If 34x = (81)-1 and (10)1y=0.0001(10)^{\dfrac{1}{y}} = 0.0001(10)y1=0.0001, find the value of 2-x × 16y.
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Given,
⇒34x=(34)−1⇒34x=3−4⇒4x=−4⇒x=−44⇒x=−1.\Rightarrow 3^{4x} = (3^4)^{-1} \\[1em] \Rightarrow 3^{4x} = 3^{-4} \\[1em] \Rightarrow 4x = -4 \\[1em] \Rightarrow x = -\dfrac{4}{4} \\[1em] \Rightarrow x = -1.⇒34x=(34)−1⇒34x=3−4⇒4x=−4⇒x=−44⇒x=−1.
⇒(10)1y=0.0001⇒(10)1y=1104⇒(10)1y=10−4⇒1y=−4⇒y=−14.\Rightarrow (10)^{\dfrac{1}{y}} = 0.0001 \\[1em] \Rightarrow (10)^{\dfrac{1}{y}} = \dfrac{1}{10^4} \\[1em] \Rightarrow (10)^{\dfrac{1}{y}} = 10^{-4} \\[1em] \Rightarrow \dfrac{1}{y} = -4 \\[1em] \Rightarrow y = -\dfrac{1}{4}.⇒(10)y1=0.0001⇒(10)y1=1041⇒(10)y1=10−4⇒y1=−4⇒y=−41.
Substituting value of x and y in 2-x × 16y, we get :
⇒2−x×16y=2−(−1)×16−14=21×(24)−14=2×24×−14=2×2−1=2×12=1.\Rightarrow 2^{-x} \times 16^y = 2^{-(-1)} \times 16^{-\dfrac{1}{4}} \\[1em] = 2^1 \times (2^4)^{-\dfrac{1}{4}} \\[1em] = 2 \times 2^{4 \times -\dfrac{1}{4}} \\[1em] = 2 \times 2^{-1} \\[1em] = 2 \times \dfrac{1}{2} \\[1em] = 1.⇒2−x×16y=2−(−1)×16−41=21×(24)−41=2×24×−41=2×2−1=2×21=1.
Hence, 2-x × 16y = 1.
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