Question
$\Big(\frac{11}{15}\Big)^{4}\times$ (_____)5 = $\Big(\frac{11}{15}\Big)^{9}$
✓
Answer
$\Big(\frac{11}{15}\Big)^{4}\times\Big(\frac{11}{15}\Big)^{5}=\Big(\frac{11}{15}\Big)^{9}$
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, $\Big(\frac{11}{15}\Big)^{4}\times\Big(\frac{11}{15}\Big)^{5}=\Big(\frac{11}{15}\Big)^{9}$
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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