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Question Number 174287 by Engr_Jidda last updated on 28/Jul/22

Answered by haf last updated on 28/Jul/22

$$uselnfunction$$

Commented by Engr_Jidda last updated on 29/Jul/22

$$plshelpmeout$$

Answered by Rasheed.Sindhi last updated on 29/Jul/22

$$2^{x+4}+4^{x-2}=8$$$$2^{x+4}+2^{2x-4}=8$$$$2^4\cdot 2^x+{\left(2^x\right)}^2/2^4=8$$$$Let{2}^x=y$$$$16y+\frac{y^2}{16}=8$$$$y^2+256y-128=0$$$$y=\frac{-256\pm \sqrt{{256}^2-4\left(1\right)(-128)}}{2}$$$$=-128\pm 8\sqrt{258}$$$$2^x=128+\sqrt{258}$$$$x\log 2=\log (-128\pm 8\sqrt{258})$$$$x=\frac{\log (-128\pm 8\sqrt{258})}{\log 2}$$

Commented by Tawa11 last updated on 31/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by Rasheed.Sindhi last updated on 31/Jul/22

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}\mathbf{miss}!$$

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