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How long will it take for a $2500 investment to grow to $4000 at an annual rate of 7.5%, compounded quarterly? Assume that no withdrawals are made. Donot round any intermediate computations, and round your answer to the nearest hundredth.If necessary, refer to the list of financial formulas.years I need help with this math problem

1 Answer

4 votes

Answer:

6.33 years

Explanation:

The formula for investment at compound interest is given below::


A(t)=P\left(1+(r)/(k)\right)^(tk)\text{ where }\begin{cases}P=\text{Principal Invested} \\ r=\text{Interest Rate} \\ k=\text{Number of compounding periods}\end{cases}

From the statement of the problem:

• The initial investment, P = $2500

,

• Annual Interest Rate, r = 7.5% = 0.075

,

• Compounding Period (Quarterly), k = 4

,

• Amount after t years, A(t) = $4000

,

• Time, t = ?

Substitute these values into the compound interest formula above:


4000=2500\left(1+(0.075)/(4)\right)^(4t)

We then solve the equation for the value of t.


\begin{gathered} \begin{equation*} 4000=2500\left(1+(0.075)/(4)\right)^(4t) \end{equation*} \\ \text{ Divide both sides by 2500} \\ (4000)/(2500)=\left(1+0.01875\right)^(4t) \\ 1.6=\left(1.01875\right)^(4t) \\ \text{ Take the log of both sides} \\ \log(1.6)=\log(1.01875)^(4t) \\ \text{ By the power law of logs, }\log a^n=n\log a \\ \log(1.6)=4t\log(1.01875) \\ \text{ Divide both sides by 4}\log(1.01875) \\ \frac{\operatorname{\log}(1.6)}{4\operatorname{\log}(1.01875)}=\frac{4t\operatorname{\log}(1.01875)}{4\operatorname{\log}(1.01875)} \\ t\approx6.33\text{ years} \end{gathered}

It will take approximately 6.33 years for a $2500 investment to grow to $4000.

User Lalit Kumar B
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