If 5^-p = 4^-q = 20^r, show that :

$\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} = 0$.

Given,

⇒ 5^-p = 4^-q = 20^r = k (let)

⇒ 5^-p = k

⇒ 5 = $k^{-\dfrac{1}{p}}$ …….(1)

⇒ 4^-q = k

⇒ 4 = $k^{-\dfrac{1}{q}}$ ……(2)

⇒ 20^r = k

⇒ 20 = $k^{\dfrac{1}{r}}$ …………(3)

We know that,

⇒ 5 × 4 = 20

From equations (1), (2) and (3), we get :

$\Rightarrow k^{-\dfrac{1}{p}} \times k^{-\dfrac{1}{q}} = k^{\dfrac{1}{r}} \\[1em] \Rightarrow k^{-\dfrac{1}{p} + \Big(-\dfrac{1}{q}\Big)} = k^{\dfrac{1}{r}} \\[1em] \Rightarrow k^{-\dfrac{1}{p} - \dfrac{1}{q}} = k^{\dfrac{1}{r}} \\[1em] \Rightarrow -\dfrac{1}{p} - \dfrac{1}{q} = \dfrac{1}{r} \\[1em] \Rightarrow \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} = 0.$

Hence, proved that $\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} = 0.$.