Question 959033
log64 32 = x
<pre>{{{log (64, 32) = x}}} ----- LOGARITHMIC form
{{{64^x = 32}}} -------- Changing from LOGARITHMIC to EXPONENTIAL form
{{{(2^6)^x = 2^5}}} ------ Changing bases to a common base of 2
{{{2^(6x) = 2^5}}}
6x = 5 --------- Bases are equal, and so are their exponents
{{{highlight_green(x = 5/6)}}}