Question

Find the value of n, if

$(\frac{5}{3}{)}^{2n+1}\times (\frac{5}{3}{)}^5=(\frac{5}{3}{)}^{n+2}.$

• A. $n=4$
• B. $n=1$
• C. $n=-4$
• D. $n=0$

Answer 1

The correct option is C $n=-4$

By the Laws of exponents , we know that $a^m\times a^n=a^{m+n}$

$\Rightarrow (\frac{5}{3}{)}^{2n+1}\times (\frac{5}{3}{)}^5=(\frac{5}{3}{)}^{2n+1+5}$

$\Rightarrow (\frac{5}{3}{)}^{2n+6}=(\frac{5}{3}{)}^{n+2}$

Since the bases are same, we can equate the powers.

$\Rightarrow 2n+6=n+2\Rightarrow 2n-n=2-6\Rightarrow n=-4$