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Question Number 172744 by Adilali last updated on 30/Jun/22
Answered by aleks041103 last updated on 30/Jun/22
obv.x=0isasolution.ifx<0→obv.nosoln.ifx>0:2xln(x)=ln(2)+ln(x)⇒(2x−1)ln(x)=ln(2)⇒ln(x)=ln(2)2x−1forx>0ln(x)isstrictlyincreasingln(2)2x−1isstrictlydecreasing⇒incr.=decr.⇒forx>0,existatmost1solutionx=1:(2x−1)ln(x)=0<ln(2)x=2:(2x−1)ln(x)=3ln(2)>ln(2)⇒∃x1∈(1,2),(2x1−1)ln(x1)=ln(2)numerically:x=0andx≈1.476
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