Question Number 103961 by MMpotulo last updated on 18/Jul/20
$${what}\:{is}\: \\ $$$$\mathrm{2}^{{log}\mathrm{2}{x}} =\mathrm{3}^{{log}\mathrm{3}{x}} \\ $$
Answered by bramlex last updated on 18/Jul/20
$$\mathrm{ln}\:\left(\mathrm{2}^{\mathrm{ln}\:\left(\mathrm{2}{x}\right)} \right)\:=\:\mathrm{ln}\:\left(\mathrm{3}^{\mathrm{ln}\:\left(\mathrm{3}{x}\right)} \right) \\ $$$$\mathrm{ln}\:\left(\mathrm{2}{x}\right).\mathrm{ln}\:\left(\mathrm{2}\right)=\:\mathrm{ln}\:\left(\mathrm{3}{x}\right).\mathrm{ln}\:\left(\mathrm{3}\right) \\ $$$$\left\{\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:{x}\right\}.\mathrm{ln}\:\mathrm{2}\:=\:\left\{\mathrm{ln}\:\mathrm{3}+\mathrm{ln}\:{x}\right\}.\mathrm{ln}\:\mathrm{3} \\ $$$$\mathrm{ln}\:^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{ln}\:^{\mathrm{2}} \left(\mathrm{3}\right)=−\left(\mathrm{ln}\:\mathrm{2}−\mathrm{ln}\:\mathrm{3}\right).\mathrm{ln}\:{x} \\ $$$$\mathrm{ln}\:\left(\mathrm{2}\right)+\mathrm{ln}\:\left(\mathrm{3}\right)=\:−\mathrm{ln}\:{x} \\ $$$$−\mathrm{ln}\:\left(\mathrm{6}\right)\:=\:\mathrm{ln}\:\left({x}\right)\:\Rightarrow\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\:\bigstar \\ $$
Commented by MMpotulo last updated on 18/Jul/20
$${there}\:{is}\:{another}\:{way}\:{of}\:{doing}\:{it}\:{right} \\ $$$${which}\:{is} \\ $$$$\mathrm{2}^{{log}\mathrm{2}{x}} =\mathrm{3}^{{log}\mathrm{3}{x}} \\ $$$${log}\mathrm{2}^{{log}\mathrm{2}{x}} ={log}\mathrm{3}^{{log}\mathrm{3}{x}} \\ $$$$\left({log}\mathrm{2}{x}\right)\left({log}\mathrm{2}\right)=\left({log}\mathrm{3}{x}\right)\left({log}\mathrm{3}\right) \\ $$$$\left({log}\mathrm{2}×{log}\right. \\ $$$$\left.{x}\right)\left({log}\mathrm{2}\right)=\left({log}\mathrm{3}\right)\left({logx}\right)\left({log}\mathrm{3}\right) \\ $$$$?{then}? \\ $$
Commented by bobhans last updated on 18/Jul/20
$${wrong}.\:\left(\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:{x}\right).\mathrm{ln}\:\mathrm{2}\:=\:\left(\mathrm{ln}\:\mathrm{3}+\mathrm{ln}\:{x}\right).\mathrm{ln}\:\mathrm{3} \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 18/Jul/20
$${log}\mathrm{2}{xlog}\mathrm{2}={log}\mathrm{3}{xlog}\mathrm{3} \\ $$$$\left({log}\mathrm{2}+{logx}\right){log}\mathrm{2}=\left({log}\mathrm{3}+{logx}\right){log}\mathrm{3} \\ $$$$\left({log}\mathrm{2}\right)^{\mathrm{2}} +{logxlog}\mathrm{2}=\left({log}\mathrm{3}\right)^{\mathrm{2}} +{log}\mathrm{3}{logx} \\ $$$$\left({log}\mathrm{2}\right)^{\mathrm{2}} −\left({log}\mathrm{3}\right)^{\mathrm{2}} ={logx}\left({log}\mathrm{3}−{log}\mathrm{2}\right) \\ $$$$−\left({log}\mathrm{6}\right)={logx} \\ $$$${log}_{{x}} \mathrm{6}=−\mathrm{1} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$$$ \\ $$
Answered by OlafThorendsen last updated on 18/Jul/20
$$\mathrm{log}\left(\mathrm{2}^{\mathrm{log2}{x}} \right)\:=\:\mathrm{log}\left(\mathrm{3}^{\mathrm{log3}{x}} \right) \\ $$$$\mathrm{log2}{x}\mathrm{log2}\:=\:\mathrm{log3}{x}\mathrm{log3} \\ $$$$\left(\mathrm{log2}+\mathrm{log}{x}\right)\mathrm{log2}\:=\:\left(\mathrm{log3}+\mathrm{log}{x}\right)\mathrm{log3} \\ $$$$\mathrm{log}{x}\left(\mathrm{log3}−\mathrm{log2}\right)\:=\:\mathrm{log}^{\mathrm{2}} \mathrm{2}−\mathrm{log}^{\mathrm{2}} \mathrm{3} \\ $$$$\mathrm{log}{x}\left(\mathrm{log3}−\mathrm{log2}\right)\:=\:\left(\mathrm{log2}−\mathrm{log3}\right)\left(\mathrm{log2}+\mathrm{log3}\right) \\ $$$$\mathrm{log}{x}\:=\:−\left(\mathrm{log2}+\mathrm{log3}\right) \\ $$$$\mathrm{log}{x}\:=\:−\mathrm{log6} \\ $$$$\mathrm{log}{x}\:=\:\mathrm{log}\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${x}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$