Solve :
(3)x−3=(34)x+1(\sqrt{3})^{x - 3} = (\sqrt[4]{3})^{x + 1}(3)x−3=(43)x+1
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Given,
⇒(3)x−3=(34)x+1⇒(312)x−3=(314)x+1⇒3x−32=3x+14⇒x−32=x+14⇒4(x−3)=2(x+1)⇒4x−12=2x+2⇒4x−2x=2+12⇒2x=14⇒x=142=7.\Rightarrow (\sqrt{3})^{x - 3} = (\sqrt[4]{3})^{x + 1} \\[1em] \Rightarrow (3^{\dfrac{1}{2}})^{x - 3} = (3^{\dfrac{1}{4}})^{x + 1} \\[1em] \Rightarrow 3^{\dfrac{x - 3}{2}} = 3^{\dfrac{x + 1}{4}} \\[1em] \Rightarrow \dfrac{x - 3}{2} = \dfrac{x + 1}{4}\\[1em] \Rightarrow 4(x - 3) = 2(x + 1) \\[1em] \Rightarrow 4x - 12 = 2x + 2 \\[1em] \Rightarrow 4x - 2x = 2 + 12 \\[1em] \Rightarrow 2x = 14 \\[1em] \Rightarrow x = \dfrac{14}{2} = 7.⇒(3)x−3=(43)x+1⇒(321)x−3=(341)x+1⇒32x−3=34x+1⇒2x−3=4x+1⇒4(x−3)=2(x+1)⇒4x−12=2x+2⇒4x−2x=2+12⇒2x=14⇒x=214=7.
Hence, x = 7.
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8 × 22x + 4 × 2x + 1 = 1 + 2x
22x + 2x + 2 - 4 × 23 = 0
Find the values of m and n if :
42m = (163)−6n=(8)2(\sqrt[3]{16})^{-\dfrac{6}{n}} = (\sqrt{8})^2(316)−n6=(8)2
Solve for x and y, if :
(32)x÷2y+1=1 and 8y−164−x2=0(\sqrt{32})^x ÷ 2^{y + 1} = 1 \text{ and } 8^y - 16^{4 - \dfrac{x}{2}} = 0(32)x÷2y+1=1 and 8y−164−2x=0
