SOLUTION: Given that {{{log (x^2y) = n}}} and {{{log (x/y^2) = p}}}, express {{{log (x/y)}}} in terms of n and p. NOTE THAT THE BASE OF THE LOGARITHM is m.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Given that {{{log (x^2y) = n}}} and {{{log (x/y^2) = p}}}, express {{{log (x/y)}}} in terms of n and p. NOTE THAT THE BASE OF THE LOGARITHM is m.      Log On


   

Question 1138426: Given that log+%28x%5E2y%29+=+n and log+%28x%2Fy%5E2%29+=+p, express log+%28x%2Fy%29 in terms of n and p.
NOTE THAT THE BASE OF THE LOGARITHM is m.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


log%28%28x%5E2y%29%29+=+n
2log%28%28x%29%29%2Blog%28%28y%29%29+=+n (1)

log%28%28x%2Fy%5E2%29%29+=+p
log%28%28x%29%29-2log%28%28y%29%29+=+p (2)

(1) and (2) are a pair of linear equation with log(x) and log(y) as the variables. Solve the pair of equations for log(x) and log(y) in terms of n and p.

4log%28%28x%29%29%2B2log%28%28y%29%29+=+2n (3) [equation (1), multiplied by 2]
5log%28%28x%29%29+=+2n%2Bp [equation (2) plus equation (3)]
log%28%28x%29%29+=+%282n%2Bp%29%2F5

-2log%28%28x%29%29%2B4log%28%28y%29%29+=+-2p (4) [equation (2), multiplied by -2]
5log%28%28y%29%29+=+n-2p [equation (1) plus equation (4)]
log%28%28y%29%29+=+%28n-2p%29%2F5

Now we have expressions for log(x) and log(y) in terms of n and p, so we can find an expression for log(x/y) in terms of n and p.

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