It's basically asking how do we ensure that:
4^x - 5(2^x) = k, has two distinct solutions?
Revision
If you recall from year 11, there was the concept of the "discriminant" (Δ).
Δ = b^2 - 4ac, where ax^2 + bx + c = 0
If Δ < 0, then there are no solutions
If Δ = 0, there is one solution
If Δ > 0, then there are two solutions.
This comes from the quadratic formula, where we have +/- root(b^2 - 4ac) = +/- root(Δ).
If Δ is negative, we can't take the square root
If Δ is zero, we get zero: one solution
If Δ is positive, we get two different solutions.
Okay, so lets actually do this:
4^x - 5(2^x) = k
=> (2^x)^2 - 5(2^x) = k [Here you must recognise that 4^x = (2^x)^2]
Let a = 2^x: a quadratic equation is evident:
=> a^2 - 5a = k
=> a^2 - 5a - k = 0
To fix 2 solutions: Δ > 0
Δ = (-5)^2 - 4(-k) = 25 + 4k
=> 25 + 4k > 0
=> 4k > -25
=> k > -25/4
This ensures there are 2 solutions to a^2 - 5a - k = 0.
(i.e.: a = u, and a = v are solutions to the equation, where u and v are some real numbers)
=> 2^x = u, and 2^x = v
=> x = log2(u), x = log2(v)
This means that we must ensure the two solutions, u and v are greater than zero (because you can't log negative numbers, or zero).
Use the quadratic formula: a^2 - 5a - k = 0
a = [ 5 +/- root(25 + 4k) ] / 2
We need to ensure:
=> [ 5 +/- root(25 + 4k) ] / 2 > 0
=> 5 - root(25 + 4k) > 0
[I removed the positive root solution, because that will always be larger than the negative root solution: remember, we just want to make sure both of the solutions are positive, so if the smallest solution is positive, both will be]
=> root(25 + 4k) < 5
=> root(25/4 + k) < 5/2
Sketch graph, with the root(x) function translated 25/4 units to the left. Convince yourself that for the function to be less than 5/2, k must be less than some value, which we will now find.
Solve: root(25/4 + k) = 5/2
=> 25/4 + k = 25/4
=> k = 0
So, k < 0.
This means the boundary is: k for (-25/4, 0)