Question Number 224069 by MirHasibulHossain last updated on 17/Aug/25

$$\mathrm{If}\:\mathrm{x}^{\mathrm{32}} =\mathrm{2}^{\mathrm{x}} \:\mathrm{then}\:\mathrm{solve}\:\mathrm{for}\:\mathrm{x}. \\ $$
Answered by mr W last updated on 17/Aug/25

$${x}=\pm\mathrm{2}^{\frac{{x}}{\mathrm{32}}} \\ $$$${x}=\pm{e}^{\frac{\left(\mathrm{ln}\:\mathrm{2}\right){x}}{\mathrm{32}}} \\ $$$$−\frac{\left(\mathrm{ln}\:\mathrm{2}\right){x}}{\mathrm{32}}{e}^{−\frac{\left(\mathrm{ln}\:\mathrm{2}\right){x}}{\mathrm{32}}} =\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}} \\ $$$$−\frac{\left(\mathrm{ln}\:\mathrm{2}\right){x}}{\mathrm{32}}={W}\left(\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}}\right) \\ $$$$\Rightarrow{x}=−\frac{\mathrm{32}}{\mathrm{ln}\:\mathrm{2}}{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}}\right) \\ $$$$=\begin{cases}{−\frac{\mathrm{32}}{\mathrm{ln}\:\mathrm{2}}\:{W}\left(\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}}\right)\approx−\frac{\mathrm{32}}{\mathrm{ln}\:\mathrm{2}}×\mathrm{0}.\mathrm{02120633838}\approx−\mathrm{0}.\mathrm{97901693}}\\{−\frac{\mathrm{32}}{\mathrm{ln}\:\mathrm{2}}\:{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{32}}\right)=\begin{cases}{−\frac{\mathrm{32}}{\mathrm{ln}\:\mathrm{2}}×\left(−\mathrm{5}.\mathrm{5451774445}\right)=\mathrm{256}}\\{−\frac{\mathrm{32}}{\mathrm{ln}\:\mathrm{2}}×\left(−\mathrm{0}.\mathrm{0221458995}\right)\approx\mathrm{1}.\mathrm{02239294}}\end{cases}}\end{cases} \\ $$