Answer: 18 years
Work Shown:
[tex]f(\text{x}) = 2.56(1.04)^\text{x}\\\\5 = 2.56(1.04)^\text{x}\\\\5/2.56 = (1.04)^\text{x}\\\\1.953125 = (1.04)^\text{x}\\\\\log(1.953125) = \log(1.04^\text{x})\\\\\log(1.953125) = \text{x}*\log(1.04)\\\\\text{x} = \frac{\log(1.953125)}{\log(1.04)}\\\\\text{x} \approx 17.0682937693249\\\\[/tex]
The steps above show f(x) replaced with 5. Then you'd use logarithms to isolate the variable x. The relevant useful log rule is [tex]\log(A^B) = B\log(A)[/tex] so we can pull down the exponent.
From here it seems your teacher wants you to round up to the nearest integer.
If we plugged in x = 17, then f(x) = 4.99 which is one cent too small.
While x = 18 leads to f(x) = 5.19
Therefore, x = 18 has the price reach or exceed $5
