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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=0 is a solution.  if x<0→obv. no soln.  if x>0:  2^x ln(x)=ln(2)+ln(x)  ⇒(2^x −1)ln(x)=ln(2)  ⇒ln(x)=((ln(2))/(2^x −1))  for x>0  ln(x) is strictly increasing  ((ln(2))/(2^x −1)) is strictly decreasing  ⇒incr.=decr.  ⇒for x>0, exist at most 1 solution  x=1:(2^x −1)ln(x)=0<ln(2)  x=2:(2^x −1)ln(x)=3ln(2)>ln(2)  ⇒∃x_1 ∈(1,2), (2^x_1  −1)ln(x_1 )=ln(2)  numerically:  x=0 and x≈1.476

obv.x=0isasolution.ifx<0obv.nosoln.ifx>0:2xln(x)=ln(2)+ln(x)(2x1)ln(x)=ln(2)ln(x)=ln(2)2x1forx>0ln(x)isstrictlyincreasingln(2)2x1isstrictlydecreasingincr.=decr.forx>0,existatmost1solutionx=1:(2x1)ln(x)=0<ln(2)x=2:(2x1)ln(x)=3ln(2)>ln(2)x1(1,2),(2x11)ln(x1)=ln(2)numerically:x=0andx1.476

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