Question

Question 69

Find $m$, so that ${\left(\frac{2}{9}\right)}^3\times {\left(\frac{2}{9}\right)}^6={\left(\frac{2}{9}\right)}^{2m-1}$

Answer 1

We have ${\left(\frac{2}{9}\right)}^3\times {\left(\frac{2}{9}\right)}^6={\frac{2}{9}}^{2m-1}\Rightarrow {\left(\frac{2}{9}\right)}^{3+6}={\left(\frac{2}{9}\right)}^{2m-1}\Rightarrow {\left(\frac{2}{9}\right)}^9={\left(\frac{2}{9}\right)}^{2m-1}\Rightarrow 9=2m-1\text{(since bases are equal, exponents can be equated)}\Rightarrow 9+1=2m\Rightarrow 10=2m\Rightarrow \frac{10}{2}=\frac{2m}{2}\Rightarrow 5=m$