Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_2 \left(25^{x+3} - 1\right) = 2 + \log_2 \left(5^{x+3} + 1\right)$

Given equation: $\;\;$ $\log_2 \left(25^{x+3} - 1\right) = 2 + \log_2 \left(5^{x+3} + 1\right)$

i.e. $\;$ $\log_2 \left[\left(5^2\right)^{x+3} - 1\right] - \log_2 \left(5^{x+3} + 1\right) = 2$

i.e. $\;$ $\log_2 \left(\dfrac{5^{2x+6} - 1}{5^{x+3} + 1}\right) = 2$

i.e. $\;$ $\dfrac{5^{2x+6} - 1}{5^{x+3} + 1} = 2^2 = 4$

i.e. $\;$ $5^{2x+6} - 1 = 4 \times 5^{x+3} + 4$

i.e. $\;$ $5^{2x} \times 5^6 - 4 \times 5^x \times 5^3 - 5 = 0$

i.e. $\;$ $\left(5^x\right)^2 \times 5^5 - 4 \times 5^x \times 5^2 - 1 = 0$

i.e. $\;$ $3125 \times \left(5^x\right)^2 - 100 \times 5^x - 1 = 0$

i.e. $\;$ $3125 \times \left(5^x\right)^2 - 125 \times 5^x + 25 \times 5^x - 1 = 0$

i.e. $\;$ $125 \times 5^x \left(25 \times 5^x - 1\right) + 1 \left(25 \times 5^x - 1\right) = 0$

i.e. $\;$ $\left(25 \times 5^x - 1\right) \left(125 \times 5^x + 1\right) = 0$

i.e. $\;$ $25 \times 5^x - 1 = 0$ $\;$ or $\;$ $125 \times 5^x + 1 = 0$

i.e. $\;$ $25 \times 5^x = 1$ $\;$ or $\;$ $125 \times 5^x = -1$

i.e. $\;$ $5^x = \dfrac{1}{25} = 5^{-2}$ $\;$ or $\;$ $5^x = \dfrac{-1}{125} = \left(\dfrac{-1}{5}\right)^3$

i.e. $\;$ $x = \log_5 5^{-2}$ $\;$ or $\;$ $x = \log_5 \left(\dfrac{-1}{5}\right)^3$

i.e. $\;$ $x = -2 \log_5 5$ $\;$ or $\;$ $x = 3 \log_5 \left(\dfrac{-1}{5}\right)$

But logarithim of a negative number is not defined.

$\therefore \;$ $x = 3 \log_5 \left(\dfrac{-1}{5}\right)$ $\;$ is not a valid solution.

Now, $\;$ $x = -2 \log_5 5$ $\implies$ $x = -2$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-2 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\left(\dfrac{1}{5}\right)^{\log^2 x - \log x} = \dfrac{1}{125} \times 5^{\log x - 1}$

Given equation: $\;\;$ $\left(\dfrac{1}{5}\right)^{\log^2 x - \log x} = \dfrac{1}{125} \times 5^{\log x - 1}$

i.e. $\;$ $\left(\dfrac{1}{5}\right)^{\log^2 x - \log x} = \dfrac{1}{5^3} \times 5^{\log x - 1}$

i.e. $\;$ $\left[\left(5^{-1}\right)\right]^{\log^2 x - \log x} = 5^{\log x - 4}$

i.e. $\;$ $5^{\log x - \log^2 x} = 5^{\log x - 4}$

i.e. $\;$ $\log x - \log^2 x = \log x - 4$

i.e. $\;$ $\log^2 x = 4$

i.e. $\;$ $\log x = 2$ $\;$ or $\;$ $\log x = -2$

i.e. $\;$ $x = 10^2$ $\;$ or $x = 10^{-2}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^2, 10^{-2} \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log \left(3^x - 2^{4-x}\right) = 2 + \dfrac{1}{4} \log 16 - \dfrac{x \log4}{2}$

