Question 205766
1. use this equation for future reference
{{{x = b^y}}} then {{{y = log(b,x)}}} 
a. {{{32=2^5}}}
= {{{5=log(2,32)}}} 
b. {{{1/125 = 25^(-3/2)}}}
= {{{-3/2 = log(25,5)}}} 
c. {{{12    = 144^(1/2)}}}
= {{{1/2 = log(144,12)}}}