Question
Find $m$, so that $\left(\frac{2}{9}\right)^3 \times\left(\frac{2}{9}\right)^6=\left(\frac{2}{9}\right)^{2 m-1}$.
✓
Answer
Given, $\left(\frac{2}{9}\right)^3 \times\left(\frac{2}{9}\right)^6=\left(\frac{2}{9}\right)^{2 m-1}$
We know that
$a^m \times a^n=a^{m+n}$
Let $a=\frac{2}{9}$
So, $\left(\frac{2}{9}\right)^3 \times\left(\frac{2}{9}\right)^6=\left(\frac{2}{9}\right)^{3+6}=\left(\frac{2}{9}\right)^{2 m-1}$
$\Rightarrow \quad\left(\frac{2}{9}\right)^9=\left(\frac{2}{9}\right)^{2 m-1}$
If $a^m=a^n$, then $m=n$
So, $\quad 9=2 m-1 \Rightarrow 9+1=2 m$
$\therefore \quad m=\frac{10}{2}=5$
We know that
$a^m \times a^n=a^{m+n}$
Let $a=\frac{2}{9}$
So, $\left(\frac{2}{9}\right)^3 \times\left(\frac{2}{9}\right)^6=\left(\frac{2}{9}\right)^{3+6}=\left(\frac{2}{9}\right)^{2 m-1}$
$\Rightarrow \quad\left(\frac{2}{9}\right)^9=\left(\frac{2}{9}\right)^{2 m-1}$
If $a^m=a^n$, then $m=n$
So, $\quad 9=2 m-1 \Rightarrow 9+1=2 m$
$\therefore \quad m=\frac{10}{2}=5$
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