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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.

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

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  • 4
    $\begingroup$ Didi you try anything? $\endgroup$ – Aqua Mar 21 '18 at 20:54
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$$2 \ln \left(\frac{1}{2} (a + b)\right) = \ln a + \ln b$$

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

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

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

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