Given equation: $\;\;$ $\log \left(3^x - 2^{4-x}\right) = 2 + \dfrac{1}{4} \log 16 - \dfrac{x \log4}{2}$

i.e. $\;$ $\log \left(3^x - 2^{4 - x}\right) - \dfrac{1}{4} \log 16 + x \times \dfrac{1}{2} \log 4 = 2$

i.e. $\;$ $\log \left(3^x - \dfrac{2^4}{2^x}\right) - \log \left(2^4\right)^{\frac{1}{4}} + x \times \log \left(2^2\right)^{\frac{1}{2}} = 2$

i.e. $\;$ $\log \left(\dfrac{6^x - 2^4}{2^x}\right) - \log 2 + \log 2^x = 2$

i.e. $\;$ $\log \left[\left(\dfrac{6^x - 2^4}{2^x}\right) \times \dfrac{1}{2} \times 2^x\right] = 2$

i.e. $\;$ $\log \left[\dfrac{6^x - 2^4}{2}\right] = 2$

i.e. $\;$ $\dfrac{6^x - 2^4}{2} = 10^2 = 100$

i.e. $\;$ $6^x - 16 = 200$

i.e. $\;$ $6^x = 216$

i.e. $\;$ $x = \log_6 216$

i.e. $\;$ $x = \log_6 6^3$

i.e. $\;$ $x = 3 \log_6 6$ $\implies$ $x = 3$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{3 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_2 \left(4^x + 4\right) = \log_2 2^x + \log_2 \left(2^{x+1} - 3\right)$

Given equation: $\;\;$ $\log_2 \left(4^x + 4\right) = \log_2 2^x + \log_2 \left(2^{x+1} - 3\right)$

i.e. $\;$ $\log_2 \left(4^x + 4\right) - \log_2 2^x - \log_2 \left(2^{x+1} - 3\right) = 0$

i.e. $\;$ $\log_2 \left[\dfrac{4^x + 4}{2^x \left(2^{x+1} - 3\right)}\right] = 0$

i.e. $\;$ $\dfrac{4^x + 4}{2^{2x + 1} - 3 \times 2^x} = 2^0 = 1$

i.e $\;$ $\left(2^2\right)^x + 4 = 2^{2x + 1} - 3 \times 2^x$

i.e. $\;$ $\left(2^x\right)^2 + 3 \times 2^x - \left(2^2\right)^x \times 2 + 4 = 0$

i.e. $\;$ $\left(2^x\right)^2 - 2 \times \left(2^x\right)^2 + 3 \times 2^x + 4 = 0$

i.e. $\;$ $\left(2^x\right)^2 - 3 \times 2^x - 4 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $2^x = p$ $\;\;\; \cdots \; (2)$

Then equation $(1)$ becomes

$p^2 - 3p -4 = 0$

i.e. $\;$ $\left(p + 1\right) \left(p - 4\right) = 0$

i.e. $\;$ $p = -1$ $\;$ or $\;$ $p = 4$

When $\;$ $p = -1$, $\;$ we have from equation $(2)$,

$2^x = -1$ $\implies$ $x = \log_2 \left(-1\right)$

But, logarithim of a negative number is not defined.

$\therefore \;$ $p = -1$ $\;$ does not give a valid solution.

When $\;$ $p = 4$, $\;$ we have from equation $(2)$,

$2^x = 4 = 2^2$ $\implies$ $x = 2$

Check

Substituting $\;$ $x = 2$ $\;$ in the given problem, we get,

$\log_2 \left(4^2 + 4\right) = \log_2 2^2 + \log_2 \left(2^{2+1} - 3\right)$

i.e. $\;$ $\log_2 20 = \log_2 4 + \log_2 5$

i.e. $\;$ $\log_2 20 = \log_2 \left(4 \times 5\right)$

i.e. $\;$ $\log_2 20 = \log_2 20$ $\;\;\;$ which is true.

