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\( \frac{8^{x}-2^{x}}{6^{x}-3^{x}}=2 \) |
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Answer» \(\frac {8^x-2^x}{6^x-3^x} = \frac {2^x(2^{2x}-1)}{3^x (2^x-1)}\) = \(\frac {2^x (2^x-1)(2^x+1)}{3^x (2^x-1)}\) = \(\frac {2^x (2^x+1)}{3^x}\) \(\therefore\) \(\frac {2^x (2^x+1)}{3^x}\) = 2 (Given) \(\Rightarrow\) \(2^{2x} +2^x = 2.3^x\) For x = 0, \(2^{2x} + 2^x = 2^0 + 2^0 = 1+1 = 2\) \(2.3^x = 2.3^0 = 2 \times 1 = 2\) Hence, \(2^{2x}+2^x = 2.3^x \) then x = 0 |
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