Question

If $5^p=7^q={35}^{-r}$ then value of $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$ is :

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{3}{2}$

Answer 1

Answer: (C)

$0$

$5^p=7^q={35}^{-r}=k$

$5^p=k\Rightarrow 5=k^{1/p}$

$7^q=k\Rightarrow 7=k^{1/q}$

${35}^{-r}k\to 35=k^{-1/r}$

Now,

$5\times 7=35$

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

$k^{(\frac{1}{p}+\frac{1}{q})}=k^{\frac{-1}{r}}$

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

$\Rightarrow \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=0$

$\therefore$ The solution is $0$