SOLUTION: A great thank you to whoever solve the following problem!!! Given that log(base b)(x y^2) = m and log(base b)(x^3 y) = n, express log(base b)(y/x) and log(base b) (x^0.5 y^0.5) i

Question 549638: A great thank you to whoever solve the following problem!!!
Given that log(base b)(x y^2) = m and log(base b)(x^3 y) = n, express log(base b)(y/x) and log(base b) (x^0.5 y^0.5) in terms of m and n.

Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
Given that log(base b)(x y^2) = m and log(base b)(x^3 y) = n, express log(base b)(y/x) and log(base b) (x^0.5 y^0.5) in terms of m and n.
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For simplification, we will write log(base b) as log
Using the rules of logarithms, we can write:
log(xy^2) = log(x) + 2log(y) = m [1]
log(x^3y) = 3log(x) + log(y) = n [2]
We can use these two equations to solve for log(x) and log(y) in terms of m and n:
Subtract 1/2 the 1st equation from the 2nd:
3log(x) - 1/2log(x) = n - m/2
5/2log(x) = n - m/2
log(x) = 2/5(n - m/2) = 1/5(2n - m)
Now use [2] to solve for log(y):
log(y) = n - 3/5(2n - m) -> log(y) = -n/5 + 3m/5 = 1/5(3m - n)
(a) log(y/x) = log(y) - log(x) = 1/5(3m - n) - 1/5(2n - m) = 4m/5 - 3m/5 = 1/5(4m - 3n)
(b) log(x^0.5 y^0.5) = (1/2)log(x) + (1/2)log(y) = 1/10(2n - m) + 1/10(3m - n) = 1/10(n + 2m)

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