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Mathematics

Solve for x :

320+23=(0.6)2−3x\sqrt{32^0+\dfrac{2}{3}} = (0.6)^{2-3x}.

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Answer

320+23=(0.6)2−3x⇒1+23=(0.6)2−3x⇒3+23=(0.6)2−3x⇒53=(0.6)2−3x⇒53=(610)2−3x⇒53=(35)2−3x⇒5312=(53)3x−2⇒12=3x−2⇒3x=12+2⇒3x=1+42⇒3x=52⇒x=56\sqrt{32^0+\dfrac{2}{3}} = (0.6)^{2-3x}\\[1em] \Rightarrow\sqrt{1+\dfrac{2}{3}} = (0.6)^{2-3x}\\[1em] \Rightarrow\sqrt{\dfrac{3+2}{3}} = (0.6)^{2-3x}\\[1em] \Rightarrow\sqrt{\dfrac{5}{3}} = (0.6)^{2-3x}\\[1em] \Rightarrow\sqrt{\dfrac{5}{3}} = \Big(\dfrac{6}{10}\Big)^{2-3x}\\[1em] \Rightarrow\sqrt{\dfrac{5}{3}} = \Big(\dfrac{3}{5}\Big)^{2-3x}\\[1em] \Rightarrow{\dfrac{5}{3}}^\dfrac{1}{2} = \Big(\dfrac{5}{3}\Big)^{3x-2}\\[1em] \Rightarrow\dfrac{1}{2} = 3x-2\\[1em] \Rightarrow 3x = \dfrac{1}{2} + 2\\[1em] \Rightarrow 3x = \dfrac{1+4}{2}\\[1em] \Rightarrow 3x = \dfrac{5}{2}\\[1em] \Rightarrow x = \dfrac{5}{6}

Hence, the value of x = 56\dfrac{5}{6}.

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