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# Applications of Logarithm and Exponential Equations Help

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## Introduction to Applications of Logarithm and Exponential Equations

Now that we can solve exponential and logarithmic equations, we can solve many applied problems. We will need the compound growth formula for an investment earning interest rate r , compounded n times per year for t years, and the exponential growth formula for a population growing at the rate of r per year for t years, n ( t ) = n 0 e rt. In the problems below, we will be looking for the time required for an investment to grow to a specified amount.

#### Examples

• How long will it take for \$1000 to grow to \$1500 if it earns 8% annual interest, compounded monthly?
• In the formula we know A ( t ) = 1500, P = 1000, r = 0.08, and n = 12. We do not know t .

We will solve this equation for t and will round up to the nearest month.

In five years and one month, the investment will grow to about \$1500.

• How long will it take an investment to double if it earns 6 % annual interest, compounded daily?
• An investment of \$P doubles when it grows to \$2P, so let A ( t ) = 2 P in the compound growth formula.

### Applications of Logarithm and Exponential Equations Practice Problems

#### Practice

1. How long will it take \$2000 to grow to \$40,000 if it earns 9% annual interest, compounded annually?
2. How long will it take for \$5000 to grow to \$7500 if it earns 6 % annual interest, compounded weekly?
3. How long will it take an investment to double if it earns 6 % annual interest, compounded quarterly?

#### Solutions

1. 40,000 = 2000(1 + 0.09) t

The \$2000 investment will grow to \$40,000 in 35 years.

In 6 years, 13 weeks (0.24 × 52 = 12.48 rounds up to 13), the \$5000 investment will grow to \$7500.

In 11 years and 3 months (0.18 rounded up to the nearest quarter is 0.25, one quarter is 3 months), the investment will double.

## Growth and Decay Formulas

This method works with population models where the population (either of people, animals, insects, bacteria, etc.) grows or decays at a certain percent every period. We will use the growth formula n ( t ) = n 0 e rt . If the population is decreasing, we will use the decay formula, n ( t ) = n 0 e rt . Because we will be working with the base e , instead of taking the log of both sides, we will be rewriting the equations as log equations (this is equivalent to taking the natural log of both sides).

#### Example 1:

• A school district estimates that its student population will grow about 5% per year for the next 15 years. How long will it take the student population to grow from the current 8000 students to 12,000? We will solve for t in the equation 12,000 = 8000 e 0.05 t.

The population is expected to reach 12,000 in about 8 years.

#### Example 2:

• The population of a certain city in the year 2004 is about 650,000. If it is losing 2% of its population each year, when will the population decline to 500,000?
• Because the population is declining, we will use the formula n ( t ) = n 0 e rt .

Solve for t in the equation 500,000 = 650,000 e −0.02 t .

The population is expected to drop to 500,000 around the year 2017.

#### Example 3:

• At 2:00 a culture contained 3000 bacteria. They are growing at the rate of 150% per hour. When will there be 5400 bacteria in the culture?
• A growth rate of 150% per hour means that r = 1.5 and that t is measured in hours.

At about 2:24 (0.39 × 60 = 23.4 minutes) there will be 5400 bacteria in the culture.

### Growth and Decay Practice Problems

#### Practice

1. In 2003 a rural area had 1800 birds of a certain species. If the bird population is increasing at the rate of 15% per year, when will it reach 3000?
2. In 2002, the population of a certain city was 2 million. If the city’s population is declining at the rate of 1.8% per year, when will it fall to 1.5 million?
3. At 9:00 a petrie dish contained 5000 bacteria. The bacteria population is growing at the rate of 160% per hour. When will the dish contain 20,000 bacteria?

#### Solutions

1. 3000 = 1800 e 0.15 t

The bird population should reach 3000 in the year 2006.

In the year 2018, the population will decline to 1.5 million.

At about 9:52 (0.87 × 60 = 52.2 minutes), there will be 20,000 bacteria in the dish.

Find practice problems and solutions for these concepts at: Exponents and Logarithms Practice Test

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