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 $\;\;$ $x = \left\{-2 \right\}$