Find x, if :
(233)x−1=278\Big(\sqrt[3]{\dfrac{2}{3}}\Big)^{x - 1} = \dfrac{27}{8}(332)x−1=827
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Given,
⇒(233)x−1=278⇒(2313)x−1=(32)3⇒(23)x−13=(23)−3⇒x−13=−3⇒x−1=3×−3⇒x−1=−9⇒x=−9+1=−8.\Rightarrow \Big(\sqrt[3]{\dfrac{2}{3}}\Big)^{x - 1} = \dfrac{27}{8} \\[1em] \Rightarrow \Big(\dfrac{2}{3}^{\dfrac{1}{3}}\Big)^{x - 1} = \Big(\dfrac{3}{2}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{2}{3}\Big)^{\dfrac{x - 1}{3}} = \Big(\dfrac{2}{3}\Big)^{-3} \\[1em] \Rightarrow \dfrac{x - 1}{3} = -3 \\[1em] \Rightarrow x - 1 = 3 \times -3 \\[1em] \Rightarrow x - 1 = -9 \\[1em] \Rightarrow x = -9 + 1 = -8.⇒(332)x−1=827⇒(3231)x−1=(23)3⇒(32)3x−1=(32)−3⇒3x−1=−3⇒x−1=3×−3⇒x−1=−9⇒x=−9+1=−8.
Hence, x = -8.
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2x+3=16\sqrt{2^{x + 3}} = 162x+3=16
(35)x+1=12527\Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27}(53)x+1=27125
Solve :
4x - 2 - 2x + 1 = 0
3x2 : 3x = 9 : 1
