Question 1132452
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Using change-of-base formula...<br>
{{{log(256,16) = log(b,16)/log(b,256) = log(2,16)/log(2,256) = 4/8 = 1/2 = 0.50}}}<br>
ANSWER: 0.50<br>
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...or, changing the log equation to exponential form...<br>
{{{log(256,16) = b}}}  -->  {{{256^b = 16}}}  -->  {{{(2^8)^b = 2^(8b) = 2^4}}}  -->  {{{8b = 4}}}  -->  b = 0.50