$\Big(\frac{-1}{4}\Big)^{3}\times\Big(\frac{-1}{4}\Big)^{\text{x}}=\Big(\frac{-1}{4}\Big)^{11}$ Solution: Let $\Big(\frac{-1}{4}\Big)^{3}\times\Big(\frac{-1}{4}\Big)=\Big(\frac{-1}{4}\Big)^{11}$ $\Rightarrow\Big(\frac{-1}{4}\Big)^{\text{x}}=\frac{\Big(\frac{-1}{4}\Big)^{11}}{\Big(\frac{-1}{4}\Big)^{3}}$ $\Rightarrow\Big(\frac{-1}{4}\Big)^{\text{x}}=\Big(\frac{-1}{4}\Big)^{\text{11-3}}$$\big[\because\text{a}^{\text{m}}+\text{a}^{\text{n}}=\text{a}^{\text{m-n}}\text{m}>\text{n}\big]$ $\Rightarrow\Big(\frac{-1}{4}\Big)^{\text{x}}=\Big(\frac{-1}{4}\Big)^{8}$ Since, base are equal do, ny equating the power the pawers, we get $\text{x}=8$ $\therefore\Big(-\frac{1}{4}\Big)^{3}\times\Big(-\frac{1}{4}\Big)^{8}=\Big(-\frac{1}{4}\Big)^{11}$
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