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Question Number 10492 by ABD last updated on 14/Feb/17
3^(logx) −2^(logx−1) =2^(logx+1) −2×3^(logx−1) ⇒x=?
$$\mathrm{3}^{{logx}} −\mathrm{2}^{{logx}−\mathrm{1}} =\mathrm{2}^{{logx}+\mathrm{1}} −\mathrm{2}×\mathrm{3}^{{logx}−\mathrm{1}} \Rightarrow{x}=? \\ $$
Answered by mrW1 last updated on 14/Feb/17
3^(logx) −(2^(log x) /2)=2×2^(logx) −2×(3^(log x) /3)  (5/3)×3^(logx) =(5/2)×2^(logx)   3^(logx−1) =2^(logx−1)   ((3/2))^(log x−1) =1  log x−1=0  log x=1  x=10
$$\mathrm{3}^{{logx}} −\frac{\mathrm{2}^{\mathrm{log}\:{x}} }{\mathrm{2}}=\mathrm{2}×\mathrm{2}^{{logx}} −\mathrm{2}×\frac{\mathrm{3}^{\mathrm{log}\:{x}} }{\mathrm{3}} \\ $$$$\frac{\mathrm{5}}{\mathrm{3}}×\mathrm{3}^{{logx}} =\frac{\mathrm{5}}{\mathrm{2}}×\mathrm{2}^{{logx}} \\ $$$$\mathrm{3}^{{logx}−\mathrm{1}} =\mathrm{2}^{{logx}−\mathrm{1}} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{log}\:{x}−\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{log}\:{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{log}\:{x}=\mathrm{1} \\ $$$${x}=\mathrm{10} \\ $$

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