Question

State true or false:

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

• A. True
• B. False

Answer 1

The correct option is B False

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

$5^{-p}=k5=k^{-1/p}$

$\Rightarrow 5={\frac{1}{k}}^{1/p}$-----------------(i)

$4^{-q}=k4=k^{-1/q}$

$\Rightarrow 4={\frac{1}{k}}^{1/q}$-----------------(ii)

Now, ${20}^r=k20=k^{1/r}$

$\Rightarrow 5\times 4=k^{1/r}$

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

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

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

$\Rightarrow k^{1/p+1/q+1/r}=k^0$

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