Answer:
(a) 14.21
(b) 13.78
(c) 13.57
(d) 13.86
Step-by-step explanation:
Compound Interest, compounded n times per year,
[tex]A = P(1 + \frac{r}{n} )^{nt}[/tex]
Compound Interest, compounded continuously,
[tex]A = Pe^{rt}[/tex]
Where
A is the new amount/balance
P is the Principal (amount invested)
r is the rate in percentage
n is the number of times it is compounded.
Given:
Principal, P = $1000
rate, r = 5% = 0.05
For the balance to double, A = 2 × $1000
A = $2000
(a) annually, n = 1
[tex]2000 = 1000(1 + \frac{0.05}{1} )^{1 * t}[/tex]
[tex]2 = (1 + 0.05)^{t}[/tex]
[tex]2 = (1.05)^{t}[/tex]
To solve for t, we take [tex]log_{10}[/tex] of both sides
[tex]log_{10}2 = log_{10}1.05^{t}[/tex]
[tex]log_{10}2 = t(log_{10}1.05)[/tex]
[tex]t = \frac{log_{10}2}{log_{10}1.05}[/tex]
[tex]t = \frac{0.3010}{0.0212}[/tex]
t = 14.21
(b) monthly, n = 12
[tex]2000 = 1000(1 + \frac{0.05}{12} )^{12 * t}[/tex]
[tex]2 = (1 + 0.0042)^{12t}[/tex]
[tex]2 = (1.0042)^{12t}[/tex]
To solve for t, we take [tex]log_{10}[/tex] of both sides
[tex]log_{10}2 = log_{10}1.0042^{12t}[/tex]
[tex]log_{10}2 = 12t(log_{10}1.0042)[/tex]
[tex]t = \frac{log_{10}2}{12 * log_{10}1.0042}[/tex]
[tex]t = \frac{0.3010}{0.0218}[/tex]
t = 13.78
(c) daily, n = 365
[tex]2000 = 1000(1 + \frac{0.05}{365} )^{365 * t}[/tex]
[tex]2 = (1 + 0.00014)^{365t}[/tex]
[tex]2 = (1.00014)^{365t}[/tex]
To solve for t, we take [tex]log_{10}[/tex] of both sides
[tex]log_{10}2 = log_{10}1.00014^{365t}[/tex]
[tex]log_{10}2 = 365t(log_{10}1.00014)[/tex]
[tex]t = \frac{log_{10}2}{365 * log_{10}1.00014}[/tex]
[tex]t = \frac{0.3010}{0.022}[/tex]
t = 13.565
t = 13.57
(d) Continuously
[tex]2000 = 1000e^{0.05t}[/tex]
[tex]2 = e^{0.05t}[/tex]
To solve for t, we take [tex]log_{e}[/tex] of both sides
[tex]log_{e}2 = log_{e}e^{0.05t}[/tex]
But log_{e}e = 1
In2 = 0.05t
[tex]t =\frac{In2}{0.05}[/tex]
[tex]t =\frac{0.693}{0.05}[/tex]
t = 13.86
