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Question Number 90762 by Ar Brandon last updated on 25/Apr/20

$$2^{\mathrm{x}}+2^{3\mathrm{x}}=16$$$$\mathrm{solve}\mathrm{for}\mathrm{x}$$

Answered by MJS last updated on 25/Apr/20

$$2^x=t$$$$t^3+t-16=0$$$$D=\frac{{\left(1\right)}^3}{27}+\frac{{(-16)}^2}{4}>0\Rightarrow \mathrm{Cardano}$$$$t_1=\sqrt[3]{8+\frac{\sqrt{5187}}{9}}-\sqrt[3]{-8+\frac{\sqrt{5187}}{9}}$$$$\Rightarrow x=\frac{1}{\ln 2}\ln (\sqrt[3]{8+\frac{\sqrt{5187}}{9}}-\sqrt[3]{-8+\frac{\sqrt{5187}}{9}})$$$$t_1\approx 2.38769$$$$x\approx 1.25561$$

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