Question 278548
{{{2^log2^5 = x}}}
If this is what the problem really looks like then, since {{{2^5 = 32}}}, this simplifies to:
{{{2^log((32))=x}}}
Now we use our calculators to find log(32) and then use the calculators again to raise to to that power.<br>
However, I suspect the the problem is actually:
{{{2^log(2, (5)) = x}}}
This equation is extremely easy if you understand what logairthms are. Logarithms are exponents. In general, {{{log(a, (b))}}} represents the exponent you put on a which results in b. This specific logarithm, {{{log(2, (5))}}}, represents the exponent you put on 2 to get 5. And where do we see {{{log(2, (5))}}}? We see it as an exponent on 2! So {{{2^log(2, (5))}}}, by the definition of logarithms, is 5!