Question

If $\log \frac{a-b}{2}=\frac{1}{2}(\log a+\log b)$, then which of the following is true?

• A. $a^2+b^2=6ab$
• B. $a^2+b^2=-6ab$
• C. $a^2-b^2=6ab$
• D. $a^2-b^2=-6ab$

Answer 1

The correct option is A $a^2+b^2=6ab$

$\log \frac{a-b}{2}=\frac{1}{2}(\log a+\log b)$

$\log \frac{a-b}{2}=\frac{1}{2}(\log a+\log b)$$\Rightarrow 2\log \frac{a-b}{2}=(\log a+\log b)$

$\Rightarrow \log {\left(\frac{a-b}{2}\right)}^2=\log ab$

$\because \log m^n=n\log m$ and $\log \left(mn\right)=\log m+\log n$
$\Rightarrow {\left(\frac{a-b}{2}\right)}^2=ab$

$\frac{a^2+b^2-2ab}{4}=ab$

$\Rightarrow a^2+b^2-2ab=4ab$
$\Rightarrow a^2+b^2=2ab+4ab$
$\Rightarrow a^2+b^2=6ab$