Question

A sum of $1000 was invested for 4 years, and the interest was compounded semiannually. If this sum amounted to $1459.54 in the given time, what was the interest rate? (Round your answer to two decimal places.)

Answer 1

$A=P(1+ \frac{r}{n})^{nt}$
A=furuter aount
P=present amount
r=rate in decimal
n=number of times per year it is compounded
t=time in years


A=1459.54
P=1000
r=r
n=2 (semianually is 2 times per year)
t=4

$1459.54=1000(1+ \frac{r}{2})^{(2)(4)}$
$1459.54=1000(1+ \frac{r}{2})^{8}$
divide both sides by 1000
$1.45954=(1+ \frac{r}{2})^{8}$
$1.45954=(\frac{2+r}{2})^{8}$
$1.45954=\frac{(2+r)^8}{2^8}$
$1.45954=\frac{(2+r)^8}{256}$
times both sides by 256
373.64224=$(2+r)^8$
take the 8th root of both sides
$\sqrt[8]{373.64224} =2+r$
minus 2 from both sides
$\sqrt[8]{373.64224} -2=r$
aprox
0.0968=r
9.67%