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Question Number 117392 by ZiYangLee last updated on 11/Oct/20

$$\mathrm{If}{\log }_{4}x+{\left({\log }_{4}x\right)}^2+{\left({\log }_{4}x\right)}^3+{\left({\log }_{4}x\right)}^4+...=1$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}x.$$

Answered by Dwaipayan Shikari last updated on 11/Oct/20

$${log}_{4}x+{\left({log}_{4}x\right)}^2+...=1$$$$\frac{{log}_{4}x}{1-{log}_{4}x}=1$$$$\frac{{log}_{4}x}{{log}_4\left(\frac{4}{x}\right)}=1\Rightarrow {log}_4\left(\frac{4}{x}\right)={log}_{4}x\Rightarrow \frac{4}{x}=x\Rightarrow x=2$$$$So$$$$itbecomes{log}_{4}2+{\left({log}_{4}2\right)}^2+...=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....=1$$

Answered by $@y@m last updated on 11/Oct/20

$$\frac{\log x}{1-\log x}=1$$$$1-\log x=\log x$$$$2\log x=1$$$$\log x=\frac{1}{2}$$$$x=4^{\frac{1}{2}}$$$$x=2$$

Commented by ZiYangLee last updated on 12/Oct/20

$$\mathrm{K}$$

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