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please help guys anything will be helpful

Please Help Guys Anything Will Be Helpful class=

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Loneguyidn

[tex]~~~~~2^{x+1} = 3^x\\\\\implies \ln \left(2^{x+1}\right) = \ln\left(3^x \right)\\\\\\\implies (x+1) \ln 2 = x \ln 3\\\\\\\implies \dfrac{x+1}x = \dfrac{\ln 3}{\ln 2}\\\\\\\implies 1+ \dfrac 1x = \dfrac{\ln 3}{\ln 2}\\\\\\\implies \dfrac 1x = \dfrac{\ln 3 }{ \ln 2}-1\\\\\\\implies \dfrac 1x = \dfrac{ \ln 3 - \ln 2}{\ln 2}\\\\\\\implies x = \dfrac{\ln 2}{\ln 3 - \ln 2}\\\\\\\implies x \approx 1.7095\\\\\\[/tex]

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