Question Number 187856 by Michaelfaraday last updated on 23/Feb/23 | ||
$${find}\:{x} \\ $$$$\mathrm{2}^{\sqrt{{x}}} =\mathrm{8}{x} \\ $$ | ||
Answered by mr W last updated on 23/Feb/23 | ||
$${t}=\sqrt{{x}}\:>\mathrm{0} \\ $$$$\mathrm{2}^{{t}} =\mathrm{8}{t}^{\mathrm{2}} \\ $$$$\mathrm{2}^{\frac{{t}}{\mathrm{2}}} =\mathrm{2}\sqrt{\mathrm{2}}{t} \\ $$$${e}^{\frac{{t}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} =\mathrm{4}\sqrt{\mathrm{2}}×\frac{{t}}{\mathrm{2}} \\ $$$$−\frac{{t}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}{e}^{−\frac{{t}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}} =−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$$$−\frac{{t}\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}={W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}}\right) \\ $$$${t}=−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}}\right) \\ $$$${x}={t}^{\mathrm{2}} =\left[−\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{2}}{W}\left(−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}}\right)\right]^{\mathrm{2}} \\ $$$$\approx\begin{cases}{\left[−\frac{\mathrm{2}×\left(−\mathrm{3}.\mathrm{29038565}\right)}{\mathrm{ln}\:\mathrm{2}}\right]^{\mathrm{2}} =\mathrm{90}.\mathrm{136912}}\\{\left[−\frac{\mathrm{2}×\left(−\mathrm{0}.\mathrm{141101}\right)}{\mathrm{ln}\:\mathrm{2}}\right]^{\mathrm{2}} =\mathrm{0}.\mathrm{165756}}\end{cases} \\ $$ | ||
Commented by Michaelfaraday last updated on 25/Feb/23 | ||
$${thanks}\:{sir} \\ $$ | ||