right answer, wrong reason.
You don't "cancel" the 2's. You evaluate the log function. By definition (using bas 2),
2^(log x) = x
That is, the log of a number is the power of the base you need to get that number.
So, log 2^x = x*log 2 = x*1 = x
Solve
(log base 2, cube root of 2)
log2(2^1/3)
You transform it into the form above, right? then you cancel the 2s to get 1/3
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