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)