Solve for x:
(35)x−1=(27125)−1\Big(\sqrt{\dfrac{3}{5}}\Big)^{x-1} = \Big(\dfrac{27}{125}\Big)^{-1}(53)x−1=(12527)−1
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Given,
(35)x−1=(27125)−1⇒[(35)12]x−1=(27125)−1⇒(35)x−12=(3353)−1⇒(35)x−12=(35)3×(−1)⇒(35)x−12=(35)−3\Big(\sqrt\dfrac{3}{5}\Big)^{x-1} = \Big(\dfrac{27}{125}\Big)^{-1} \\[1em] \Rightarrow \Big[\Big(\dfrac{3}{5}\Big)^{\dfrac{1}{2}}\Big]^{x-1} = \Big(\dfrac{27}{125}\Big)^{-1} \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x - 1}{2}} = \Big(\dfrac{3^3}{5^3}\Big)^{-1} \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x - 1}{2}} = \Big(\dfrac{3}{5}\Big)^{3 \times (-1)} \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x - 1}{2}} = \Big(\dfrac{3}{5}\Big)^{-3}(53)x−1=(12527)−1⇒[(53)21]x−1=(12527)−1⇒(53)2x−1=(5333)−1⇒(53)2x−1=(53)3×(−1)⇒(53)2x−1=(53)−3
Equating the exponents,
⇒x−12=−3⇒x−1=−3×2⇒x−1=−6⇒x=−6+1=−5.\Rightarrow \dfrac{x - 1}{2} = -3 \\[1em] \Rightarrow x - 1 = -3 \times 2 \\[1em] \Rightarrow x - 1 = -6 \\[1em] \Rightarrow x = -6 + 1 = -5.⇒2x−1=−3⇒x−1=−3×2⇒x−1=−6⇒x=−6+1=−5.
Hence, x = -5.
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ab=(ba)1−3x\sqrt{\dfrac{a}{b}} = \Big(\dfrac{b}{a}\Big)^{1-3x}ba=(ab)1−3x
(233)x−1=278\Big(\sqrt[3]{\dfrac{2}{3}}\Big)^{x-1} = \dfrac{27}{8}(332)x−1=827
9 × 3x = (27)2x - 5
2(5x - 1) = 4 × 2(3x + 1)
