Question
Solve the following equations:$4^{2\text{x}}=(\sqrt[3]{16})^{\frac{-6}{\text{y}}}=(\sqrt{8})^2$
✓
Answer
$4^{2\text{x}}=\big(\sqrt[3]{16}\big)^{\frac{-6}{\text{y}}}=\big(\sqrt{8}\big)^2$
Hence, $\text{x}=\frac{3}{4}$and $\text{y}=-\frac{8}{3}$
| Consider, $4^2=\big(\sqrt{8}\big)^2$ | Now, consider, $\big(\sqrt[3]{16}\big)^{\frac{-6}{?\text{y}?}}=\big(\sqrt{8}\big)^2$ |
| $\Rightarrow\big(2^2\big)^{2\text{x}}=\Big(\sqrt{2^3}\Big)^2$ | $\Rightarrow\Big(\sqrt[3]{2^4}\Big)^{\frac{-6}{?\text{y}?}}=\Big(\sqrt{2^3}\Big)^2$ |
| $\Rightarrow2^{4\text{x}}=\Big(2^{3\times\frac{1}{2}}\Big)$ | $\Rightarrow\Big(2^{4\times\frac{1}{3}}\Big)^{\frac{-6}{\text{y}}}=\Big(2^{3\times\frac{1}{2}}\Big)^2$ |
| $\Rightarrow2^{4\text{x}}=2^{3\times\frac{1}{2}\times2}$ | $\Rightarrow2^{4\times\frac{1}{3}\times\Big(-\frac{6}{\text{y}}\Big)}=2^{3\times\frac{1}{2}\times2}$ |
| $\Rightarrow2^{4\text{x}}=2^3$ | $\Rightarrow2^{-\frac{8}{\text{y}}}=2^3$ |
| $\Rightarrow4\text{x}=3$ | $\Rightarrow-\frac{8}{\text{y}}=3$ |
| $\Rightarrow\text{x}=\frac{3}{4}$ | $\Rightarrow\text{y}=-\frac{8}{3}$ |
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