SOLUTION: How do I solve this simultaneous logarithm. log (x-4)+ 2 log y = log 16 log x + log y = 16

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Question 787559: How do I solve this simultaneous logarithm.
log (x-4)+ 2 log y = log 16
log x + log y = 16
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log (x-4)+ 2 log y = log 16
log x + log y = 16
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Modify for elimination:
log(x-4) + 2log(y) = log(16)
2log(x) + 2log(y) = 32
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Subtract and solve for "x":
2log(x) - log(x-4) = 32-log(16)
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log(x^2) - log(x-4) + log(16) = 32
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log[16x^2/(x-4)] = 32
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16x^2/(x-4) = 10^32
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16x^2 -10^32x -4*10^32 = 0
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Solve for "x"; then solve for "y".
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Cheers,
Stan H.
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