# Help with a logarithm proof [closed]

If $$\log \frac {1} {2} (a + b) = \frac {1} {2}(\log (a) + \log (b) )$$ Prove that $$(a + b)^2 = 4ab$$

Can anyone show me how to do this?

Thanks.

## closed as off-topic by GNUSupporter 8964民主女神 地下教會, futurebird, Arnaud D., hardmath, Trevor GunnMar 21 '18 at 23:53

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• Didi you try anything? – Aqua Mar 21 '18 at 20:54

## 1 Answer

$$2 \ln \left(\frac{1}{2} (a + b)\right) = \ln a + \ln b$$

$$\ln \left(\frac{1}{4} (a+b)^2\right) = \ln (ab)$$

$$(a+b)^2 = 4ab$$

• I'll just add that you can go from the second line to the third because $ln(x)$ is an injective function. – F.A. Mar 21 '18 at 21:00
• @F.A. yes, a useful remark – D F Mar 21 '18 at 21:02
• How did you go from $\frac {1} {2} (a+b)$ to $\frac {1} {4} (a+b)^2$ ? – Jimmy Mar 21 '18 at 21:09
• @Jimmy because of the identity $a \ln b = \ln (b^a)$ – D F Mar 21 '18 at 21:10
• Okay I see. Thank you. – Jimmy Mar 21 '18 at 21:12