Nathaniel invested $37,000 in an account paying an interest rate of 5.7% compounded daily. Assuming no deposits or withdrawals are made, how long would it take, to the nearest year, for the value of the account to reach $70,900?

Answers

Answer 1

Answer:

11

Step-by-step explanation:

\text{Compounded Daily:}

Compounded Daily:

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

A=P(1+  

n

r

​  

)  

nt

 

Compound interest formula

A=70900\hspace{35px}P=37000\hspace{35px}r=0.057\hspace{35px}n=365

A=70900P=37000r=0.057n=365

Given values

70900=

70900=

\,\,37000\left(1+\frac{0.057}{365}\right)^{365t}

37000(1+  

365

0.057

​  

)  

365t

 

Plug in values

70900=

70900=

\,\,37000(1.0001562)^{365t}

37000(1.0001562)  

365t

 

Simplify

\frac{70900}{37000}=

37000

70900

​  

=

\,\,\frac{37000(1.0001562)^{365t}}{37000}

37000

37000(1.0001562)  

365t

 

​  

 

Divide by 37000

1.9162162=

1.9162162=

\,\,1.0001562^{365t}

1.0001562  

365t

 

\log\left(1.9162162\right)=

log(1.9162162)=

\,\,\log\left(1.0001562^{\color{blue}{365t}}\right)

log(1.0001562  

365t

)

Take the log of both sides

\log\left(1.9162162\right)=

log(1.9162162)=

\,\,\color{blue}{365t}\log\left(1.0001562\right)

365tlog(1.0001562)

Bring exponent to the front

\frac{\log\left(1.9162162\right)}{\log\left(1.0001562\right)}=

log(1.0001562)

log(1.9162162)

​  

=

\,\,\frac{365t\log\left(1.0001562\right)}{\log\left(1.0001562\right)}

log(1.0001562)

365tlog(1.0001562)

​  

 

Divide both sides by log(1.0001562)

4164.8632402=

4164.8632402=

\,\,365t

365t

Use calculator

\frac{4164.8632402}{365}=

365

4164.8632402

​  

=

\,\,\frac{365t}{365}

365

365t

​  

 

Divide by 365

11.4105842=

11.4105842=

\,\,t

t

t\approx

t≈

\,\,11

11

Answer 2

Using compound interest, it is found that it would take 11 years for the value of the account to reach $70,900.

What is compound interest?

The amount of money earned, in compound interest, after t years, is given by:

[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]

In which:

A(t) is the amount of money after t years. P is the principal(the initial sum of money). r is the interest rate(as a decimal value). n is the number of times that interest is compounded per year.

Hence, the parameters are given as follows:

[tex]A(t) = 70900, P = 37000, n = 365, r = 0.057[/tex]

We have to solve for t, then:

[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]

[tex]70900 = 37000\left(1 + \frac{0.057}{365}\right)^{365t}[/tex]

[tex]\left(1 + \frac{0.057}{365}\right)^{365t} = \frac{70900}{37000}[/tex]

[tex]\left(1 + \frac{0.057}{365}\right)^{365t} = 1.91621622[/tex]

[tex](1.00015616)^{365t} = 1.91621622[/tex]

[tex]\log{(1.00015616)^{365t}} = \log{1.91621622}[/tex]

[tex]365t\log{1.00015616} = \log{1.91621622}[/tex]

[tex]t = \frac{\log{1.91621622}}{365\log{1.00015616}}[/tex]

t = 11.41

Rounding to the nearest year, it would take 11 years for the value of the account to reach $70,900.

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