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Question Number 160981 by mathlove last updated on 10/Dec/21

Commented by cortano last updated on 10/Dec/21

 x^(log _3 (2))  = (√(x+1))    log _3 (2) log _3 (x)= log _3 ((√(x+1)))   log _3 (2)= ((log _3 ((√(x+1)) ))/(log _3 (x)))    log _3 (2)= log _x ((√(x+1)))     { ((3=x)),((2=(√(x+1)) ; 4=x+1 ; x=3)) :}

$$\:\mathrm{x}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)} \:=\:\sqrt{\mathrm{x}+\mathrm{1}}\: \\ $$$$\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)=\:\mathrm{log}\:_{\mathrm{3}} \left(\sqrt{\mathrm{x}+\mathrm{1}}\right) \\ $$$$\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)=\:\frac{\mathrm{log}\:_{\mathrm{3}} \left(\sqrt{\mathrm{x}+\mathrm{1}}\:\right)}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{x}\right)} \\ $$$$\:\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}\right)=\:\mathrm{log}\:_{\mathrm{x}} \left(\sqrt{\mathrm{x}+\mathrm{1}}\right) \\ $$$$\:\:\begin{cases}{\mathrm{3}=\mathrm{x}}\\{\mathrm{2}=\sqrt{\mathrm{x}+\mathrm{1}}\:;\:\mathrm{4}=\mathrm{x}+\mathrm{1}\:;\:\mathrm{x}=\mathrm{3}}\end{cases}\: \\ $$

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