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Question Number 194613 by cortano12 last updated on 11/Jul/23

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Answered by MM42 last updated on 11/Jul/23

$$\frac{1}{{log}_{x}4x}+\frac{1}{{log}_x\frac{x}{2}}=2$$$$\frac{1}{2{log}_{x}2+1}+\frac{1}{1-{log}_{x}2}=2$$$$let{log}_{x}2=y$$$$4y^2-y=0\Rightarrow y=0={log}_{x}2\times$$$$ory=\frac{1}{4}={log}_{x}2\Rightarrow x=16✓$$$$$$

Answered by mahdipoor last updated on 11/Jul/23

$${log}_{ab}c=\frac{1}{{log}_{c}ab}=\frac{1}{{log}_{c}a+{log}_{c}b}$$$${log}_{4x}x+{log}_{x/2}x=\frac{1}{{log}_{x}4+{log}_{x}x}+\frac{1}{{log}_{x}x+{log}_{x}1/2}=$$$$=\frac{1}{2{log}_{x}2+1}+\frac{1}{1-{log}_{x}2}=2\Rightarrow {log}_{x}2=u\ne 1,-.5\Rightarrow$$$$(1-u)+(2u+1)=2(1-u)(2u+1)\Rightarrow$$$${log}_{x}2=\frac{1}{{log}_{2}x}=\frac{1}{4},0\Rightarrow x=2^4=16$$

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