Solve the equation: $\;$ $3^{2x+1} + 10 \times 3^x + 3 = 0$
Given equation: $\;\;$ $3^{2x+1} + 10 \times 3^x + 3 = 0$
i.e. $\;$ $3^x \left(3^{x+1} + 10\right) = -3^1$
i.e. $\;$ $3^{x+1} + 10 = -3^{1-x}$
i.e. $\;$ $3^{x+1} + 3^{1-x} = -10$
i.e. $\;$ $3 \left(3^x + 3^{-x}\right) = -10$
i.e. $\;$ $3^x + \dfrac{1}{3^x} = \dfrac{-10}{3}$ $\;\;\; \cdots \; (1)$
Let $\;$ $3^x = p$ $\;\;\; \cdots \; (2)$
Then, equation $(1)$ becomes
$p + \dfrac{1}{p} = \dfrac{-10}{3}$
i.e. $\;$ $3p^2 + 10p + 3 = 0$
i.e. $\;$ $\left(3p + 1\right) \left(p + 3\right) = 0$
i.e. $\;$ $p = \dfrac{-1}{3}$ $\;\;$ or $\;\;$ $p = -3$
Substituting the values of $p$ in equation $(2)$ gives
When $\;$ $p = \dfrac{-1}{3}$ $\implies$ $3^x = \dfrac{-1}{3}$ $\implies$ $x = \log_3 \left(\dfrac{-1}{3}\right)$
But logarithm of a negative number is not defined.
$\implies$ $p = \dfrac{-1}{3}$ $\;$ is not a valid solution.
When $\;$ $p = -3$ $\implies$ $3^x = -3$ $\implies$ $x = \log_3 \left(-3\right)$
But logarithm of a negative number is not defined.
$\implies$ $p = -3$ $\;$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \phi$ $\;$ (null)