Question

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

• A. 1
• B. 0
• C. 5
• D. 4

Answer 1

The correct option is B

0

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

$\Rightarrow 5^{-p}=k$ $\Rightarrow 5=k^{\frac{-1}{p}}$

$\therefore 5=\frac{1}{k^{\frac{1}{p}}}$.....(i)

Similarly $4=\frac{1}{k^{\frac{1}{q}}}$.......(ii)

20 = $k^{\frac{1}{r}}$

4 x 5 = $k^{\frac{1}{r}}$

From (i) and (ii)

$\frac{1}{k^{\frac{1}{p}}}\times \frac{1}{k^{\frac{1}{q}}}$ = $k^{\frac{1}{r}}$

$\Rightarrow 1=k^{\frac{1}{r}}\times k^{\frac{1}{p}}\times k^{\frac{1}{q}}$

$\Rightarrow 1=k^{\frac{1}{r}+\frac{1}{p}+\frac{1}{q}}$

$\Rightarrow \frac{1}{r}+\frac{1}{p}+\frac{1}{q}=0$ (since $k^0$ = 1)