An investment initially worth 4500 earns 79 annual interest

An investment initially worth $4500 earns 7.9% annual interest, and an investment initially worth $8100 earns 5.3% annual interest, both compounded annually. How long will it take for the smaller investment to catch up with the larger one? It will take (log(4500/8100) years.

Solution

A be the amount
A1 must equal A2

A1 = A2
P1*(1+R1)^t = P2*(1+R2)^t
4500*(1+0.079)^t = 8100*(1+0.053)^t
4500*(1.079)^t = 8100*(1.053)^t
(1.079/1.053)^t = 8100/4500
1.025^t = 1.8
take ln on both sides
ln (1.025^t) = ln(1.8)
t*ln(1.025) = ln(1.8)
t = ln (1.8) / ln (1.025)
   =24 years
Answer: 24 years