[tex]\text{Goal: Find } x \text{ such that:} \\ 2^x > 4x + 12[/tex]
[tex]2^x > 4x + 12 \\ 2^x > 4(x + 3) \\ 2^{x - 2} > x + 3[/tex]
[tex]\text{A lemma: } \\ \text{It is given that: } \\ 2^n > 2n + 1, \text{ } n > 3[/tex]
[tex]\text{This can be proven by Mathematical Induction.}[/tex]
[tex]\text{Using this lemma, we can manipulate the expression:} \\ 2^{x - 2} > 2(x - 2) + 1 = 2x - 3, \text{ } x > 5 \\ 2\cdot 2^{x - 2} > 2(2x - 3) \\ 2^{x - 1} > 4x - 12 \\ 2^{x - 1} + 24 > 4x + 12[/tex]
[tex]\text{Observe the following: } 2^x \text{ will always grow faster than } 2^{x - 1} \\ \text{This implies that } 2^{x} > 2^{x - 1} + 24 > 4x + 12, \text{ given that } x > 5[/tex]
[tex]\text{This means that the first month } f(x) > g(x) \text{ is Month 6.}[/tex]
