Checking Mistake: Am I Wrong with $\ln\frac{1}{2}$?

[jiasyuen]

\(\displaystyle \int_{0}^{1}\frac{1}{x-2}dx\)

\(\displaystyle =\left [ \ln\left | x-2 \right | \right ]_{0}^{1}\)

\(\displaystyle =\ln1-\ln2\)

\(\displaystyle -\ln2\)

But the answer given is \(\displaystyle \ln\frac{1}{2}\). Am I wrong?

 

[Siron]

You're correct. Recall that $\ln(a)-\ln(b) = \ln \left(\frac{a}{b}\right)$ and $\ln(a^b) = b \ln(a)$.

 

[jiasyuen]

I've forgotten that. Thanks for reminding.