If a^2+b^2=23 a b, show that : a+b 5 =1 2 ( a+ b) - Vidyadip

Question

If $a^2+b^2=23 a b$, show that :$\log \frac{a+b}{5}=\frac{1}{2}(\log a+\log b)$

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Answer

Given that
$a^2+b^2=23 a b$
$ \Rightarrow a^2+b^2+2 a b=23 a b+2 a b$
$ \Rightarrow a^2+b^2+2 a b=25 a b$
$ \Rightarrow(a+b)^2=25 a b$
$ \Rightarrow \frac{(a+b)^2}{25}=a b$
$ \Rightarrow\left(\frac{a+b}{5}\right)^2=a b$
$ \Rightarrow \log \left(\frac{a+b}{5}\right)^2=\mathrm{ab}$
$ \Rightarrow 2 \log \left(\frac{a+b}{5}\right)=\log \mathrm{ab}$
$ \Rightarrow \log \left(\frac{a+b}{5}\right)=\log \mathrm{ab}$
$ \Rightarrow \log \left(\frac{a+b}{5}\right)=\frac{1}{2}(\log a+\log b)$

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