Simplify the following:
(16164)−23\Big(1\dfrac{61}{64}\Big)^{-\dfrac{2}{3}}(16461)−32
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Given,
⇒(16164)−23=(12564)−23=(64125)23=(4353)23=43×2353×23=4252=1625.\Rightarrow \Big(1\dfrac{61}{64}\Big)^{-\dfrac{2}{3}} = \Big(\dfrac{125}{64}\Big)^{-\dfrac{2}{3}} \\[1em] = \Big(\dfrac{64}{125}\Big)^{\dfrac{2}{3}} = \Big(\dfrac{4^3}{5^3}\Big)^{\dfrac{2}{3}} \\[1em] = \dfrac{4^{3 \times \dfrac{2}{3}}}{5^{3 \times \dfrac{2}{3}}} \\[1em] = \dfrac{4^2}{5^2} = \dfrac{16}{25}.\\[1em]⇒(16461)−32=(64125)−32=(12564)32=(5343)32=53×3243×32=5242=2516.
Hence, (16164)−23=1625\Big(1\dfrac{61}{64}\Big)^{-\dfrac{2}{3}} = \dfrac{16}{25}(16461)−32=2516.
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Solve the following equation :
3x + 1 = 27.34
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3x - 1 × 52y - 3 = 225
8x + 1 = 16y + 2, (12)3+x=(14)3y\Big(\dfrac{1}{2}\Big)^{3 + x} = \Big(\dfrac{1}{4}\Big)^{3y}(21)3+x=(41)3y
