Solve :
3x2 : 3x = 9 : 1
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Given,
⇒ 3x2 : 3x = 9 : 1
⇒3x23x=91⇒3x23x=3230⇒3x2−x=32−0⇒x2−x=2−0⇒x2−x=2⇒x2−x−2=0⇒x2−2x+x−2=0⇒x(x−2)+1(x−2)=0⇒(x+1)(x−2)=0⇒x+1=0 or x−2=0⇒x=−1 or x=2.\Rightarrow \dfrac{3^{x^2}}{3^x} = \dfrac{9}{1} \\[1em] \Rightarrow \dfrac{3^{x^2}}{3^x} = \dfrac{3^2}{3^0} \\[1em] \Rightarrow 3^{x^2 - x} = 3^{2- 0} \\[1em] \Rightarrow x^2 - x = 2 - 0 \\[1em] \Rightarrow x^2 - x = 2 \\[1em] \Rightarrow x^2 - x - 2 = 0 \\[1em] \Rightarrow x^2 - 2x + x - 2 = 0 \\[1em] \Rightarrow x(x - 2) + 1(x - 2) = 0 \\[1em] \Rightarrow (x + 1)(x - 2) = 0 \\[1em] \Rightarrow x + 1 = 0 \text{ or } x - 2 = 0 \\[1em] \Rightarrow x = -1 \text{ or } x = 2.⇒3x3x2=19⇒3x3x2=3032⇒3x2−x=32−0⇒x2−x=2−0⇒x2−x=2⇒x2−x−2=0⇒x2−2x+x−2=0⇒x(x−2)+1(x−2)=0⇒(x+1)(x−2)=0⇒x+1=0 or x−2=0⇒x=−1 or x=2.
Hence, x = -1 or x = 2.
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Find x, if :
(233)x−1=278\Big(\sqrt[3]{\dfrac{2}{3}}\Big)^{x - 1} = \dfrac{27}{8}(332)x−1=827
4x - 2 - 2x + 1 = 0
8 × 22x + 4 × 2x + 1 = 1 + 2x
22x + 2x + 2 - 4 × 23 = 0
