Step-by-step explanation:
[tex]A=P(1+\frac{R}{n})^{nt}[/tex]
A = Amount after t time period
P = Principal amount
R = Interest rate
n = Number of times interest applied per time period
We have:
a) P = $1,500
R =5%=0.05
n = 1
t = 3 years
[tex]A=\$1,500\times (1+\frac{0.05}{1})^{1\times 3}[/tex]
[tex]A=\$1,736.43[/tex]
The amount after 3 years in the bank will be $1,736.43.
b) P = $1,500
R =5%=0.05
n = 1
t = 18 years
[tex]A=\$1,500\times (1+\frac{0.05}{1})^{1\times 18}[/tex]
[tex]A=\$3,609.92[/tex]
The amount after 3 years in the bank will be $3,609.92.
c) P = $1,500
R =5%=0.05
n = 1
t = ?
A= $2,000
[tex]\$2,000=\$1,500\times (1+\frac{0.05}{1})^{1\times t}[/tex]
[tex]\frac{\$2,000}{\$1,500}= (1+\frac{0.05}{1})^{1\times t}[/tex]
[tex]\log \frac{\$2,000}{\$1,500}=t\times\log(1.05)[/tex]
[tex]0.1249=t\times 0.02119[/tex]
[tex]t=\frac{0.1249}{0.02119}=5.89 years[/tex]
5.89 years will it take for the account to contain $2,000.
d) P = $1,500
R =5%=0.05
n = 1
t = ?
A= $2,500
[tex]\$2,500=\$1,500\times (1+\frac{0.05}{1})^{1\times t}[/tex]
[tex]\frac{\$2,500}{\$1,500}= (1+\frac{0.05}{1})^{1\times t}[/tex]
[tex]\log \frac{\$2,500}{\$1,500}=t\times\log(1.05)[/tex]
[tex]0.2218=t\times 0.02119[/tex]
[tex]t=\frac{0.2218}{0.02119}=10.47years[/tex]
10.47 years will it take for the account to contain $2,000.
