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Question Number 94456 by Abdulrahman last updated on 18/May/20

$${\log }_{\mathrm{x}}16+{\log }_{16}\mathrm{x}={\log }_{512}\mathrm{x}+{\log }_{\mathrm{x}}512$$$$\mathrm{x}=?$$

Answered by john santu last updated on 19/May/20

$$4\left(\log {}_x\left(2\right)\right)+\frac{1}{4}\left({\log }_2\left(x\right)\right)=$$$$9\left(\log {}_x\left(2\right)\right)+\frac{1}{9}\left(\log {}_2\left(x\right)\right)$$$$\Rightarrow 5\left(\log {}_x\left(2\right)\right)=\frac{5}{36}\left(\log {}_2\left(x\right)\right)$$$$36\left(\log {}_x\left(2\right)\right)=\frac{1}{\log {}_x\left(2\right)}$$$$\{\begin{array}{l}\log {}_x\left(2\right)=\frac{1}{6}\Rightarrow \frac{1}{\log {}_2\left(x\right)}=\frac{1}{6};x=64\\ \log {}_x\left(2\right)=-\frac{1}{6}\Rightarrow \frac{1}{\log {}_2\left(x\right)}=-\frac{1}{6}\end{array}$$$$x=\frac{1}{64}$$

Commented by Abdulrahman last updated on 19/May/20

$$\mathrm{very}\mathrm{good}$$$$\mathrm{but}\mathrm{i}\mathrm{didnt}\mathrm{know}\mathrm{in}\mathrm{two}\mathrm{final}\mathrm{lines}$$

Commented by i jagooll last updated on 19/May/20

$$\mathrm{nice}\mathrm{solution}$$

Commented by i jagooll last updated on 19/May/20

$$\mathrm{why}\mathrm{sir}?$$$${\left(\log {}_x\left(2\right)\right)}^2=\frac{1}{36}.let\log {}_x\left(2\right)=p$$$$p^2-\frac{1}{36}=0$$$$(p-\frac{1}{6})(p+\frac{1}{6})=0sir$$

Commented by Abdulrahman last updated on 19/May/20

$$\mathrm{thanks}\mathrm{alot}\mathrm{now}\mathrm{i}\mathrm{got}\mathrm{it}$$

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