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Question Number 163060 by tounghoungko last updated on 03/Jan/22

$$2\log {}_3\left(\frac{x^2}{27}\right)=2+\frac{\log {}_3\left(\frac{1}{x}\right)}{\log {}_5\left(\sqrt{x}\right)}$$

Commented by tounghoungko last updated on 04/Jan/22

Answered by aleks041103 last updated on 04/Jan/22

$${log}_3\left(x^2\right)-{log}_3\left(27\right)=1-\frac{{log}_{3}x}{2{log}_5\sqrt{x}}$$$${log}_3\left(x^2\right)-3=1-\frac{{log}_{3}x}{{log}_{5}x}$$$${log}_{5}x=\frac{{log}_{3}x}{{log}_{3}5}$$$$\Rightarrow {log}_3\left(x^2\right)-4=-\frac{{log}_{3}x}{\frac{{log}_{3}x}{{log}_{3}5}}=-{log}_{3}5$$$$\Rightarrow {log}_3\left(x^2\right)=4-{log}_{3}5=$$$$={log}_3{3}^4-{log}_{3}5={log}_3\left(\frac{81}{5}\right)$$$$\Rightarrow x=\sqrt{\frac{81}{5}}=\frac{9}{\sqrt{5}}$$$$Ans.x=\frac{9}{\sqrt{5}}$$

Commented by mkam last updated on 03/Jan/22

$$how{log}_{5}x=\frac{{log}_{3}x}{{log}_{3}5}?$$

Commented by aleks041103 last updated on 03/Jan/22

$$b=a^{{log}_{a}b}={\left(c^{{log}_{c}a}\right)}^{{log}_{a}b}=c^{{log}_{c}b}$$$$\Rightarrow {log}_{c}b={log}_{c}a\times {log}_{a}b$$$$\Rightarrow {log}_{a}b=\frac{{log}_{c}b}{{log}_{c}a}$$

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