Question 358333
{{{log(64, (128))}}}
The key to a simple solution to this is recognizing that both 64 and 128 are both powers of 2. So this expression will simplify quickly if we use the base conversion formula, {{{log(a, (p)) = log(b, (p))/log(b, (a))}}}, to convert the base 64 logarithm into a base 2 logarithms:
{{{log(64, (128)) = log(2, (128))/log(2, (64))}}}
Since {{{2^6 = 64}}} and {{{2^7 = 128}}} the two logarithms are 6 and 7, respectively:
{{{log(64, (128)) = log(2, (128))/log(2, (64)) = 7/6}}}<br>
If 64 and 128 were not both powers of the same number (or if we don't recognize that they are), then an expression like this can be simplified using the base conversion formula to change {{{log(64, (128))}}} into logarithms your calculator "knows" (like base 10 or base e (aka ln)).