If $2^{a}=3^{b}=6^{c}$ then show that $ c=\frac{a b}{a+b} $

Given: $2^{a}=3^{b}=6^{c}$ 

To find:  Show that $ c=\frac{a b}{a+b} $

Solution:

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

So,$2^a=d$

⇒$k^{\frac{1}{a}}=2$---------------[i].

and $3^b=d$

⇒$k{\frac{1}{b}}=3$----------------[ii]

.

and also $6^c=k$

⇒$k{\frac{1}{c}}=6$-----------------[iii].

we know that $6=2\times3$

Now substituting , 2,3 and 6

$d^{\frac{1}{a}+\frac{1}{b}} = d^{\frac{1}{c}}$

So, $\frac{1}{c} = \frac{1}{a} + \frac{1}{b}$

⇒$\frac{1}{c} = \frac{b+a}{ ab}$

⇒$c=\frac{ab }{ a+b}$ Hence proved