Question Number 143438 by Rankut last updated on 14/Jun/21
$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\boldsymbol{{x}}} =\boldsymbol{{log}}_{\frac{\mathrm{1}}{\mathrm{2}}} \boldsymbol{{x}} \\ $$$$\boldsymbol{{find}}\:\:\:\boldsymbol{{x}} \\ $$
Answered by mr W last updated on 14/Jun/21
$${x}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} } \\ $$$${x}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} ={e}^{−{x}\mathrm{ln}\:\mathrm{2}} \\ $$$${xe}^{{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{1} \\ $$$${x}\mathrm{ln}\:\mathrm{2}{e}^{{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{ln}\:\mathrm{2} \\ $$$$\Rightarrow{x}=\frac{{W}\left(\mathrm{ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}}\approx\frac{\mathrm{0}.\mathrm{444439091}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{0}.\mathrm{64119} \\ $$