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

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

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) ★

Commented byMMpotulo 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?

Commented bybobhans last updated on 18/Jul/20

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

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)