Find x, if :
(35)x+1=12527\Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27}(53)x+1=27125
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Given,
⇒(35)x+1=12527⇒(3512)x+1=(53)3⇒(35)x+12=(35)−3⇒x+12=−3⇒x+1=−6⇒x=−6−1=−7.\Rightarrow \Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27} \\[1em] \Rightarrow \Big(\dfrac{3}{5}^{\dfrac{1}{2}}\Big)^{x + 1} = \Big(\dfrac{5}{3}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x + 1}{2}} = \Big(\dfrac{3}{5}\Big)^{-3} \\[1em] \Rightarrow \dfrac{x + 1}{2} = -3 \\[1em] \Rightarrow x + 1 = -6 \\[1em] \Rightarrow x = -6 - 1 = -7.⇒(53)x+1=27125⇒(5321)x+1=(35)3⇒(53)2x+1=(53)−3⇒2x+1=−3⇒x+1=−6⇒x=−6−1=−7.
Hence, x = -7.
Answered By
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42x = 132\dfrac{1}{32}321
2x+3=16\sqrt{2^{x + 3}} = 162x+3=16
(233)x−1=278\Big(\sqrt[3]{\dfrac{2}{3}}\Big)^{x - 1} = \dfrac{27}{8}(332)x−1=827
Solve :
4x - 2 - 2x + 1 = 0
