Find t for a corporation that needs $800,000 and it can invest $75,000 per year in a fund earning...

Question:

Find t for a corporation that needs $800,000 and it can invest $75,000 per year in a fund earning 13% interest compounded continuously.

Exponential and Logarithmic Functions, Continuously Compounded Interest, Future Value of Annuities:

Suppose that an amount A is invested for a period of {eq}t {/eq} years at continuously compounded interest at the rate of {eq}r {/eq} %.

Then at the end of the period the accumulated value of this investment would be,

{eq}\displaystyle A(t) = Ae^{rt} {/eq}

Suppose that an amount {eq}A {/eq} is invested every year for {eq}t {/eq} years and the interest on this annuity is compounded continuously at the rate of {eq}r {/eq}%.

Then the future value of this annuity is given by the formula,

{eq}\displaystyle FV = Ae^r \bigg(\frac{e^{r(t-1)} - 1}{e^r - 1} \bigg) {/eq}

The above formula is due to the following result:

The sum of {eq}n {/eq} terms of the geometric series, {eq}a, ar, ar^2, ar^3, \cdots ar^n {/eq} is given by

{eq}\displaystyle \frac{a(1 - r^n)}{1 - r} \; r \ne1 {/eq}

If the yearly deposit is A, the accumulation over {eq}t {/eq} years with continuously compounded interest would be,

{eq}\displaystyle Ae^r + Ae^{2r} + \cdots Ae^{tr} {/eq}

assuming the deposits are made at the beginning of each year.

Then the future value of the annuity at the end of {eq}t {/eq} years would be,

{eq}\begin{align*} A(t) &= Ae^r + Ae^{2r} + \cdots Ae^{tr} \\ &= Ae^r( 1 + e^r + \cdots + e^{r(t-1)}) \\ &= Ae^r \bigg( \frac{1 - e^{r(t-1)}}{1 - e^r} \bigg)\\ &= Ae^r \bigg( \frac{e^{r(t-1) } - 1}{e^r - 1} \bigg) \end{align*} {/eq}

Answer and Explanation:

We are given that the accumulated future value of an annuity of $75,000 per year in a fund earning 13% interest compounded continuously is to be $, 800,000. We are asked to find the time {eq}t {/eq} for the investment.

We use the formula,

{eq}\begin{align*} FV &= Ae^r \bigg(\frac{e^{r(t-1)} - 1}{e^r - 1} \bigg) \\ 800,000 &= 75000 \times e^{0.13} \times \bigg(\frac{e^{0.13(t-1)} - 1}{e^{0.13} - 1} \bigg) \\ e^{0.13(t-1)} - 1 &= \frac{800,000}{75,000} \times \frac{e^{0.13} - 1}{e^{0.13}}\\ e^{0.13(t-1)} &= 1 + \frac{800,000}{75,000} \times \frac{e^{0.13} - 1}{e^{0.13}}\\ e^{0.13(t-1)} &\approx. 2.3\\ 0.13( t - 1) &\approx \ln 2.3 \\ t -1 &\approx \frac{\ln 2.3}{0.13} \\ t &\approx 7.41 \end{align*} {/eq}

The time {eq}t {/eq} required for the annuity of $75,000 invested at the beginning of every year, to mature to $ 800,000, is approximately

{eq}\displaystyle \boxed{ \Longrightarrow t \approx 7.41 } {/eq} years.


Learn more about this topic:

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Exponential Growth: Definition & Examples

from High School Algebra I: Help and Review

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