1). Solve the system of x^(log y) + (y^(log x))^(1/2) = 10 and x.y =1000
2). Prove that if a + (b/a) – (1/b) is integer, then it is a perfect square, where a, b element of N
1) x^(log y) + (y^(log x))^(1/2) = 10,
xy = 1000
Let’s use identity 10^(log a) = a, here logarithm is with a base 10, in the first equation:
10^[log(x^(logy))] + {10^[log(y^(log x))]}^(1/2) = 10
10^(logy * logx) + {10^(logx * logy)}^(1/2) = 10
10^(logy * logx) + 10^(logx * logy/2) – 10 = 0
Substitution t = 10^(logx * logy/2) (t > 0) , equation becomes
t² + t – 10 = 0
D = 1 + 40 = 41
t1 = ( – 1 + √41)/2
t2 = ( – 1 – √41)/2 < 0 rejected
So 10^(logx * logy/2) = (√41 – 1)/2 or to look shorter set (√41 – 1)/2 = T
10^(logx * logy/2) = T
take logarithm of both sides
log(10^(logx * logy/2)) = logT
logx *logy /2 = logT
logx * logy = 2logT
Now we have system of equations
logx * logy = 2logT
xy = 1000
logx * logy = 2logT
y = 1000/x
logx * log(1000/x) = 2logT
logx * (3 – logx) = 2logT
3logx – log²x – 2logT = 0
log²x – 3logx + 2logT = 0
Set logx = X
X² – 3X + 2logT = 0
D = 9 – 8logT
X1 = (3 – √(9 – 8logT))/2 ===> x1 = 10^X1 = 10^[(3 - √(9 - 8logT))/2]
X2 = (3 + √(9 – 8logT))/2 ===> x2 = 10^X2 = 10^[(3 + √(9 - 8logT))/2]
Ultimately
y1 = 1000/x1 = 10^[(3 + √(9 - 8logT))/2]
y2 = 1000/x2 = 10^[(3 - √(9 - 8logT))/2].
So, the solutions are
x1 = 10^[(3 - √(9 - 8logT))/2] ; y1 = 10^[(3 + √(9 - 8logT))/2] and
x2 = 10^[(3 + √(9 - 8logT))/2] ; y2 = 10^[(3 - √(9 - 8logT))/2],
where T = (√41 – 1)/2.
The values approximately
(476; 2.1) and (2.1; 476).
Thanks a ton!