Exponential Absolute Values

Complex numbers are incredible, but in order to use them we have to remember some definitions. Here is a question involving both complex numbers and exponential inequalities. It's not hard, but it requires some thinking.

Solve for x (x is real):
abs(2 + 3^{x2 - x - 1} - 12i) > 13

To do that, here is a reminder for the definition of absolute value:
Let z be a complex number.
z = a + bi
abs(z) = √(a² + b²)

In the question:
a = 2 + 3^{x2 - x - 1}
b = -12

The absolute value is:
√((2 + 3^{x2 - x - 1})² + 12²)
√(4 + 4*3^{x2 - x - 1} + (3^{x2 - x - 1})² + 144)

Now return to the inequality:
√(4 + 4*3^{x2 - x - 1} + (3^{x2 - x - 1})² + 144) > 13

Since both sides are always positive, we can square both sides without changing the inequality sign:
4 + 4*3^{x2 - x - 1} + (3^{x2 - x - 1})² + 144 > 169

Let t = 3^{x2 - x - 1}. This gives:
4 + 4t + t² + 144 > 169
t² + 4t - 21 > 0
(t + 7)(t - 3) > 0
t > 3 or t < -7 Since t is an exponent of 3, it can't be negative, so we're left with only one solution: t > 3.

Switch t back:
3^{x2 - x - 1} > 3

The base is bigger than 1, so the inequality sign remains the same:
x² - x - 1 > 1
x² - x - 2 > 0
(x - 2)(x + 1) > 0
x > 2 or x < -1

And that's it.

Hope you liked it,
Nadav

nadavs