Find the term t of each continuously compounded account below rounded to the nearest
tenth of a year. In the chart, B is the ending balance, P is the principal, and r is the
interest rate expressed as a percent.
B
P
r
a.
$2,124.00
$2,000.00
2%
b.
$957.60
$900.00
1.25%
C.
$25,700.00
$25,000.00
0.9%
d.
$185.50
$175.00
3%
e.
$1,084,000.00
$1,000,000.00
1.6%
a.
b.
C.
d.
e.



Answer :

Answer:

a) 3.0 years

b) 5.0 years

c) 3.1 years

d) 1.9 years

e) 5.0 years

Step-by-step explanation:

To find the term (t) of each continuously compounded account rounded to the nearest tenth of a year, we can use the Continuous Compounding Interest formula:

[tex]\boxed{\begin{array}{l}\underline{\textsf{Continuous Compounding Interest Formula}}\\\\B=Pe^{rt}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\;\textsf{$B$ is the ending balance.}\\\phantom{ww}\bullet\;\;\textsf{$P$ is the principal amount.}\\\phantom{ww}\bullet\;\;\textsf{$e$ is Euler's number (constant).}\\\phantom{ww}\bullet\;\;\textsf{$r$ is the interest rate (in decimal form).}\\\phantom{ww}\bullet\;\;\textsf{$t$ is the time (in years).}\end{array}}[/tex]

As we need to solve for t, first rearrange the formula to isolate t:

[tex]B=Pe^{rt}\\\\\\\ln(B)=\ln(Pe^{rt})\\\\\\\ln(B)=\ln(P)+\ln(e^{rt})\\\\\\\ln(B)-\ln(P)=\ln(e^{rt})\\\\\\\ln\left(\dfrac{B}{P}\right)=rt\ln(e)\\\\\\\ln\left(\dfrac{B}{P}\right)=rt\\\\\\t=\dfrac{\ln\left(\dfrac{B}{P}\right)}{r}[/tex]

Now, we can use this formula to find the term (t) for each account.

[tex]\dotfill[/tex]

Term a

Given values:

  • B = $2,124.00
  • P = $2,000.00
  • r = 2% = 0.02

Substitute the given values into the formula and solve for t:

[tex]t=\dfrac{\ln\left(\dfrac{2124}{2000}\right)}{0.02}[/tex]

[tex]t=\dfrac{\ln\left(1.062\right)}{0.02}[/tex]

[tex]t=3.00769614...[/tex]

[tex]t=3.0\; \rm years[/tex]

Therefore, the term is:

[tex]\Large\boxed{\boxed{3.0\; \rm years}}[/tex]

[tex]\dotfill[/tex]

Term b

Given values:

  • B = $957.60
  • P = $900.00
  • r = 1.25% = 0.0125

[tex]t=\dfrac{\ln\left(\dfrac{957.6}{900}\right)}{0.0125}[/tex]

[tex]t=\dfrac{\ln\left(1.064\right)}{0.0125}[/tex]

[tex]t=4.96283127...[/tex]

[tex]t=5.0\; \rm years[/tex]

Therefore, the term is:

[tex]\Large\boxed{\boxed{5.0\; \rm years}}[/tex]

[tex]\dotfill[/tex]

Term c

Given values:

  • B = $25,700.00
  • P = $25,000.00
  • r = 0.9% = 0.009

[tex]t=\dfrac{\ln\left(\dfrac{25700}{25000}\right)}{0.009}[/tex]

[tex]t=\dfrac{\ln\left(1.028\right)}{0.009}[/tex]

[tex]t=3.068351892...[/tex]

[tex]t=3.1\; \rm years[/tex]

Therefore, the term is:

[tex]\Large\boxed{\boxed{3.1\; \rm years}}[/tex]

[tex]\dotfill[/tex]

Term d

Given values:

  • B = $185.50
  • P = $175.00
  • r = 3% = 0.03

[tex]t=\dfrac{\ln\left(\dfrac{185.5}{175}\right)}{0.03}[/tex]

[tex]t=\dfrac{\ln\left(1.06\right)}{0.03}[/tex]

[tex]t=1.942296937...[/tex]

[tex]t=1.9\; \rm years[/tex]

Therefore, the term is:

[tex]\Large\boxed{\boxed{1.9\; \rm years}}[/tex]

[tex]\dotfill[/tex]

Term e

Given values:

  • B = $1,084,000.00
  • P = $1,000,000.00
  • r = 1.6% = 0.016

[tex]t=\dfrac{\ln\left(\dfrac{1084000}{1000000}\right)}{0.016}[/tex]

[tex]t=\dfrac{\ln\left(1.084\right)}{0.016}[/tex]

[tex]t=5.041118938...[/tex]

[tex]t=5.0\; \rm years[/tex]

Therefore, the term is:

[tex]\Large\boxed{\boxed{5.0\; \rm years}}[/tex]