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 $\;\;$ $x = \left\{48 \right\}$