Question

NO LINKS!! URGENT HELP PLEASE!!

41. Sophie invested $5000 into an account that will increase in value by 2.3% each year. Write a function to model this situation, then find when will the account will have $15,000?

42. A baseball that was valued in 1980 has increased by 7% each year. Write a function to model this situation then find when the card will be worth $2000?

Answer 1

41. 49 years

42. 34 years and 12 days

Explanation:

As the account increases by a constant percentage each year, we can use the exponential growth formula to write a function to model the situation.

`\boxed{\begin{minipage}{8 cm}\underline{Exponential Growth Formula}\\\\$ f(t)=a\left(1+r\right)^(t)$\\\\where:\\\\ \phantom{ww}$\bullet$ $f(t)$ is the account balance. \\ \phantom{ww}$\bullet$ $a$ is the principal amount.\\ \phantom{ww}$\bullet$ $r$ is the interest rate (in decimal form). \\ \phantom{ww}$\bullet$ $t $ is the time (in years). \\ \end{minipage}}`

Given values:

• f(t) = $15,000
• a = $5,000
• r = 2.3% = 0.023

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

`\begin{aligned}f(t)&=a(1+r)^t\\\\\implies 15000&=5000(1+0.023)^t\\15000&=5000(1.023)^t\\3&=(1.023)^t\\\\\textsf{Take natural logs:} \quad \ln (3)&=\ln (1.023)^t\\\ln (3)&=t\ln (1.023)\\t&=(\ln (3))/(\ln (1.023))\\t&=48.3129760...\end{aligned}`

The account balance will reach $15,000 during the 48th year. As the interest is applied annually, we need to round up to the nearest whole year. Therefore, the account balance will reach $15,000 after 49 years.

Note: At the end of 48 years, the account balance will be $14,893.63, and after 49 years it will be $15,236.18.

`\hrulefill`

As the baseball's value increases by a constant percentage each year, we can use the exponential growth formula to write a function to model the situation.

`\boxed{\begin{minipage}{8 cm}\underline{Exponential Growth Formula}\\\\$ V(t)=a\left(1+r\right)^(t)$\\\\where:\\\\ \phantom{ww}$\bullet$ $V(t)$ is the value of the baseball card. \\ \phantom{ww}$\bullet$ $a$ is the initial value of the card in 1980.\\ \phantom{ww}$\bullet$ $r$ is the interest rate (in decimal form). \\ \phantom{ww}$\bullet$ $t $ is the time (in years). \\ \end{minipage}}`

Given values:

• V(t) = $2,000
• a = $200
• r = 7% = 0.07

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

`\begin{aligned}V(t)&=a(1+r)^t\\\\\implies 2000&=200(1+0.07)^t\\2000&=200(1.07)^t\\10&=(1.07)^t\\\\\textsf{Take natural logs:} \quad \ln (10)&=\ln (1.07)^t\\\ln (10)&=t\ln (1.07)\\t&=(\ln (10))/(\ln (1.07))\\t&=34.0323838...\end{aligned}`

The value of the baseball card will reach $2,000 during the 34th year - after 34 years and 12 days. As this is not an investment account where interest is applied annually, we don't need to round up. Therefore, the baseball card will be worth $2,000 after 34 years and 12 days.

Answer 2

41. 48 years and 3 months.

Explanation:

41.

let's use the compound interest formula. The formula for compound interest is:

`\bold{A = P * (1 + r)^n}`

Where:

A is the final amount

P is the principal (initial investment)

r is the interest rate per period

n is the number of periods

we have,

P= $5000

A=$15000

r=2.3%

Now substituting value

`15000=5000(1+(2.3)/(100))^n\\15000=5000(1.023)^n\\(15000)/(5000)=(1.023)^n\\3=(1.023)^n`

doing logarithms of both sides of the equation.

We can use the natural logarithm (ln) to solve for n. The equation becomes:

`ln(3) = ln(1.023)^n`

Using the logarithmic property
`lna^b = b * ln(a),` we can rewrite the equation as:

ln(3) = n * ln(1.023)

n=
`(ln \:3)/(ln\:1.023)`

n=48.313 years approximately 48 years and 3 months.