Question

if
${2}^{a } = {3}^{b} = {6}^{c}$
then show that
$c \times \frac{ab}{a + b}$
​

Answer 1

Let $2^a = 3^b = 6^c = k$

Then, a = $log_{2}k$, b = $log_{3}k$ and c = $log_{6}k$.

ab = $log_{2}k \times log_{3}k$

= $\frac{log^2k}{log2 \times log3}$, since $log_{a}b = \frac{logb}{loga}$

And a + b = $log_{2}k + log_{3}k = \frac{logk}{log2} + \frac{logk}{log3} = logk \times \frac{log2 + log3}{log2 \times log3}$

= $logk \times \frac{log6}{log2 \times log3}$

Now, $\frac{ab}{a + b} = \frac{\frac{log^2k}{log2 \times log3}}{logk \times \frac{log6}{log2 \times log3}}$

= $\frac{logk}{log6}$ (after cancellation)

= $log_{6}k$ = c.

So, $c = \frac{ab}{a + b}$