$\therefore \;$ $x = 2$ $\;$ is a valid solution.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_{\sqrt{5}} \left(4^x - 6\right) - \log_{\sqrt{5}} \left(2^x - 2\right) = 2$

Given equation: $\;\;$ $\log_{\sqrt{5}} \left(4^x - 6\right) - \log_{\sqrt{5}} \left(2^x - 2\right) = 2$ $\;\;\; \cdots \; (1)$

i.e. $\;$ $\log_{\sqrt{5}} \left(\dfrac{4^x - 6}{2^x - 2}\right) = 2$

i.e. $\;$ $\dfrac{4^x - 6}{2^x - 2} = \left(\sqrt{5}\right)^2 = 5$

i.e. $\;$ $4^x - 6 = 5 \times 2^x - 10$

i.e. $\;$ $\left(2^2\right)^x - 5 \times 2^x + 4 = 0$

i.e. $\;$ $\left(2^x\right)^2 - 5 \times 2^x + 4 = 0$ $\;\;\; \cdots \; (2)$

Let $\;$ $2^x = p$ $\;\;\; \cdots \; (3)$

Then equation $(2)$ becomes

$p^2 - 5p + 4 = 0$

i.e. $\;$ $\left(p - 1\right) \left(p - 4\right) = 0$

i.e. $\;$ $p = 1$ $\;$ or $\;$ $p = 4$

Substituting the value of $p$ in equation $(3)$ gives

when $\;$ $p = 1$, $\;$ we have $\;$ $2^x = 1$ $\implies$ $x = \log_2 1 = 0$

when $\;$ $p = 4$, $\;$ we have $\;$ $2^x = 4 = 2^2$ $\implies$ $x = 2$

Substituting $\;$ $x = 0$ $\;$ in the term $\;$ $\log_{\sqrt{5}} \left(4^x - 6\right)$ $\;$ in equation $(1)$ gives

$\log_{\sqrt{5}} \left(4^0 - 6\right) = \log_{\sqrt{5}} \left(1 - 6\right) = \log_{\sqrt{5}} \left(-5\right)$

But, logarithim of a negative number is not defined.

$\therefore \;$ $x = 0$ $\;$ is not a valid solution.

Substituing $\;$ $x = 2$ $\;$ in equation $(1)$ gives

$\log_{\sqrt{5}} \left(4^2 - 6\right) - \log_{\sqrt{5}} \left(2^2 - 2\right) = 2$

i.e. $\;$ $\log_{\sqrt{5}} \left(\dfrac{10}{2}\right) = 2$

i.e. $\;$ $\log_{\sqrt{5}} 5 = 2$

i.e. $\;$ $\log_{\sqrt{5}} \left(\sqrt{5}\right)^2 = 2$

i.e. $\;$ $2 \log_{\sqrt{5}} \sqrt{5} = 2$

i.e. $\;$ $2 = 2$ $\;\;\;$ which is true.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log \left(\log x\right) + \log \left(\log x^3 - 2\right) = 0$

Given equation: $\;\;$ $\log \left(\log x\right) + \log \left(\log x^3 - 2\right) = 0$

i.e. $\;$ $\log \left(\log x\right) = - \log \left(3 \log x - 2\right)$

i.e. $\;$ $\log \left(\log x\right) = \log \left(3 \log x - 2\right)^{-1}$ $\;\;\; \cdots \; (1)$

Taking antilog on both sides of equation $(1)$ gives

$\log x = \left(3 \log x - 2\right)^{-1}$

i.e. $\;$ $\log x = \dfrac{1}{3 \log x - 2}$

i.e. $\;$ $3 \left(\log x\right)^2 - 2 \log x - 1 = 0$ $\;\;\; \cdots \; (2)$

Let $\;$ $\log x = p$ $\;\;\; \cdots \; (3)$

Then equation $(2)$ becomes

$3p^2 - 2p -1 = 0$

i.e. $\;$ $\left(3p - 1\right) \left(p - 1\right) = 0$

i.e. $\;$ $p = \dfrac{1}{3}$ $\;$ or $\;$ $p = 1$

Substituting the value of $p$ in equation $(3)$ gives

when $\;$ $p = \dfrac{1}{3}$, $\;$ then $\;$ $\log x = \dfrac{1}{3}$ $\implies$ $x = 10^{\frac{1}{3}}$

