Proving of exponential equation

Question

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, $3^a=7^b$.

Answer 1

It's wrong, of course. Try $a=b=c=0$.

For $abc\neq0$ we obtain:

$3^{ac}=21^{bc}$ and $7^{ac}=21^{ab}$.

Thus, $$21^{ac}=21^{bc+ab}$$ or $$ac=ab+bc$$ or $$b=\frac{ac}{a+c}$$ and we are done!

Answer 2

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

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

$$\implies b=\frac {ac}{a+c} $$