meredith48034
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NO LINKS!!! Kim invested $5000 in Mutual fund at 6% compounded quarterly. Write an equation to represent this situation. When will her investment be worth more than $10,000​

Respuesta :

Answer:

Equation is  A = 5000(1.015)^(4t)

Her investment will be worth $10,000 in about 11.63888 years

Rounding up to the nearest whole number gets to 12 years

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Explanation:

Part 1) Finding the equation

The compound interest formula is

A = P(1+r/n)^(n*t)

Here are the variables

  • A = final amount
  • P = starting amount, or deposit, or principal
  • r = interest rate in decimal form
  • n = number of times money is compounded per year
  • t = number of years

In this case,

  • P = 5000
  • r = 0.06 from the 6% annual interest
  • n = 4 times a year is the compounding frequency
  • t = unknown amount of time

Therefore, the equation is

A = P(1+r/n)^(n*t)

A = 5000(1+0.06/4)^(4t)

A = 5000(1.015)^(4t)

The decimal value is exact.

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Part 2) Let's plug in A = 10,000 and solve for t.

You'll need to use logarithms to isolate the exponent.

A = 5000(1.015)^(4t)

10,000 = 5000(1.015)^(4t)

10,000/5000 = (1.015)^(4t)

2 = (1.015)^(4t)

Log[ 2 ] = Log[ (1.015)^(4t) ]

Log(2) = 4t*Log( 1.015 )

4t = Log(2)/Log(1.015)

4t = 46.5555256308062

t = 46.5555256308062/4

t = 11.6388814077015

t = 11.63888

It takes about 11.63888 years for the investment to reach $10,000.

Therefore, at the 12 year mark is when the investment is more than $10,000.

semsee45

Answer:

11 years and 9 months

Step-by-step explanation:

Compound Interest Formula

[tex]\large \text{$ \sf A=P(1+\frac{r}{n})^{nt} $}[/tex]

where:

  • A = final amount
  • P = principal amount
  • r = interest rate (in decimal form)
  • n = number of times interest applied per time period
  • t = number of time periods elapsed

Given:

  • A = $10000
  • P = $5000
  • r = 6% = 0.06
  • n = 4 (as compounded quarterly)
  • t = years

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

[tex]\implies 10000=5000\left(1+\dfrac{0.06}{4}\right)^{4t}[/tex]

[tex]\implies \dfrac{10000}{5000}=\left(1+0.015\right)^{4t}[/tex]

[tex]\implies 2=\left(1.015\right)^{4t}[/tex]

Take natural logs of both sides:

[tex]\implies \ln 2=\ln \left(1.015\right)^{4t}[/tex]

Apply the power log law:

[tex]\implies \ln 2=4t\ln \left(1.015\right)[/tex]

Simplify:

[tex]\implies t=\dfrac{\ln 2}{4 \ln 1.015}[/tex]

[tex]\implies t=11.63888141[/tex]

11.6388141... years ≈ 11 years and 7.7 months

As the interest is earned quarterly, round this to the nearest quarter

⇒ 11 years and 9 months.

Therefore, Kim's investment will be worth more than $10,000 at 11 years and 9 months.

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