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

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The integral evaluated is $$\int_{0}^{1}\frac{1}{x-2}dx$$, leading to the expression $$-\ln2$$. The confusion arises from the relationship between logarithmic properties, specifically that $$\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$$. Thus, $$-\ln2$$ can be rewritten as $$\ln\frac{1}{2}$$, confirming the answer given. The discussion highlights the importance of remembering logarithmic identities in solving integrals.
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$$\int_{0}^{1}\frac{1}{x-2}dx$$

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

$$=\ln1-\ln2$$

$$-\ln2$$

But the answer given is $$\ln\frac{1}{2}$$. Am I wrong?
 
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You're correct. Recall that $\ln(a)-\ln(b) = \ln \left(\frac{a}{b}\right)$ and $\ln(a^b) = b \ln(a)$.
 
I've forgotten that. Thanks for reminding.
 

Thread 'Proving that convexity implies second order derivative being positive'

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