when $\;$ $p = 1$, $\;$ then $\;$ $\log x = 1$ $\implies$ $x = 10^1 = 10$

Substituting $\;$ $\log x = \dfrac{1}{3}$ $\;$ in the term $\;$ $\log \left(\log x^3 - 2\right)$ $\;$ i.e. in $\;$ $\log \left(3 \log x - 2\right)$ $\;$ in the given question, the term becomes

$\log \left(3 \times \dfrac{1}{3} - 2\right) = \log \left(-1\right)$

But, logarithim of a negative number is not defined.

$\therefore \;$ $p = \dfrac{1}{3}$ $\;$ is not a valid solution.

Substituting $\;$ $\log x = 1$ $\;$ in the given question gives

$\log 1 + \log \left(3 \times 1 - 2\right) = 0$

i.e. $\;$ $\log 1 + \log 1 = 0$

i.e. $\;$ $0 + 0 = 0$ $\;\;$ which is true.

$\therefore \;$ $\log x = 1$ $\;$ is a valid solution.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_3 \left(9^x + 9\right) = x + \log_3 \left(28 - 2 \times 3^x\right)$

Given equation: $\;\;$ $\log_3 \left(9^x + 9\right) = x + \log_3 \left(28 - 2 \times 3^x\right)$

i.e. $\;$ $\log_3 \left(9^x + 9\right) - \log_3 \left(28 - 2 \times 3^x\right) = x$

i.e. $\;$ $\log_3 \left(\dfrac{9^x + 9}{28 - 2 \times 3^x}\right) = x$

i.e. $\;$ $\dfrac{9^x + 9}{28 - 2 \times 3^x} = 3^x$

i.e. $\;$ $9^x + 9 = 28 \times 3^x - 2 \times 3^{2x}$

i.e. $\;$ $3^{2x} + 9 = 28 \times 3^x - 2 \times 3^{2x}$

i.e. $\;$ $3 \times \left(3^x\right)^2 - 28 \times 3^x + 9 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $3^x = p$ $\;\;\; \cdots \; (2)$

Then, equation $(1)$ becomes

$3p^2 - 28p + 9 = 0$

i.e. $\;$ $\left(3p - 1\right) \left(p - 9\right) = 0$

i.e. $\;$ $p = \dfrac{1}{3}$ $\;$ or $\;$ $p = 9$

Substituting the value of $\;$ $p$ $\;$ in equation $(2)$ gives

when $\;$ $p = \dfrac{1}{3}$, $\;$ we have $\;$ $3^x = \dfrac{1}{3} = 3^{-1}$ $\implies$ $x = -1$

when $\;$ $p = 9$, $\;$ we have $\;$ $3^x = 9 = 3^2$ $\implies$ $x = 2$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-1, 2 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log \left(5^{x-2} + 1\right) = x + \log 13 - 2 \log 5 + \left(1-x\right) \log 2$

Given equation: $\;\;$ $\log \left(5^{x-2} + 1\right) = x + \log 13 - 2 \log 5 + \left(1-x\right) \log 2$

i.e. $\;$ $\log \left(5^{x-2} + 1\right) - \log 2^{1-x} - \log 13 + \log 5^2 = x$

i.e. $\;$ $\log \left[\dfrac{\left(5^{x-2} + 1\right) \times 5^2}{2^{1-x} \times 13}\right] = x$

