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Question Number 48735 by Pk1167156@gmail.com last updated on 28/Nov/18

The roots of the equation 2^(x+2) ∙ 3^((3x)/(x−1)) = 9 are given by

$$\mathrm{The}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\mathrm{2}^{{x}+\mathrm{2}} \centerdot\:\mathrm{3}^{\frac{\mathrm{3}{x}}{{x}−\mathrm{1}}} =\:\mathrm{9}\:\mathrm{are}\:\mathrm{given}\:\mathrm{by} \\ $$

Answered by Smail last updated on 28/Nov/18

2^(x+2) .3^((3x)/(x−1)) =9= (x+2)ln(2)+((3x)/(x−1))ln3=2ln3 (x^2 +x−2)ln2+3xln3−2xln3+2ln3=0 x^2 ln2+(ln2+ln3)x+2(ln3−ln2)=0 x^2 +((ln2+ln3)/(ln2))x+((2(ln3−ln2))/(ln2))=0 x(x+2)+((ln3−ln2)/(ln2))(x+2)=0 (x+2)(x+((ln3−ln2)/(ln2)))=0 x=−2 or x=1−((ln3)/(ln2))

$$\mathrm{2}^{{x}+\mathrm{2}} .\mathrm{3}^{\frac{\mathrm{3}{x}}{{x}−\mathrm{1}}} =\mathrm{9}= \\ $$$$\left({x}+\mathrm{2}\right){ln}\left(\mathrm{2}\right)+\frac{\mathrm{3}{x}}{{x}−\mathrm{1}}{ln}\mathrm{3}=\mathrm{2}{ln}\mathrm{3} \\ $$$$\left({x}^{\mathrm{2}} +{x}−\mathrm{2}\right){ln}\mathrm{2}+\mathrm{3}{xln}\mathrm{3}−\mathrm{2}{xln}\mathrm{3}+\mathrm{2}{ln}\mathrm{3}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} {ln}\mathrm{2}+\left({ln}\mathrm{2}+{ln}\mathrm{3}\right){x}+\mathrm{2}\left({ln}\mathrm{3}−{ln}\mathrm{2}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\frac{{ln}\mathrm{2}+{ln}\mathrm{3}}{{ln}\mathrm{2}}{x}+\frac{\mathrm{2}\left({ln}\mathrm{3}−{ln}\mathrm{2}\right)}{{ln}\mathrm{2}}=\mathrm{0} \\ $$$${x}\left({x}+\mathrm{2}\right)+\frac{{ln}\mathrm{3}−{ln}\mathrm{2}}{{ln}\mathrm{2}}\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)\left({x}+\frac{{ln}\mathrm{3}−{ln}\mathrm{2}}{{ln}\mathrm{2}}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{2}\:\:{or}\:\:{x}=\mathrm{1}−\frac{{ln}\mathrm{3}}{{ln}\mathrm{2}} \\ $$

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