## anonymous 4 years ago the exact value of this logarithm? ln√e

1. anonymous

$\frac{1}{2}$

2. anonymous

how :S

3. anonymous

since $\ln(\sqrt{e})=\ln(e^{\frac{1}{2}}$ w

4. anonymous

typo i mean $\ln(\sqrt{e})=\ln(e^{\frac{1}{2}})$

5. anonymous

when you have logs good to write in exponential notation

6. anonymous

$\ln(x)=y\iff e^y=x$ so so you are solving $e^y=e^{\frac{1}{2}}$