​$$\text{Find the value of } m \text{ which satisfies} \\\left( \frac{11}{10} \right)^7 \times \left( \frac{10}{11} \right)^{10} \times \left( \frac{11}{10} \right)^9 = \left( \frac{10}{11} \right)^{3m+17}$$​​

Correct option is B

Given:

$$\left(\frac{11}{10}\right)^7\times\left(\frac{10}{11}\right)^{10}\times\left(\frac{11}{10}\right)^9= \left(\frac{10}{11}\right)^{3m+17}$$​​

Formula Used:

Same-base rule: $$a^p\times a^q=a^{p+q}$$​​

Solution:

​$$\left(\frac{10}{11}\right)^{10}\times\left(\frac{11}{10}\right)^{7+9}=\left(\frac{10}{11}\right)^{3m+17}\\ \ \\ \left(\frac{10}{11}\right)^{10}\times \left(\frac{11}{10}\right)^{16}=\left(\frac{10}{11}\right)^{3m+17}$$​​
​​​
$$\left(\frac{10}{11}\right)^{10}\times\left(\frac{10}{11}\right)^{-16}= \left(\frac{10}{11}\right)^{3m+17}$$​​

​$$\left(\frac{10}{11}\right)^{-6}$$​=$$\left(\frac{10}{11}\right)^{3m+17}$$​​
Equate exponents:
​$$-6=3m+17 \implies 3m=-23 \implies m=-\frac{23}{3}$$​​