Question
If $\log \frac{a-b}{2}=\frac{1}{2}(\log a+\log b)$, Show that: $a^2+b^2=6 a b$.
✓
Answer
$\log \left(\frac{a-b}{2}\right)=\frac{1}{2}(\log a+\log b)$
$ \Rightarrow \log \left(\frac{a-b}{2}\right)=\frac{1}{2}(\log a b)$
$ \Rightarrow \log \left(\frac{a-b}{2}\right)=\log (a b)^{\frac{1}{2}}$
$ \Rightarrow\left(\frac{a-b}{2}\right)=(a b)^{\frac{1}{2}}$
Squaring both sides we have,
$\left(\frac{a-b}{2}\right)^2=a b$
$ \Rightarrow \frac{(a-b)^2}{4}=a b$
$ \Rightarrow(a-b)^2=4 a b$
$ \Rightarrow a^2+b^2-2 a b=4 a b$
$ \Rightarrow a^2+b^2=4 a b+2 a b$
$ \Rightarrow a^2+b^2=6 a b .$
$ \Rightarrow \log \left(\frac{a-b}{2}\right)=\frac{1}{2}(\log a b)$
$ \Rightarrow \log \left(\frac{a-b}{2}\right)=\log (a b)^{\frac{1}{2}}$
$ \Rightarrow\left(\frac{a-b}{2}\right)=(a b)^{\frac{1}{2}}$
Squaring both sides we have,
$\left(\frac{a-b}{2}\right)^2=a b$
$ \Rightarrow \frac{(a-b)^2}{4}=a b$
$ \Rightarrow(a-b)^2=4 a b$
$ \Rightarrow a^2+b^2-2 a b=4 a b$
$ \Rightarrow a^2+b^2=4 a b+2 a b$
$ \Rightarrow a^2+b^2=6 a b .$
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