Question

If $5^{-p}=4^{-q}={20}^r$, then $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=$

Answer 1

Let $5^{-p}=4^{-q}={20}^r=k$
$5^{-p}=k$
$\Rightarrow 5=k^{-\frac{1}{p}}$
$\Rightarrow 5=\frac{1}{k^{\frac{1}{p}}}$ ------------(1)
$4^{-q}=k$
$\Rightarrow 4=k^{-\frac{1}{q}}$
$\Rightarrow 4=\frac{1}{k^{\frac{1}{q}}}$ -------------(2)
Now ${20}^r=k$

$\Rightarrow 20=k^{\frac{1}{r}}$
$\Rightarrow 4\times 5=k^{\frac{1}{r}}$
$\Rightarrow \frac{1}{k^{\frac{1}{p}}}\times \frac{1}{k^{\frac{1}{q}}}=k^{\frac{1}{r}}$
$\Rightarrow k^{\frac{1}{p}}{k}^{\frac{1}{q}}{k}^{\frac{1}{r}}=1$
$\Rightarrow k^{\frac{1}{p}+\frac{1}{q}+\frac{1}{r}}=k^0$
$\Rightarrow \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=0$