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

[jiasyuen]

$$\int_{0}^{1}\frac{1}{x-2}dx$$

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

$$=\ln1-\ln2$$

$$-\ln2$$

But the answer given is $$\ln\frac{1}{2}$$. 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.