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Question Number 208623 by MWSuSon last updated on 19/Jun/24
solve for x, 3^x −2^x =65
$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x},\:\mathrm{3}^{\mathrm{x}} −\mathrm{2}^{\mathrm{x}} =\mathrm{65} \\ $$
Answered by Berbere last updated on 19/Jun/24
if x<0  3^x <1&2^x <1⇒3^x −2^x <1 no solution  if x≥0  x→^(f(x)) 3^x −2^x ⇒f′(x)=ln(3)3^x −ln(2)2^x   3^x >2^x ;ln(3)>ln(2)⇒3^x ln(3)−2^x ln(2)=f′(x)>0  f(4)=81−16=65⇒3^x −2^x =65 has unique solution x=4
$${if}\:{x}<\mathrm{0} \\ $$$$\mathrm{3}^{{x}} <\mathrm{1\&2}^{{x}} <\mathrm{1}\Rightarrow\mathrm{3}^{{x}} −\mathrm{2}^{{x}} <\mathrm{1}\:{no}\:{solution} \\ $$$${if}\:{x}\geqslant\mathrm{0} \\ $$$${x}\overset{{f}\left({x}\right)} {\rightarrow}\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \Rightarrow{f}'\left({x}\right)={ln}\left(\mathrm{3}\right)\mathrm{3}^{{x}} −{ln}\left(\mathrm{2}\right)\mathrm{2}^{{x}} \\ $$$$\mathrm{3}^{{x}} >\mathrm{2}^{{x}} ;{ln}\left(\mathrm{3}\right)>{ln}\left(\mathrm{2}\right)\Rightarrow\mathrm{3}^{{x}} {ln}\left(\mathrm{3}\right)−\mathrm{2}^{{x}} {ln}\left(\mathrm{2}\right)={f}'\left({x}\right)>\mathrm{0} \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{81}−\mathrm{16}=\mathrm{65}\Rightarrow\mathrm{3}^{{x}} −\mathrm{2}^{{x}} =\mathrm{65}\:{has}\:{unique}\:{solution}\:{x}=\mathrm{4} \\ $$

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