$\Big(\frac{11}{15}\Big)^{4}\times\Big(\frac{11}{15}\Big)^{5}=\Big(\frac{11}{15}\Big)^{9}$ Solution: $\Rightarrow(\text{x})^{5}=\frac{\Big(\frac{11}{15}\Big)^{9}}{\Big(\frac{11}{15}\Big)^{4}}$ $\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{9}\Big(\frac{11}{15}\Big)^{-4}$ $\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{9-4}$ $\big[\because\text{a}^{\text{m}}\times\text{a}^{\text{-n}}=\text{a}^{\text{m+(-n)}}=\text{a}^{\text{m-n}}\big]$ $\Rightarrow(\text{x})^{5}=\Big(\frac{11}{15}\Big)^{5}$ Since, the powers are same $\therefore\text{x}=\frac{11}{15}$ Hence, $\Big(\frac{11}{15}\Big)^{4}\times\Big(\frac{11}{15}\Big)^{5}=\Big(\frac{11}{15}\Big)^{9}$
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