which statement is true?\na $\frac{3^{-15}}{3^{7}}<(3^{-8})cdot(3^{-9})$\nb $\frac{3^{-15}}{3^{7}}>(3^{-8})cd…

Explanation:

Step1: Simplify the left - hand side

Using the rule $\frac{a^m}{a^n}=a^{m - n}$, for $\frac{3^{-15}}{3^{7}}$, we have $3^{-15-7}=3^{-22}$.

Step2: Simplify the right - hand side

Using the rule $a^m\times a^n=a^{m + n}$, for $(3^{-8})\times(3^{-9})$, we get $3^{-8+( - 9)}=3^{-17}$.

Step3: Compare the two results

Since for a non - zero base $a$ and exponents $m$ and $n$, if $m\lt n$, then $a^m\lt a^n$ when $a > 1$. Here $a = 3$, and $-22\lt - 17$, so $3^{-22}<3^{-17}$, which means $\frac{3^{-15}}{3^{7}}<(3^{-8})\times(3^{-9})$.

Answer:

A. $\frac{3^{-15}}{3^{7}}<(3^{-8})\cdot(3^{-9})$