log_x \left(6 - x\right) = 2$
i.e. $\;$ $6 - x = x^2$
i.e. $\;$ $x^2 + x - 6 = 0$
i.e. $\;$ $\left(x + 3\right) \left(x - 2\right) = 0$
i.e. $\;$ $x = -3$ $\;$ or $\;$ $x = 2$
But $\;$ $x = -3$ $\;$ makes the base of the term $\;$ $\log_x \left(6 - x\right)$, $\;$ in the given question, negative.
$\therefore \;$ $x = -3$ $\;$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2 \right\}$