## Related questions with answers

A sum of $1000 was invested for 4 years, and the interest was compounded semiannually. If this sum amounted to$1435.77 in the given time, what was the interest rate?

Solutions

VerifiedWe can solve for the interest rate $(r)$ using logarithms as shown below.

Consider the following given:

$P= 1000, \ \ \ t = 4, \ \ \ A(t) = 1435.77, \ \ \ n = 2$

$\begin{align*} A(t) &= P \left( 1 + \dfrac{r}{n} \right)^{nt} \tag{Substitute the values} \\ 1435.77 &= 1000 \left( 1 + \dfrac{r}{2} \right)^{(2)(4)} \tag{Simplify} \\ 1435.77 &= 1000 \left( 1 + \dfrac{r}{2} \right)^{8} \tag{Divide both sides by $1000$} \\ 1.43577 &= \left( 1 + \dfrac{r}{2} \right)^8 \tag{Apply $\log $ both sides} \\ \log 1.43577 &= \log \left( 1 + \dfrac{r}{2} \right)^8 \tag{Simplify; Law 3: $\log b^x = x \log b$ } \\ 0.1571 &= 8 \log \left( 1 + \dfrac{r}{2} \right) \tag{Divide both sides by $8$} \\ 0.0196 &= \log \left( 1 + \dfrac{r}{2} \right) \tag{$y = \log_b x \Leftrightarrow b^y = x$} \\ 10^{0.0196} &= 1+\dfrac{r}{2} \tag{Simplify} \\ 1.0463 &= 1+\dfrac{r}{2} \tag{Subtract $1$ both sides} \\ 0.0463 &= \dfrac{r}{2} \tag{Multiply both sides by $2$} \\ r &= 0.0925 \\ r&= 9.25 \% \end{align*}$

$\therefore$ The interest rate is $9.25 \%$.

$\text{Use the formula } \color{#c34632} A(t)=P\left(1+\frac{r}{n}\right)^{nt}$ where:

$\begin{align*} A(t)&=\text{value of investment after t years}\\ t&=\text{time in years}\\ n&=\text{number of times compounded in a year}\\ P&=\text{principal amount invested}\\ r&=\text{interest rate per year}\\ \end{align*}$

$\text{Substitute }P=1000, \quad t=4 \quad n=2, \text{ and }\quad A(t)=1435.77 \text{ to obtain:}$

$\begin{align*} 1435.77&=1000\left(1+\frac{r}{2}\right)^{2(4)}\\ 1435.77&=1000\left(1+\frac{r}{2}\right)^{8}\\ \end{align*}$

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