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]
