Question Statement

Approximately two-thirds of all aluminum cans distributed are recycled each year. A beverage company distributed $250,000$ cans. The number of cans still in use after $t$ years is given by the function:

$N(t)=250,000{\left(\frac{2}{3}\right)}^t$

(a) After how many years will 60,000 cans be in use? (b) After what amount of time will only 1,000 cans be in use?

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Background and Explanation

This problem involves an exponential decay function, where the quantity decreases over time by a constant percentage. To solve for the time variable $t$ located in the exponent, we must use logarithms to "bring down" the exponent and isolate the variable.

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Solution

The given decay function is: $N(t)=250,000{\left(\frac{2}{3}\right)}^t$

Part (a)

We are asked to find the time $t$ when the number of cans $N$ is 60,000.

1. Substitute the value into the equation: $60,000=250,000{\left(\frac{2}{3}\right)}^t$

2. Isolate the exponential term: Divide both sides by 250,000: $\frac{60,000}{250,000}={\left(\frac{2}{3}\right)}^t$ $\frac{6}{25}={\left(\frac{2}{3}\right)}^t$

3. Apply logarithms to both sides: Taking the log of both sides allows us to use the power rule: $\log \left(\frac{6}{25}\right)=\log {\left(\frac{2}{3}\right)}^t$ $\log \left(\frac{6}{25}\right)=t\log \left(\frac{2}{3}\right)$

4. Solve for $t$: $t=\frac{\log \frac{6}{25}}{\log \frac{2}{3}}$ Using the values from the raw data: $t=\frac{-0.6198}{-0.17609}$ $t\approx 3.52\text{ years}$

Part (b)

We are asked to find the time $t$ when the number of cans $N$ is 1,000.

1. Substitute the value into equation (1): $1,000=250,000{\left(\frac{2}{3}\right)}^t$

2. Isolate the exponential term: $\frac{1,000}{250,000}={\left(\frac{2}{3}\right)}^t$ $\frac{1}{250}={\left(\frac{2}{3}\right)}^t$

3. Apply logarithms to both sides: $\log \left(\frac{1}{250}\right)=\log {\left(\frac{2}{3}\right)}^t$ Using the quotient rule ($\log \frac{a}{b}=\log a-\log b$): $\log 1-\log 250=t\log \left(\frac{2}{3}\right)$

4. Solve for $t$: Since $\log 1=0$: $t=\frac{0-\log 250}{\log \frac{2}{3}}$ Using the values from the raw data: $t=\frac{-2.3979}{-0.17609}$ $t\approx 13.62\text{ years}$

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Key Formulas or Methods Used

• Exponential Decay Function: $N(t)=N_0(r{)}^t$
• Logarithm Power Rule: $\log (a^b)=b\log a$
• Logarithm Quotient Rule: $\log (\frac{a}{b})=\log a-\log b$
• Logarithm of One: $\log 1=0$

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Summary of Steps

1. Set the function $N(t)$ equal to the target number of cans.
2. Divide by the initial amount (250,000) to isolate the base and exponent.
3. Take the common logarithm ($\log$) of both sides of the equation.
4. Use the power rule of logarithms to move the variable $t$ from the exponent to a multiplier.
5. Divide the resulting values to find the number of years.