x and y are positive real numbers that satisfy logxy+logyx=174 and xy=2883√. If x+y=a+bc√, 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?
To solve this problem, we need to use properties of logarithms and simplify the given equations.
Let's start with the first equation: logxy + logyx = 174.
We can rewrite this equation using the property of logarithms that says log(a) + log(b) = log(ab):
log(xy) + log(yx) = log(xy * yx).
Since logarithms are inverses of exponentials, we can simplify xy * yx using the property (ab)^n = a^n * b^n:
xy * yx = (xy)^2.
Therefore, the equation becomes:
log(xy * yx) = log((xy)^2),
which further simplifies to:
log((xy)^2) = 174.
Now, applying the property of logarithms that says log(a^b) = b * log(a), we get:
2 * log(xy) = 174.
Dividing both sides of the equation by 2:
log(xy) = 87.
Now, let's move on to the second equation: xy = 2883√.
We can rewrite this using the property of square roots that says √(ab) = √a * √b:
xy = 288 * 3√3.
To find the values of x and y, we can equate the exponents of both sides of the equation. That is:
x * y = 288,
and
3√(x * y) = 3√3.
Now, we can rewrite the second equation as:
√(xy)^3 = 3√3.
Taking the cube on both sides:
(xy)^3 = 27.
Since we know that xy = 288, we can substitute this value back into the equation:
(288)^3 = 27.
Cubing 288:
2985984 = 27.
Taking the cube root:
xy = 12.
Now, we have the values of xy and log(xy). We can solve for x and y by solving the simultaneous equations:
xy = 12 (from xy = 288) and log(xy) = 87.
We find that x = 4 and y = 3.
Finally, we can find the values a, b, and c using the equations x+y = a+bc√:
4 + 3 = a + bc√.
Since a, b, and c are positive integers, we can rewrite the equation as:
7 = a + bc√.
Since √3 is not divisible by the square of any prime, we can take b = 1 and c = 3. Therefore, the equation becomes:
7 = a + 3√.
Comparing the coefficients, we find that a = 7 and b + c = 3.
Hence, the value of a + b + c is 7 + 1 + 3 = 11.