Determine the total number of ordered pairs(x,y) of positive integers with 2≤ x,y ≤ 2023 that satisfy this equation:
xlogy(x-4) = ylogx(yx-3)
x^(log_y(x^-4)) = y^(log_x(yx^-3))
I started by applying ln to each side and change of base to make all logs into ln:
(ln x)*ln(x^-4)/(ln y) = (ln y)*ln(yx^-3)/(ln x)
Then some more log rules and multiplying to clear denominators:
(ln y)^3 - 3(ln y)^2(ln x) + 4(ln x)^3 = 0
This is a factorable cubic:
(ln y - 2ln x)^2 * (ln y + ln x) = 0
Then ln y - 2ln x = 0 or ln y + ln x = 0.
For the first branch we get y=x^2 and for the second branch we get y=1/x.
In the first branch note sqrt(2023)=44.98, so we have 43 solutions where y=x^2 is between 2 and 2023: (x,y)=(2,4), (3,9), ..., (44,1936).
The second branch has no solutions where both x and y are greater than 2.
Thus there are 43 ordered pairs (x,y) of positive integers within the given bounds that satisfy the equation.
Analytic Solution
