Question 4988
Sure, say {{{X = 4^Y}}} that means that {{{Y=log( 4, X )}}}

{{{X = 4^Y = (2*2)^Y = 2^Y*2^Y}}}
{{{X = 2^Y*2^Y}}}

take a log with base 2 of the right part of the formula above

{{{log( 2, X ) = log( 2, 2^Y*2*Y) 
= log( 2, 2^Y) + log( 2, 2^Y) 
= 2*log( 2, 2^Y )}}}

Since {{{log( 2, 2^Y )}}} is Y, we have

{{{log( 2, X ) = 2Y}}}

Y, as we know is {{{log( 4, X )}}}, so 

{{{log( 2, X ) = 2*log( 4, X )}}}