i.e. $\;$ $\log \left[\dfrac{\left(5^x + 25\right) \times 2^x}{26}\right] = x$

i.e. $\;$ $\log \left[\dfrac{10^x + 2^x \times 25}{26}\right] = x$

i.e. $\;$ $\dfrac{10^x + 2^x \times 25}{26} = 10^x$

i.e. $\;$ $10^x + 2^x \times 25 = 26 \times 10^x$

i.e. $\;$ $25 \times 2^x = 25 \times 10^x$

i.e. $\;$ $2^x = 10^x$

i.e. $\;$ $2^x = 2^x \times 5^x$

i.e. $\;$ $2^x \left(5^x - 1\right) = 0$

i.e. $\;$ $2^x = 0$ $\;$ or $\;$ $5^x - 1 = 0$

When $\;$ $2^x = 0$,

$\implies$ $x = \log_2 0$

But $\;$ $\log_2 0$ $\;$ is not defined.

$\therefore \;$ $2^x = 0$ $\;$ does not give a valid solution.

When $\;$ $5^x - 1 = 0$,

$\implies$ $5^x = 1$

i.e. $\;$ $x = \log_5 1 = 0$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log \left(3x^2 + 7\right) - \log \left(3x - 2\right) = 1$

Given equation: $\;\;$ $\log \left(3x^2 + 7\right) - \log \left(3x - 2\right) = 1$

i.e. $\;$ $\log \left(\dfrac{3x^2 + 7}{3x - 2}\right) = 1$

i.e. $\;$ $\dfrac{3x^2 + 7}{3x - 2} = 10^1 = 10$

i.e. $\;$ $3x^2 + 7 = 30x - 20$

i.e. $\;$ $3x^2 - 30x + 27 = 0$

i.e. $\;$ $x^2 - 10x + 9 = 0$

i.e. $\;$ $\left(x - 9\right) \left(x - 1\right) = 0$

i.e. $\;$ $x = 9$, $\;$ or $\;$ $x = 1$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{1, 9 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\dfrac{1}{2} \log x + 3 \log \sqrt{2 + x} = \log \sqrt{x \left(x + 2\right)} + 2$

Given equation: $\;\;$ $\dfrac{1}{2} \log x + 3 \log \sqrt{2 + x} = \log \sqrt{x \left(x + 2\right)} + 2$

i.e. $\;$ $\log \sqrt{x} - \log \sqrt{x \left(x + 2\right)} + \log \left(\sqrt{2 + x}\right)^3 = 2$

i.e. $\;$ $\log \left[\dfrac{\sqrt{x} \left(2 + x\right) \sqrt{2 + x}}{\sqrt{x \left(x + 2\right)}}\right] = 2$

i.e. $\;$ $\log \left(x + 2\right) = 2$

i.e. $\;$ $x +2 = 10^2 = 100$

i.e. $\;$ $x = 98$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{98 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\dfrac{\log \sqrt{x + 7} - \log 2}{\log 8 - \log \left(x - 5\right)} = -1$

Given equation: $\;\;$ $\dfrac{\log \sqrt{x + 7} - \log 2}{\log 8 - \log \left(x - 5\right)} = -1$

i.e. $\;$ $\log \sqrt{x+7} - \log 2 = \log \left(x-5\right) - \log 8$

i.e. $\;$ $\log \sqrt{x+7} - \log \left(x - 5\right) - \log 2 + 3 \log 2 = 0$

i.e. $\;$ $\log \sqrt{x+7} - \log \left(x-5\right) + 2 \log 2 = 0$

i.e. $\;$ $\log \sqrt{x + 7} - \log \left(x - 5\right) + \log 4 = 0$

i.e. $\;$ $\log \left(\dfrac{4 \sqrt{x + 7}}{x - 5}\right) = 0$

i.e. $\;$ $\dfrac{4 \sqrt{x + 7}}{x - 5} = 10^0 = 1$

i.e. $\;$ $4 \sqrt{x + 7} = x - 5$

i.e. $\;$ $16 \left(x + 7\right) = x^2 - 10 x + 25$

i.e. $\;$ $x^2 - 26 x - 87 = 0$

i.e. $\;$ $\left(x - 29\right) \left(x + 3\right) = 0$

i.e. $\;$ $x = 29$ $\;$ or $\;$ $x = -3$

But, when $\;$ $x = -3$, $\;$ the term $\;$ $\log \left(x - 5\right)$ $\;$ in the given problem becomes

