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Question Number 175849 by Linton last updated on 08/Sep/22

$$2^{2a}+2^a=10$$$$whatisa?$$

Answered by MJS_new last updated on 08/Sep/22

$$2^{2a}+2^a-10=0$$$${\left(2^a\right)}^2+2^a-10=0$$$$2^a=-\frac{1}{2}\pm \sqrt{\frac{1}{4}+10}=-\frac{1}{2}\pm \frac{\sqrt{41}}{2}$$$$2^a>0\mathrm{\forall }a\in \mathbb{R}\Rightarrow 2^a=\frac{-1+\sqrt{41}}{2}$$$$a=\frac{\ln \frac{-1+\sqrt{41}}{2}}{\ln 2}=\frac{\ln (-1+\sqrt{41})}{\ln 2}-1$$

Answered by BaliramKumar last updated on 08/Sep/22

$$$$$$2^{2a}+2^a=10$$$$2^a(2^a+1)-10=0put{2}^a=x$$$$x(x+1)-10=0$$$$x^2+x-10=0$$$$x=\frac{-1\pm \sqrt{1+40}}{2}=\frac{-1+\sqrt{41}}{2}(+\mathrm{ve})$$$$2^a=\left(\frac{\sqrt{41}-1}{2}\right)$$$$a\cdot log\left(2\right)=log\left(\frac{\sqrt{41}-1}{2}\right)$$$$a=\frac{log\left(\frac{\sqrt{41}-1}{2}\right)}{log\left(2\right)}$$$$a={\log }_2\left(\frac{\sqrt{41}-1}{2}\right)\mathrm{A}nswer$$

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