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Question Number 208623 by MWSuSon last updated on 19/Jun/24

$$\mathrm{solve}\mathrm{for}\mathrm{x},3^{\mathrm{x}}-2^{\mathrm{x}}=65$$

Answered by Berbere last updated on 19/Jun/24

$$ifx<0$$$$3^x<{1\mathrm{\&}2}^x<1\Rightarrow 3^x-2^x<1nosolution$$$$ifx⩾0$$$$x\stackrel{f\left(x\right)}{\to }{3}^x-2^x\Rightarrow f^{\prime }\left(x\right)=ln\left(3\right)3^x-ln\left(2\right)2^x$$$$3^x>2^x;ln\left(3\right)>ln\left(2\right)\Rightarrow 3^{x}ln\left(3\right)-2^{x}ln\left(2\right)=f^{\prime }\left(x\right)>0$$$$f\left(4\right)=81-16=65\Rightarrow 3^x-2^x=65hasuniquesolutionx=4$$

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