$\log \left(-3 - 5\right) = \log \left(-8\right)$

$\because \;$ logarithm of a negative number is not defined,

$\therefore \;$ $x = -3$ $\;$ is not a valid solution.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{29 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_2 \left(4^x + 1\right) = x + \log_2 \left(2^{x+3} - 6\right)$

Given equation: $\;\;$ $\log_2 \left(4^x + 1\right) = x + \log_2 \left(2^{x+3} - 6\right)$

i.e. $\;$ $\log_2 \left(4^x + 1\right) - \log_2 \left(2^{x+3} - 6\right) = x$

i.e. $\;$ $\log_2 \left[\dfrac{4^x + 1}{2^{x+3} - 6}\right] = x$

i.e. $\;$ $\dfrac{4^x + 1}{2^{x+3} - 6} = 2^x$

i.e. $\;$ $4^x + 1 = 2^x \left(2^{x+3} - 6\right)$

i.e. $\;$ $4^x + 1 = 2^{2x+3} - 6 \times 2^x$

i.e. $\;$ $\left(2^2\right)^x + 1 = \left(2^2\right)^x \times 2^3 - 6 \times 2^x$

i.e. $\;$ $8 \times \left(2^x\right)^2 - \left(2^x\right)^2 - 6 \times 2^x - 1 = 0$

i.e. $\;$ $7 \times \left(2^x\right)^2 - 6 \times 2^x - 1 = 0$

i.e. $\;$ $7 \times \left(2^x\right)^2 - 7 \times 2^x + 2^x - 1 = 0$

i.e. $\;$ $7 \times 2^x \left(2^x - 1\right) + 1 \left(2^x - 1\right) = 0$

i.e. $\;$ $\left(7 \times 2^x + 1\right) \left(2^x - 1\right) = 0$

i.e. $\;$ $2^x = \dfrac{-1}{7}$ $\;$ or $\;$ $2^x = 1$

i.e. $\;$ $x = \log_2 \left(\dfrac{-1}{7}\right)$ $\;$ or $\;$ $x = \log_2 1$

Logarithim of a negative number is not defined.

$\therefore \;$ $x = \log_2 \left(\dfrac{-1}{7}\right)$ $\;$ is not a valid solution.

When $\;$ $x = \log_2 1$ $\implies$ $x = 0$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $1 - \dfrac{1}{2} \log \left(2x - 1\right) = \dfrac{1}{2} \log \left(x - 9\right)$

Given equation: $\;\;$ $1 - \dfrac{1}{2} \log \left(2x - 1\right) = \dfrac{1}{2} \log \left(x - 9\right)$

i.e. $\;$ $\dfrac{1}{2} \left[\log \left(x - 9\right) + \log \left(2x - 1\right)\right] = 1$

i.e. $\;$ $\log \left[\left(x - 9\right) \left(2x - 1\right)\right] = 2$ $\;\;\; \cdots \; (1)$

i.e. $\;$ $\left(x - 9\right) \left(2x - 1\right) = 10^2 = 100$

i.e. $\;$ $2 x^2 - 19 x + 9 = 100$

i.e. $\;$ $2 x^2 - 19x - 91 = 0$

i.e. $\;$ $\left(2x + 7\right) \left(x - 13\right) = 0$

i.e. $\;$ $x = \dfrac{-7}{2}$ $\;$ or $\;$ $x = 13$

When $\;$ $x = \dfrac{-7}{2}$, $\;$ the term $\;$ $\log \left(2x - 1\right)$ $\;$ in the given problem becomes

$\log \left[2 \times \left(\dfrac{-7}{2}\right) - 1\right] = \log \left(-8\right)$

But logarithm of a negative number is not defined.

$\therefore \;$ $x = \dfrac{-7}{2}$ $\;$ is not a valid solution.

