Logarithums!!!!

Question

A $5000 investment earns 8.4% annual interest, and a $9000 investment earns 6.3%, both compounded annually. How long will it take for the smaller investment to catch up to the larger one?

Enter your answer in years correct to three decimal places.

Answer 1

For this problem, you will plug into the equation: A = P(1 + r/n)^nt, where A is the final amount (A will be the same for each equation, so you can set the two equations equal to each other), P is the principal or initial amount, r is the interest rate (in a fractional/decimal form, not a percentage, i.e. 6.3% = 6.3/100 in fractional form or 0.063 in decimal form), n is the number of times the interest will be applied in a time period, and t is the number of time periods elapsed, where the time period will be set to 1 year:

A = 5000 (1 + 0.084/1)^1·t and A = 9000 (1 + 0.063/1)^1·t ∴ 5000 (1.084)^t = 9000 (1.063)^t

After this, in order to get the variable, t, out of the exponent, you will use logarithms. Take the log of both sides and apply the logarithm rules:

log_b(M·N) = log_b(M) + log_b(N) (where the base can be 10)

log_b(M^k) = k·log_b(M)

Using this, you should get a number close to 30. Let me know if you have trouble applying the rules.

Answer 2

5000(1 + 0.084)^t = 9000(1 + 0.63)^t

Yes use logarithms.