Solve for x :
4x−1×(0.5)3−2x=(18)−x4^{x - 1} \times (0.5)^{3 - 2x} = \Big(\dfrac{1}{8}\Big)^{-x}4x−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⇒22(x−1)×(12)3−2x=(2−3)−x⇒22x−2×(2−1)3−2x=23x⇒22x−2×2−1(3−2x)=23x⇒2(2x−2)+[−1(3−2x)]=23x⇒22x−2−3+2x=23x⇒24x−5=23x⇒4x−5=3x⇒4x−3x=5⇒x=5.\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^{2(x - 1)} \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^{-1(3 - 2x)} = 2^{3x} \\[1em] \Rightarrow 2^{(2x - 2) + [-1(3 - 2x)]} = 2^{3x} \\[1em] \Rightarrow 2^{2x - 2 - 3 + 2x} = 2^{3x} \\[1em] \Rightarrow 2^{4x - 5} = 2^{3x} \\[1em] \Rightarrow 4x - 5 = 3x \\[1em] \Rightarrow 4x - 3x = 5 \\[1em] \Rightarrow x = 5.⇒4x−1×(0.5)3−2x=(81)−x⇒(22)x−1×(105)3−2x=(231)−x⇒22(x−1)×(21)3−2x=(2−3)−x⇒22x−2×(2−1)3−2x=23x⇒22x−2×2−1(3−2x)=23x⇒2(2x−2)+[−1(3−2x)]=23x⇒22x−2−3+2x=23x⇒24x−5=23x⇒4x−5=3x⇒4x−3x=5⇒x=5.
Hence, x = 5.
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If 5x + 1 = 25x - 2; find the value of :
3x - 3 × 23 - x.
If 4x + 3 = 112 + 8 × 4x; find (18x)3x.
(a3x + 5)2.(ax)4 = a8x + 12
(81)34−(132)−25+x(12)−1.20=27(81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + x\Big(\dfrac{1}{2}\Big)^{-1}.2^0 = 27(81)43−(321)−52+x(21)−1.20=27
