CA Foundation Exam > CA Foundation Questions > If x2+ y2= 7xy then log({1over 3}) (x + y ) =...
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If x2+ y2= 7xy then log({1over 3}) (x + y ) = A: log x + log y B: ({1over 2})(log x+ log y) C: ({1over 3})(log x+ log y) D: ({1over 3})(log x.log y)
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If x2+ y2= 7xy then log({1over 3}) (x + y ) = A: log x + log y B: ({1o...
Problem
Given: x^2 y^2 = 7xy
To find: log(1/3)(xy)
Solution
Let's start by simplifying the given equation:
x^2 y^2 = 7xy
Dividing both sides by xy:
xy = 7/(x*y)
Taking the logarithm of both sides:
log(xy) = log(7/(x*y))
Using the logarithmic identity log(a/b) = log(a) - log(b):
log(xy) = log(7) - log(x*y)
Substituting log(1/3) for log(x*y) (as given in the problem):
log(xy) = log(7) - log(1/3)
Simplifying:
log(xy) = log(7) + log(3)
log(xy) = log(21)
Using the logarithmic identity log(ab) = log(a) + log(b):
log(xy) = log(x) + log(y)
Substituting log(xy) = log(21):
log(21) = log(x) + log(y)
Dividing both sides by 3:
log(21)/3 = (log(x) + log(y))/3
Using the logarithmic identity log(a^n) = n*log(a):
log(21)/3 = log(x*y) / 3
Substituting log(1/3) for log(x*y):
log(21)/3 = log(1/3)
Simplifying:
log(21)/3 = -log(3)
Multiplying both sides by 3:
log(21) = -3*log(3)
Therefore, the answer is (C) ({1/3})(log x log y).
Community Answer
If x2+ y2= 7xy then log({1over 3}) (x + y ) = A: log x + log y B: ({1o...
That is X^2+Y^2=7xy
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If x2+ y2= 7xy then log({1over 3}) (x + y ) = A: log x + log y B: ({1over 2})(log x+ log y) C: ({1over 3})(log x+ log y) D: ({1over 3})(log x.log y)
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