A person invests 3000 dollars in a bank. The bank pays 4% interest compounded quarterly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 6500 dollars?

p=amount put in
t=amount of time in years
%=growth rate
f(t)=total

f(t)=p(1+%)^t
6500=3000(1+.04)^t
Solve for t
6500/3000=(1.04)^t
*Logs*
Log (13/6) = 19.7138...
1.04
Rounded to nearest 10th

t = 19.7

Answer Link

victorawesomeawesome victorawesomeawesome · Answer 2 · 2021-05-28T04:53:21+02:00

Answer:19.4

Step-by-step explanation:

A=P(1+\frac{r}{n})^{nt}

A=P(1+

n

r

)

nt

A=6500 \hspace{15px} P=3000 \hspace{15px} r=0.04 \hspace{15px} t=?

A=6500P=3000r=0.04t=?

n=4\text{ (quarterly)}

n=4 (quarterly)

6500=3000(1+\frac{0.04}{4})^{4t}

6500=3000(1+

4

0.04

)

4t

6500=3000(1.01)^{4t}

6500=3000(1.01)

4t

\frac{6500}{3000}=\frac{3000(1.01)^{4t}}{3000}

3000

6500

=

3000

3000(1.01)

4t

2.1666667=(1.01)^{4t}

2.1666667=(1.01)

4t

\log(2.1666667)=\log((1.01)^{4t})

log(2.1666667)=log((1.01)

4t

)

\log(2.1666667)=4t\log(1.01)

log(2.1666667)=4tlog(1.01)

Power Rule.

\frac{\log(2.1666667)}{4\log(1.01)}=\frac{4t\log(1.01)}{4\log(1.01)}

4log(1.01)

log(2.1666667)

=

4log(1.01)

4tlog(1.01)

t=\frac{0.3357921}{0.0172855}

t=

0.0172855

0.3357921

t=19.42624\approx 19.4 \text{ years}

t=19.42624≈19.4 years