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Question
Solve the following equation for x:
`4^(x-1)xx(0.5)^(3-2x)=(1/8)^x`
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Solution
`4^(x-1)xx(0.5)^(3-2x)=(1/8)^x`
`rArr(2^2)^(x-1)xx(1/2)^(3-2x)=(1/2^3)^x`
`rArr(2^2)^(x-1)xx(2^-1)^(3-2x)=(2^-3)^x`
`rArr2^(2x-2)xx2^(2x-3)=2^(-3x)`
`rArr2^(2x-2+2x-3)=2^(-3x)`
`rArr2^(4x-5)=2^(-3x)`
⇒ 4x - 5 = -3x
⇒ 4x + 3x = 5
⇒ 7x = 5
`rArr x = 5/7`
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