Question

If $5^{-p}=4^{-q}={20}^r$; show that:
$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=0$

Answer 1

$5^{-p}=4^{-q}={20}^r=k$

$\Rightarrow {\log }_5{5}^{-p}={\log }_{5}k\Rightarrow p=-{\log }_{5}k$

Similarly, $q=-{\log }_{4}k$ and $r={\log }_{20}k$

$\therefore \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{1}{-{\log }_{5}k}+\frac{1}{-{\log }_{4}k}+\frac{1}{{\log }_{20}k}$

$={\log }_k\left(\frac{1}{5}\right)+{\log }_k\left(\frac{1}{4}\right)+{\log }_{k}20$

$={\log }_{k}1=0$