Asked by Mathslover
x and y are positive real numbers that satisfy log(base)x y + log(base)y x = 17/4 and xy=288√3. If x+y=a+b√c, where a, b and c are positive integers and c is not divisible by the square of any prime, what is the value of a+b+c?
Answers
Answered by
Steve
since log_x(y) = 1/log_y(x), if we call log_x(y)=u, we have
u + 1/u = 2
u^2 - 2u + 1 = 0
u=1
so, x=y, and
x^2 = 288√3
x = 12√2∜3
This does not fit the paradigm of the proposed answer, so I gotta think there's a typo. Especially since the "14/7" is odd to encounter.
I expect you can fix that, and then follow the logic to a solution.
u + 1/u = 2
u^2 - 2u + 1 = 0
u=1
so, x=y, and
x^2 = 288√3
x = 12√2∜3
This does not fit the paradigm of the proposed answer, so I gotta think there's a typo. Especially since the "14/7" is odd to encounter.
I expect you can fix that, and then follow the logic to a solution.
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