Problem 145 - Algebra and Trigonometry

Question

If \(\$ 4000\) is deposited into an account paying \(3 \%\) interest compounded annually and at the same time \(\$ 2000\) is deposited into an account paying \(5 \%\) interest compounded annually, after how long will the two accounts have the same balance? Round to the nearest year.

Step-by-Step Solution

Answer

It will take approximately 29 years (rounded up) for both accounts to have the same balance.

Step 1: Establish the equations

Let \(t\) be the time in years when the two accounts have the same balance, and \(A\) be the final balance of either account. Then we can set up the two equations: \(A = 4000(1 + 0.03) ^ t\) (for the first account) and \(A = 2000(1 + 0.05) ^ t\) (for the second account).

Step 2: Make the equations equal

Since both expressions equal to \(A\), we can equate them and solve for \(t\): \(4000(1 + 0.03) ^ t = 2000(1 + 0.05) ^ t\).

Step 3: Simplify the equation

We simplify the equation: \((1 + 0.03) ^ t = 2(1 + 0.05) ^ t\). Then, we isolate \(t\) by applying ln function to both sides, which gives \(t * ln(1.03) = ln(2) + t * ln(1.05)\).

Step 4: Solve for \(t\)

To get \(t\), we further simplify the equation: \(t * (ln(1.05) - ln(1.03)) = ln(2)\). Now, we can solve, yielding \(t = ln(2) / (ln(1.05) - ln(1.03))\).

Step 5: Calculate \(t\)

Substitute the natural log values to get the approximation for \(t\): \(t = 0.693147 /\ (0.04879 - 0.02956)\), which gives t approximately equal to 28.45 years.