Question

If ${\log }_5{\log }_5{\log }_{2}x=0$, find the value of $x$.

Answer 1

Given,

${\log }_5{\log }_5{\log }_{2}x=0$

or, ${\log }_5{\log }_{2}x=5^0$ [As if $N={\log }_{a}x,x>0$ then $x=a^N$, from definition of logarithm]

or, ${\log }_5{\log }_{2}x=1$

or, ${\log }_{2}x=5^1$

or, ${\log }_{2}x=5$

or, $x=2^5=32$.

So, value of $x$ is $32$.