Question 550166: 3logbase5(Y) - logbaseY(5) = 2 (solve for Y)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 3logbase5(Y) - logbaseY(5) = 2 (solve for Y)
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3log5(y)-logy(5) = 2
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log5(y^3)-logy(5) = 2
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log(y^3)/log(5) - log(5)/log(y) = 2
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multiply thru by log(5)(log(y)) to get:
log(y)*3log(y) -log(5)*log(5) = 2log(y)*(log(5))
3[log(y)]^2 -2log(5)*log(y) - [log(5)]^2 = 0
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This is a quadratic in log(y).
Let log(y) = "m"
3m^2 - log(25)m - (log(5))^2 = 0
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Solve for "m":
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Graphing I get m = -0.233
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So, log(y) = -0.233
Then y = 10^(-0.233) = 0.5848
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Cheers,
Stan H.
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