If $a^2+b^2=23ab$ then show that $\log a+ \log b=2 \log \left(\frac{a+b}{5}\right)$
If $a^2+b^2=23ab$ then show that $\log a+ \log b=2 \log \left(\frac{a+b}{5}\right)$
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Given,
$a^2+b^2=23ab$
Or, $\left(a+b\right)^2-2ab=23ab$
Or, $\left(a+b\right)^2=23ab+2ab$
Or, $\left(a+b\right)^2=25ab$
Or, $ab=\frac{\left(a+b\right)^2}{25}$
Or, $ab=\frac{\left(a+b\right)^2}{(5)^2}$
Or, $ab=\left(\frac{a+b}{5}\right)^2$
Or, $\log ab=\log \left(\frac{a+b}{5}\right)^2$
[Taking logarithm of both sides]
$\therefore \log a + \log b=2\log \left(\frac{a+b}{5}\right)$ [Proved]
$a^2+b^2=23ab$
Or, $\left(a+b\right)^2-2ab=23ab$
Or, $\left(a+b\right)^2=23ab+2ab$
Or, $\left(a+b\right)^2=25ab$
Or, $ab=\frac{\left(a+b\right)^2}{25}$
Or, $ab=\frac{\left(a+b\right)^2}{(5)^2}$
Or, $ab=\left(\frac{a+b}{5}\right)^2$
Or, $\log ab=\log \left(\frac{a+b}{5}\right)^2$
[Taking logarithm of both sides]
$\therefore \log a + \log b=2\log \left(\frac{a+b}{5}\right)$ [Proved]