Question 42997
{{{x^log(5,3)=3}}}


Taking the logarithm of both sides w.r.t. base 3
{{{log(3,x^log(5,3))=log(3,3)}}}
or {{{log(5,3)*log(3,x)=1}}}


We can write for a>0 and b>0 {{{log(a,b) = log(c,a)/log(c,b)}}} where c is any number.


or {{{(log(10,3)/log(10,5))*(log(10,x)/log(10,3))=1}}}


Cancelling {{{log(10,3)}}} from both numerator and denominator we have
{{{log(10,x)/log(10,5)=1}}}
or {{{log(5,x)=1}}}
or {{{x = 5^1}}} [since for a>0 {{{log(a,M) = b}}} means {{{M = a^b}}}]
or {{{x=5}}}


So the reqd. value of 'x' is 5.