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Question Number 103961 by MMpotulo last updated on 18/Jul/20

what is 2^(log2x) =3^(log3x)

$${what}\:{is}\: \\ $$$$\mathrm{2}^{{log}\mathrm{2}{x}} =\mathrm{3}^{{log}\mathrm{3}{x}} \\ $$

Answered by bramlex last updated on 18/Jul/20

ln (2^(ln (2x)) ) = ln (3^(ln (3x)) ) ln (2x).ln (2)= ln (3x).ln (3) {ln 2+ln x}.ln 2 = {ln 3+ln x}.ln 3 ln^2 (2)−ln^2 (3)=−(ln 2−ln 3).ln x ln (2)+ln (3)= −ln x −ln (6) = ln (x) ⇒ x = (1/6) ★

$$\mathrm{ln}\:\left(\mathrm{2}^{\mathrm{ln}\:\left(\mathrm{2}{x}\right)} \right)\:=\:\mathrm{ln}\:\left(\mathrm{3}^{\mathrm{ln}\:\left(\mathrm{3}{x}\right)} \right) \\ $$$$\mathrm{ln}\:\left(\mathrm{2}{x}\right).\mathrm{ln}\:\left(\mathrm{2}\right)=\:\mathrm{ln}\:\left(\mathrm{3}{x}\right).\mathrm{ln}\:\left(\mathrm{3}\right) \\ $$$$\left\{\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:{x}\right\}.\mathrm{ln}\:\mathrm{2}\:=\:\left\{\mathrm{ln}\:\mathrm{3}+\mathrm{ln}\:{x}\right\}.\mathrm{ln}\:\mathrm{3} \\ $$$$\mathrm{ln}\:^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{ln}\:^{\mathrm{2}} \left(\mathrm{3}\right)=−\left(\mathrm{ln}\:\mathrm{2}−\mathrm{ln}\:\mathrm{3}\right).\mathrm{ln}\:{x} \\ $$$$\mathrm{ln}\:\left(\mathrm{2}\right)+\mathrm{ln}\:\left(\mathrm{3}\right)=\:−\mathrm{ln}\:{x} \\ $$$$−\mathrm{ln}\:\left(\mathrm{6}\right)\:=\:\mathrm{ln}\:\left({x}\right)\:\Rightarrow\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\:\bigstar \\ $$

Commented by MMpotulo last updated on 18/Jul/20

there is another way of doing it right which is 2^(log2x) =3^(log3x) log2^(log2x) =log3^(log3x) (log2x)(log2)=(log3x)(log3) (log2×log x)(log2)=(log3)(logx)(log3) ?then?

$${there}\:{is}\:{another}\:{way}\:{of}\:{doing}\:{it}\:{right} \\ $$$${which}\:{is} \\ $$$$\mathrm{2}^{{log}\mathrm{2}{x}} =\mathrm{3}^{{log}\mathrm{3}{x}} \\ $$$${log}\mathrm{2}^{{log}\mathrm{2}{x}} ={log}\mathrm{3}^{{log}\mathrm{3}{x}} \\ $$$$\left({log}\mathrm{2}{x}\right)\left({log}\mathrm{2}\right)=\left({log}\mathrm{3}{x}\right)\left({log}\mathrm{3}\right) \\ $$$$\left({log}\mathrm{2}×{log}\right. \\ $$$$\left.{x}\right)\left({log}\mathrm{2}\right)=\left({log}\mathrm{3}\right)\left({logx}\right)\left({log}\mathrm{3}\right) \\ $$$$?{then}? \\ $$

Commented by bobhans last updated on 18/Jul/20

wrong. (ln 2+ln x).ln 2 = (ln 3+ln x).ln 3

$${wrong}.\:\left(\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:{x}\right).\mathrm{ln}\:\mathrm{2}\:=\:\left(\mathrm{ln}\:\mathrm{3}+\mathrm{ln}\:{x}\right).\mathrm{ln}\:\mathrm{3} \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 18/Jul/20

log2xlog2=log3xlog3 (log2+logx)log2=(log3+logx)log3 (log2)^2 +logxlog2=(log3)^2 +log3logx (log2)^2 −(log3)^2 =logx(log3−log2) −(log6)=logx log_x 6=−1 x=(1/6)

$${log}\mathrm{2}{xlog}\mathrm{2}={log}\mathrm{3}{xlog}\mathrm{3} \\ $$$$\left({log}\mathrm{2}+{logx}\right){log}\mathrm{2}=\left({log}\mathrm{3}+{logx}\right){log}\mathrm{3} \\ $$$$\left({log}\mathrm{2}\right)^{\mathrm{2}} +{logxlog}\mathrm{2}=\left({log}\mathrm{3}\right)^{\mathrm{2}} +{log}\mathrm{3}{logx} \\ $$$$\left({log}\mathrm{2}\right)^{\mathrm{2}} −\left({log}\mathrm{3}\right)^{\mathrm{2}} ={logx}\left({log}\mathrm{3}−{log}\mathrm{2}\right) \\ $$$$−\left({log}\mathrm{6}\right)={logx} \\ $$$${log}_{{x}} \mathrm{6}=−\mathrm{1} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$$$ \\ $$

Answered by OlafThorendsen last updated on 18/Jul/20

log(2^(log2x) ) = log(3^(log3x) ) log2xlog2 = log3xlog3 (log2+logx)log2 = (log3+logx)log3 logx(log3−log2) = log^2 2−log^2 3 logx(log3−log2) = (log2−log3)(log2+log3) logx = −(log2+log3) logx = −log6 logx = log(1/6) x = (1/6)

$$\mathrm{log}\left(\mathrm{2}^{\mathrm{log2}{x}} \right)\:=\:\mathrm{log}\left(\mathrm{3}^{\mathrm{log3}{x}} \right) \\ $$$$\mathrm{log2}{x}\mathrm{log2}\:=\:\mathrm{log3}{x}\mathrm{log3} \\ $$$$\left(\mathrm{log2}+\mathrm{log}{x}\right)\mathrm{log2}\:=\:\left(\mathrm{log3}+\mathrm{log}{x}\right)\mathrm{log3} \\ $$$$\mathrm{log}{x}\left(\mathrm{log3}−\mathrm{log2}\right)\:=\:\mathrm{log}^{\mathrm{2}} \mathrm{2}−\mathrm{log}^{\mathrm{2}} \mathrm{3} \\ $$$$\mathrm{log}{x}\left(\mathrm{log3}−\mathrm{log2}\right)\:=\:\left(\mathrm{log2}−\mathrm{log3}\right)\left(\mathrm{log2}+\mathrm{log3}\right) \\ $$$$\mathrm{log}{x}\:=\:−\left(\mathrm{log2}+\mathrm{log3}\right) \\ $$$$\mathrm{log}{x}\:=\:−\mathrm{log6} \\ $$$$\mathrm{log}{x}\:=\:\mathrm{log}\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${x}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$

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