When $\;$ $x = 13$, $\;$ the given problem [equation (1)] becomes

$\log \left[\left(13-9\right) \left(26-1\right) \right] = 2$

i.e. $\;$ $\log \left[4 \times 25\right] = 2$

i.e. $\log 100 = 2$ $\;$ which is true.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{13 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\dfrac{\log \left(\sqrt{x + 1} + 1\right)}{\log \left(\sqrt[3]{x - 40}\right)} = 3$

Given equation: $\;\;$ $\dfrac{\log \left(\sqrt{x + 1} + 1\right)}{\log \left(\sqrt[3]{x - 40}\right)} = 3$

i.e. $\;$ $\log \left(\sqrt{x + 1} + 1\right) = 3 \log \left(\sqrt[3]{x - 40}\right)$

i.e. $\;$ $\log \left(\sqrt{x + 1} + 1\right) = \log \left(\sqrt[3]{x - 40}\right)^3$

i.e. $\;$ $\log \left(\sqrt{x + 1} + 1\right) = \log \left(x - 40\right)$ $\;\;\; \cdots \; (1)$

Taking antilog on both sides of equation $(1)$ gives

$\sqrt{x + 1} + 1 = x - 40$

i.e. $\;$ $\sqrt{x + 1} = x - 41$

i.e. $\;$ $x +1 = x^2 - 82x + 1681$

i.e. $\;$ $x^2 - 83x + 1680 = 0$

i.e. $\;$ $\left(x - 48\right) \left(x - 35\right) = 0$

i.e. $\;$ $x = 48$ $\;$ or $\;$ $x = 35$

When $\;$ $x = 35$, $\;$ the term $\;$ $\log \left(\sqrt[3]{x - 40}\right)$ $\;$ in the given problem becomes

$\log \left(\sqrt[3]{35 - 40}\right) = \log \left(\sqrt[3]{-5}\right)$

But logarithim of a negative number is not defined.

$\therefore \;$ $x = 35$ $\;$ is not a valid solution.

When $\;$ $x = 48$, $\;$ the given problem becomes

$\dfrac{\log \left(\sqrt{48+1} + 1\right)}{\log \left(\sqrt[3]{48 - 40}\right)} = 3$

i.e. $\;$ $\dfrac{\log \left(\sqrt{49} + 1\right)}{\log \left(\sqrt[3]{8}\right)} = 3$

i.e. $\;$ $\dfrac{\log 8}{\log 2} = 3$

i.e. $\;$ $\dfrac{\log 2^3}{\log 2} = 3$

i.e. $\;$ $\dfrac{3 \log 2}{\log 2} = 3$

i.e. $\;$ $3 = 3$ $\;$ which is true.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{48 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_4 \left(2 \times 4^{x-2} - 1\right) + 4 = 2x$

Given equation: $\;\;$ $\log_4 \left(2 \times 4^{x-2} - 1\right) + 4 = 2x$

i.e. $\;$ $\log_4 \left(2 \times 4^{x-2} - 1\right) = 2x - 4$

i.e. $\;$ $\log_4 \left(2 \times 4^{x-2} - 1\right) = 2\left(x - 2\right)$ $\;\;\; \cdots \; (1)$

Let $\;$ $x - 2 = p$ $\;\;\; \cdots \; (2)$

In view of equation $(2)$, equation $(1)$ becomes

$\log_4 \left(2 \times 4^p - 1\right) = 2p$

i.e. $\;$ $2 \times 4^p - 1 = 4^{2p}$

i.e. $\;$ $\left(4^p\right)^2 - 2 \times 4^p + 1 = 0$

i.e. $\;$ $\left(4^p - 1\right)^2 = 0$

i.e. $\;$ $4^p - 1 = 0$

i.e. $\;$ $4^p = 1 = 4^0$

$\implies$ $p = 0$

Substituting the value of $\;$ $p$ $\;$ in equation $(2)$ gives

$x - 2 = 0$ $\implies$ $x = 2$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2 \right\}$