SOLUTION: Solve for x. Log2x � Log4x+ Log8x = 5

Question 293954: Solve for x. Log2x � Log4x+ Log8x = 5
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Solve for x. Log2x � Log4x+ Log8x = 5
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Assuming that the 2,4, and 8 are base numbers you get:
Log2(x) - (1/2)Log2(x) + (1/3)log2(x) = 5
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log2(x) - log2(x^(1/2)) + log2(x^(1/3)) = 5
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log2[(x*x^(1/3))/x^(1/2)] = 5
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log2[x^(4/3-1/2)] = 5
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x^(5/8) = 5
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x = 5^(8/5)
x = 13.13..
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Assuming the 2x, 4x, and 8x are all inverse logs you get:
log(2x) - log(4x) + log(8x) = 5
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log[(2x)(8x)/(4x)] = 5
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log(4x) = 5
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4x = 10^5
x = (1/4)(100,000)
x = 250,000
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Cheers,
Stan H.
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