Question Number 163110 by ampomahlosotho last updated on 03/Jan/22
Answered by mahdipoor last updated on 03/Jan/22
$$\mathrm{2}^{{x}−\mathrm{2}} =\mathrm{8}{x}\Rightarrow\frac{\mathrm{1}}{\mathrm{32}}\mathrm{2}^{{x}} ={x}\Rightarrow−\frac{\mathrm{1}}{\mathrm{32}×{ln}\mathrm{2}}=\left(−\frac{{x}}{{ln}\mathrm{2}}\right)\mathrm{2}^{\left(−{x}\right)} = \\ $$$$\left(−\frac{{x}}{{ln}\mathrm{2}}\right)\mathrm{2}^{{ln}\mathrm{2}×\frac{−{x}}{{ln}\mathrm{2}}} =\left(−\frac{{x}}{{ln}\mathrm{2}}\right){e}^{\left(\frac{−{x}}{{ln}\mathrm{2}}\right)} \Rightarrow \\ $$$$\frac{−{x}}{{ln}\mathrm{2}}={w}\left(−\frac{\mathrm{1}}{\mathrm{32}.{ln}\mathrm{2}}\right)\Rightarrow{x}=−{ln}\mathrm{2}{w}\left(\frac{−\mathrm{1}}{\mathrm{32}.{ln}\mathrm{2}}\right) \\ $$