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Question Number 202116 by Calculusboy last updated on 21/Dec/23

$$\sqrt{3^{\mathit{x}}}+1=2^{\mathit{x}}\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{x}$$

Answered by Frix last updated on 21/Dec/23

$$\mathrm{Obviously}x=2$$$$\sqrt{3^2}+1=\sqrt{9}+1=3+1=4=2^2$$

Commented by Calculusboy last updated on 21/Dec/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s},\mathit{b}\mathit{u}\mathit{t}\mathit{s}\mathit{i}\mathit{r}\mathit{c}\mathit{a}\mathit{n}\mathit{i}\mathit{g}\mathit{e}\mathit{t}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{f}\mathit{o}\mathit{r}\mathit{i}\mathit{t}$$

Answered by MATHEMATICSAM last updated on 22/Dec/23

$$\sqrt{3^{\mathit{x}}}+1=2^{\mathit{x}}\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{x}$$$$\mathrm{let}f\left(x\right)=\sqrt{3^x}+1-2^x$$$$f\left(2\right)=\sqrt{3^2}+1-2^2=3+1-4=0$$$$\mathrm{As}f\left(2\right)=0\mathrm{so}\mathrm{that}\mathrm{we}\mathrm{can}\mathrm{say}x=2$$

Commented by Calculusboy last updated on 23/Dec/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

Commented by Frix last updated on 23/Dec/23

$$\mathrm{This}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{solution}\mathrm{method}.$$$$\mathrm{Try}\sqrt{5^x}+1=3^x$$$$\mathrm{You}\mathrm{can}\mathrm{only}\mathrm{approximate}.$$$$\sqrt{3^x}+1=2^x\mathrm{is}\mathrm{constructed}\mathrm{to}\mathrm{have}\mathrm{the}$$$$\mathrm{obvious}\mathrm{solution}x=2.\mathrm{The}\mathrm{principle}\mathrm{is}$$$$a+b=c^n\Rightarrow \sqrt[n]{a^n}+b=c^n$$

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