Find the value of m which satisfies $$\left( \dfrac{29}{4} \right)^3 \times \left( \dfrac{4}{29} \right)^2 \times \left( \dfrac{29}{4} \right)^{11} = \left( \dfrac{4}{29} \right)^{9m+7}$$​

Correct option is A

Given:

$$\left(\frac{29}{4}\right)^3 \times \left(\frac{4}{29}\right)^2 \times \left(\frac{29}{4}\right)^{11}= \left(\frac{4}{29}\right)^{9m+7}$$​​

Formula Used:

Laws of indices: $$a^p a^q=a^{p+q}, \left(\frac{a}{b}\right)^{-n}=\left(\frac{b}{a}\right)^n$$​​

Solution:

​$$\left(\frac{29}{4}\right)^3 \times \left(\frac{29}{4}\right)^{-2 }\times \left(\frac{29}{4}\right)^{11}= \left(\frac{4}{29}\right)^{9m+7}$$​​

​$$\left(\frac{29}{4}\right)^{11+3-2}=\left(\frac{4}{29}\right)^{9m+7}$$​​

​$$\left(\frac{29}{4}\right)^{12}=\left(\frac{29}{4}\right)^{-(9m+7)}$$​​

Equate powers:
​$$12 = -(9m+7)\implies 9m=-19 \implies m= -\frac{19}{9}.$$​​