Solve the following equation for x:
4x−1×(0.5)3−2x=(18)x4^{x - 1} \times (0.5)^{3 - 2x} = \Big(\dfrac{1}{8}\Big)^x4x−1×(0.5)3−2x=(81)x
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Given,
⇒4x−1×(0.5)3−2x=(18)x⇒(22)x−1×(510)3−2x=(123)x⇒22x−2×(12)3−2x=(2−3)x⇒22x−2×(2−1)3−2x=2−3x⇒22x−2×22x−3=2−3x⇒22x−2+2x−3=2−3x⇒24x−5=2−3x⇒4x−5=−3x⇒4x+3x=5⇒7x=5⇒x=57.\Rightarrow 4^{x - 1} \times (0.5)^{3 - 2x} = \Big(\dfrac{1}{8}\Big)^x \\[1em] \Rightarrow (2^2)^{x - 1} \times \Big(\dfrac{5}{10}\Big)^{3 - 2x} = \Big(\dfrac{1}{2^3}\Big)^x \\[1em] \Rightarrow 2^{2x - 2} \times \Big(\dfrac{1}{2}\Big)^{3 - 2x} = (2^{-3})^x \\[1em] \Rightarrow 2^{2x - 2} \times (2^{-1})^{3 - 2x} = 2^{-3x} \\[1em] \Rightarrow 2^{2x - 2} \times 2^{2x - 3} = 2^{-3x} \\[1em] \Rightarrow 2^{2x - 2 + 2x - 3} = 2^{-3x} \\[1em] \Rightarrow 2^{4x - 5} = 2^{-3x} \\[1em] \Rightarrow 4x - 5 = -3x \\[1em] \Rightarrow 4x + 3x = 5 \\[1em] \Rightarrow 7x = 5 \\[1em] \Rightarrow x = \dfrac{5}{7}.⇒4x−1×(0.5)3−2x=(81)x⇒(22)x−1×(105)3−2x=(231)x⇒22x−2×(21)3−2x=(2−3)x⇒22x−2×(2−1)3−2x=2−3x⇒22x−2×22x−3=2−3x⇒22x−2+2x−3=2−3x⇒24x−5=2−3x⇒4x−5=−3x⇒4x+3x=5⇒7x=5⇒x=75.
Hence, x = 57\dfrac{5}{7}75.
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(43)2x+12=132(\sqrt[3]{4})^{2x + \dfrac{1}{2}} = \dfrac{1}{32}(34)2x+21=321
pq=(qp)1−2x\sqrt{\dfrac{p}{q}} = \Big(\dfrac{q}{p}\Big)^{1 - 2x}qp=(pq)1−2x
If 53x = 125 and (10)y = 0.001, find x and y.
If 9n.32.3n−(27)n33m.23=127\dfrac{9^n.3^2.3^n - (27)^n}{3^{3m}.2^3} = \dfrac{1}{27}33m.239n.32.3n−(27)n=271, prove that m = 1 + n.
