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?

Question

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?

Answer 1

Having posted the same problem three times and not

getting a response should have given you a hint, that there might be something wrong with the question, or the way you typed it.
e.g. 2883√ makes no sense, the √ is not written behind the number
which makes x+y = a + bc√ completely ambiguous.

I will start it off:

logxy+logyx=174
the way you typed it .....

2 log xy = 174
log xy = 72
xy = 10^72 ---- huge!!
but assuming xy = √2883
so √2883 = 10^72 ??? , not likely

I have a feeling you meant:

log_xy + log_yx = 174

Answer 2

logxy + logyx = 17/4

Answer 3

log y base x + log x base y =17/4

sorry for the typo.

Answer 4

To solve this problem, we will use the properties of logarithms and the given equations.

First, let's simplify the equation log(xy) + log(yx) = 174 using the logarithmic identities. Since log(xy) = log(x) + log(y), we can rewrite the equation as log(x) + log(y) + log(x) + log(y) = 174. Simplifying further, we get 2(log(x) + log(y)) = 174.

Next, let's simplify the equation xy = 2883√. We can rewrite √(2883) as √(3 * 13 * 73) = √3 * √13 * √73. Therefore, xy = (x * y) = √3 * √13 * √73.

Now, let's use the logarithmic property log(a * b) = log(a) + log(b) to simplify the equation 2(log(x) + log(y)) = 174. We get log(x^2 * y^2) = 174.

Using the logarithmic property log(a * b) = log(a) + log(b) again, we can rewrite the equation as 2(log(x) + log(y)) = log(x^2 * y^2) = 174. Taking the logarithm of both sides, we get log(xy) = 87.

Since we know that xy = √3 * √13 * √73, we can rewrite the equation as log(√3 * √13 * √73) = 87.

Using the logarithmic property log(ab) = b * log(a), we get (1/2) * (log(3) + log(13) + log(73)) = 87.

Now, we can solve for log(3 * 13 * 73) by multiplying both sides by 2: log(3 * 13 * 73) = 174.

Since log(3 * 13 * 73) = log(3^1 * 13^1 * 73^1) = log(3^1) + log(13^1) + log(73^1) = log(3) + log(13) + log(73), we get log(3) + log(13) + log(73) = 174.

At this point, we have simplified the given equations to two separate equations:
1) log(√3 * √13 * √73) = 87
2) log(3) + log(13) + log(73) = 174

Now, let's focus on equation 2) log(3) + log(13) + log(73) = 174. We can add the logarithms since they have the same base, which is 10. Therefore, 3) log(3 * 13 * 73) = 174.

Solving equation 3), we find that log(2853) = 174. Now, to find the value of a+b+c, we need to express x+y in the form a+bc√.

Since xy = 2883√, we have xy = √3 * √13 * √73. Therefore, xy = √(2853), but we already know that xy = 2883√. This implies that √(2853) = 2883√.

By comparing these two equations, we find that a = 2853, b = 0, and c = 2883. Therefore, the value of a+b+c is 2853 + 0 + 2883 = 5736.

Hence, the value of a+b+c is 5736.