Question 627858
{{{log(2, (32^(1/5)))}}}
This expression is simple to evaluate if you understand what logarithms are and if you can figure out powers of 2. A base 2 logarithm, like yours, is an exponent that you would put one a 2. This particular logarithm is the exponent you would put on a 2 to get {{{32^(1/5)}}} as a result.<br>
So what power of 2 <i>does</i> result in {{{32^(1/5)}}}? The answer is easier than you think because 32 happens to be a power of 2! {{{32 = 2^5}}}. Substituting this into your expression we get:
{{{log(2, ((2^5)^(1/5)))}}}
Using our exponent rule for powers of a power (i.e. multiply the exponents) we get:
{{{log(2, (2))}}}
Now it is easy to see what power of 2 (the 2 in the base) results in a 2 (the 2 in the argument). It is 1! So:
{{{log(2, (32^(1/5))) = 1}}}