How do I solve ln(1/2)^m = ln2?

Apr 1, 2017

Two ways...

One is simply recognizing that by algebra, ${2}^{- 1} = \frac{1}{2}$. Hence, $m = - 1$ by inspection, i.e. ${\left({2}^{- 1}\right)}^{- 1} = 2 = {\left(\frac{1}{2}\right)}^{- 1}$.

Another way is that anytime you see an exponent, try taking the $\ln$ of both sides.

$\ln {\left(\frac{1}{2}\right)}^{m} = \ln 2$

Use the property that

$\ln {a}^{b} = b \ln a$

Thus, we get:

$m \ln \left(\frac{1}{2}\right) = \ln 2$

$m \ln {\left(2\right)}^{- 1} = \ln 2$

$- m \ln 2 = \ln 2$

$\implies m = - 1$