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Find $a$, if

$\left(\frac{1}{9}\right)^{-4} \times\left(\frac{1}{27}\right)^{-6}=\left(\frac{1}{9}\right)^{a}$

(a) 13

(b) $-13$

(c) 15

(d) $-15$
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Solution :

 $\left(\frac{1}{9}\right)^{a}=\left(\frac{1}{9}\right)^{-4} \times\left(\frac{1}{27}\right)^{-6}$

$\Rightarrow\left(\frac{1}{3}\right)^{2 a}=\left(\frac{1}{3}\right)^{-8} \times\left(\frac{1}{3}\right)^{-18}$

$\Rightarrow\left(\frac{1}{3}\right)^{2 a}=\left(\frac{1}{3}\right)^{-26}$

$\Rightarrow 2 a=-26$

$\therefore \quad a=-\frac{26}{2}=-13 ;$ Ans.

Therefore, the correct option $is(\mathrm{b})$

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