SOLUTION: How do you solve: log2(3x+2)-log2(x)/log2(4)=3 3log2(x-1)+log2(4)=5 Logx(1/27)=3 5^(1-2x)=1/5

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How do you solve: log2(3x+2)-log2(x)/log2(4)=3 3log2(x-1)+log2(4)=5 Logx(1/27)=3 5^(1-2x)=1/5      Log On


   

Question 970669: How do you solve:
log2(3x+2)-log2(x)/log2(4)=3
3log2(x-1)+log2(4)=5
Logx(1/27)=3
5^(1-2x)=1/5
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
solve:
log2(3x+2)-log2(x)/log2(4) = 3
log2[(3x+2)/x] = log2(64)
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(3x+2)/x = 64
61x = 2
x = 2/61
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3log2(x-1)+log2(4) = 5
log2[(x-1)^3*4] = 5
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4(x-1)^3 = 2^5
(x-1)^3 = 2^3
x-1 = 2
x = 3
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Logx(1/27) = 3
x^3 = 1/27
x = 1/3
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5^(1-2x) = 1/5
5^(1-2x) = 5^-1
1-2x = -1
2x = 2
x = 1
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Cheers,
Stan H.
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