Question

1. Find the inverse of the logarithmic function f defined by f(y) = 2 log5 (2y-8) + 3.

2. If x > y > 1, what is the largest possible value of logx (x/y) + logy (y/x)?

Answer 1

f = 2log5(2y-8) + 3
f = log5(2y-8)^2 + log5 125
f-3 = log5((2y-8)^2)
5^(f-3) = (2y-8)^2
5^((f-3)/2) = 2y-8
2y = 8 + 5^((f-3)/2)
y = 4 + 1/2 √(5^(f-3))

so, the inverse function is

4 + 1/2 √(5^(y-3))

logx(x/y) = logx(x) - logx(y) = 1 - logx(y)
logy(y/x) = 1 - logy(x)

adding them, we have

2 - logx(y) - logy(x)
2 - logy(x) - 1/logy(x)

we know logy(x) > 1, so if we call that z, we want the maximum value of 2 - z - 1/z

That function achieves a maximum of 0 when z=1. That is, as y->x. So, technically, 0 is the least upper bound, but there is no maximum number less than zero.