Question

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

Answer 1

Let 5^-P = 4^-q = 20^r = k
5^-P = k ⇒ 5 = ${\displaystyle k^{-\frac{1}{p}} [\because a^p=b^q\Rightarrow a=b^{\frac{q}{p}}]}$

4^-q = k ⇒ 4 = ${\displaystyle k^{-\frac{1}{q}} [\because a^p=b^q\Rightarrow a=b^{\frac{q}{p}}]}$

20^r = k ⇒ 20 = ${\displaystyle k^{\frac{1}{r}} [\because a^p=b^q\Rightarrow a=b^{\frac{q}{p}}]}$

5 x 4 = 20
⇒ ${\displaystyle k^{-\frac{1}{p}}\times k^{-\frac{1}{q}}=k^{\frac{1}{r}}}$ 

⇒ ${\displaystyle k^{-\frac{1}{p}-\frac{1}{q}}=k^{\frac{1}{r}}}$

⇒ ${\displaystyle k^0=k^{\frac{1}{p}+\frac{1}{q}+\frac{1}{r}}}$

If bases are equal, powers are also equal.
⇒  ${\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=0}$