Question Number 148048 by puissant last updated on 25/Jul/21
$$\mathrm{2}^{\mathrm{x}} −\mathrm{3}^{\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} =\mathrm{1} \\ $$
Answered by mindispower last updated on 25/Jul/21
$$\mathrm{2}^{{x}} −\left(\sqrt{\mathrm{3}}\right)^{{x}} =\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{1}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{x}} \\ $$$${x}=\mathrm{2}\:\:{solution} \\ $$$${if}\:{x}\leqslant\mathrm{0}\:,\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{x}} \geqslant\mathrm{2} \\ $$$${x}\rightarrow\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{x}} \:{decrease}\:{function}\:{for}\:{x}>\mathrm{0} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{1},\Rightarrow{x}=\mathrm{2} \\ $$$$ \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 25/Jul/21
![2^x −3^(((x/2))) =1 2^(2y) −3^y =1 [x/2→y] 4^y −3^y =1 y>1⇒4^y −3^y >1 y>1⇒x/2>1→No solution y=1⇒x=2](https://www.tinkutara.com/question/Q148067.png)
$$\mathrm{2}^{\mathrm{x}} −\mathrm{3}^{\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} =\mathrm{1} \\ $$$$\mathrm{2}^{\mathrm{2}{y}} −\mathrm{3}^{{y}} =\mathrm{1}\:\:\:\:\:\left[{x}/\mathrm{2}\rightarrow{y}\right] \\ $$$$\mathrm{4}^{{y}} −\mathrm{3}^{{y}} =\mathrm{1} \\ $$$${y}>\mathrm{1}\Rightarrow\mathrm{4}^{{y}} −\mathrm{3}^{{y}} >\mathrm{1} \\ $$$${y}>\mathrm{1}\Rightarrow{x}/\mathrm{2}>\mathrm{1}\rightarrow{No}\:{solution} \\ $$$${y}=\mathrm{1}\Rightarrow{x}=\mathrm{2} \\ $$