Proving of exponential equation If $3^a=21^b$ and $7^c=21^b$
 prove that $b=\frac{ac}{(a+c)}$ 
Can someone please help me prove this ? 
Already tried: 
$3^a=3^b \times 7^b$ and $7^c=3^b \times 7^b$ 
Also, $3^a=7^b$. 

from the first, You have
$$3=(21)^{\frac {b}{a}} $$
and from the second,
$$7=(21 )^{\frac {b}{c} }$$
thus by product,
$$21=(21)^{\frac {b}{a}+\frac {b}{c}} $$
hence
$$1=\frac {b}{a}+\frac {b}{c} $$
and
$$\frac {1}{b}=\frac {1}{a}+\frac {1}{c} $$
$$\implies b=\frac {ac}{a+c} $$
