Here's a Pre-Calculus math question I can't do (but hopefully you can?)?

Question

Two populations of bacteria are growing exponentially in separate petri dishes. The population in the first dish has growth constant 0.13 and initial population 2000. The population in the second dish has growth constant 0.08 and initial population 5000.

Part 1. Find the time at which the popluations are equal. Your answer must be correct to two decimal places.

Part 2. What is the common value of the two populations at that time? Round your answer to the nearest whole number.

If you can do this problem, let God be with you (lol)

Answer 1

Let
t = time

2000e^(.13t) = 5000e^(.08t)
e^(.13t) = 2.5e^(.08t)
e^(.05t) = 2.5
.05t = ln 2.5
t = (ln 2.5)/.05 = 20(ln 2.5) ≈ 18.325815

To the nearest integer t = 18.

Answer 2

exponential growth is given by

Q(t) = Qi e^(kt)

Q(t) = quantity at time t

Qi = intial quantity

k = growth constant

1)

Q(t1) = 2000e^(0.13t) ----------eqn(1)

Q(t2) = 5000e^(0.08t) -----------eqn(2)

when population are equal

2000e^(0.13t) = 5000e^(0.08t)

e^(0.13t)/e^(0.08t) = 5000/2000

e^(0.13t-0.08t) = 5/2

e^(0.05t) = 5/2

0.05 t = ln(5/2)

0.05 t = 0.916

t = 0.916/0.05

= 18.32 time units

2)

Qt = 2000e^(0.13(18.32)

Qt = 2000e^(2.3816)

= 2000(10.82)

= 21644