Question

If $\sqrt{{13}^2-{12}^2}=\sqrt[n]{n125}$, then $n$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4
• E. 5

Answer 1

Answer: (C)

3

Given, $\sqrt{{13}^2-{12}^2}=\sqrt[n]{n125}$
We have $\sqrt{{13}^2-{12}^2}=\sqrt{169-144}$

$=\sqrt{25}$

$=5$

And $\sqrt[n]{n125}=5^{3/n}$
When we equate those two, we get $5^1=5^{3/n}$
Therefore by comparing powers, we get $\frac{3}{n}=1$ $\Rightarrow n=3$