Question

Assuming that log(mn) = log m + log n prove that $log{x}^n$ = $nlogx$

Answer 1

Let P (n) be true for $n=m$
$\therefore log{X}^m=mlogX$
$\therefore P(m+1)=log{X}^{m+1}$
$=log{X}^m\cdot X=log{X}^m+logX$
$mlogX+logX=(m+1)logX$
Above relation shows that P(n) is true for
$n=m+1$
Now when $n=1,log{X}^1=1logX$
$n=2,log{X}^2=logX\cdot X=logX+logX=2logX$
Above relation show that P(n) is true for $n=1and2$
Hence P(n) is universally true.