Latest Elitmus Aptitude Question SOLUTION: logX(base e) + logYcube(base e) = a logXsquare(base e) + logY(base e) = b Find the value of Y.

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(#M40038293) ELITMUS Numerical Ability question Logerithm Keep an EYE

logX(base e) + logYcube(base e) = a
logXsquare(base e) + logY(base e) = b
Find the value of Y.

Asked In Elitmus Rahul Bhaskar (9 years ago)

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Mukesh Nayal  (9 years ago)

y=e^((2a-b)/5)

Is this solution Helpfull? Yes (16) | No (2)  

Sayak Das  (9 years ago)

logX(base e) + logYcube(base e) = a
or, log{(X*y^3)base e}=a
which can be rewritten as e^a =X*Y^3 ----- (1)

logXsquare(base e) + logY(base e) = b
or, log{(X^2*y)base e}=b
which can be rewritten as E^b = X^2*Y----- (2)

now from expression 1:-
X = e^a / Y^3
Substituting this value of X in expression 2:-
e^b = ((e^a/Y^3)^2)*Y
= e^2a / Y^5
or Y^5 = e^2a/e^b
therefore, Y = e ^((2a-b)/5)

Is this solution Helpfull? Yes (8) | No (2)  

ashish  (9 years ago)

i think it should be the ans, y=e^((2a-b))
can u explain @mukesh nayal how is it y=e^((2a-b)/5)

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kumar saurav  (9 years ago)

i think it should be
y=e^(6b-3a)/5

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nishant  (9 years ago)

this question is not able for elitmus

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