Mastering Logarithms, Exponentials, and Their Applications

Fundamental Logarithm Properties

These are key cancellation properties of logarithms and exponentials:

• e^ln8 = 8
• ln(e⁴) = 4
• log₈(8³) = 3
• 5^log5(2) = 2

Essential Logarithm Rules

• Product Rule: log_a(xy) = log_a(x) + log_a(y)
• Quotient Rule: log_a(x/y) = log_a(x) - log_a(y)
• Power Rule: log_a(x^p) = p log_a(x)
• Change of Base Formula: log_B(D) = log(D) / log(B) (using common log) or ln(D) / ln(B) (using natural log)
• Logarithmic to Exponential Form: log_b(n) = a is equivalent to n = b^a.

Logarithm Definitions and Undefined Cases

• A logarithm answers the question: "What is the power?" For example, log_b(n) asks "To what power must b be raised to get n?"
• Logarithms are undefined for:
  • log(0)
  • log(-#) (logarithm of a negative number)
• The natural logarithm: log_e(x) = ln(x).
• Exponential and logarithmic graphs are inverse functions of each other.

Transformations of Logarithmic Functions

For a function of the form f(x) = a log_b(x - c) + k:

• If a is negative (-a), the graph flips over the x-axis.
• If x is replaced by -x (e.g., log_b(-x)), the graph flips over the y-axis.
• log_b(x + c): Shifts the graph c units to the left.
• log_b(x - c): Shifts the graph c units to the right.
• +k: Shifts the graph k units up.
• -k: Shifts the graph k units down.

Example: ln(1/e) = ln(e^-1) = -1.

One-to-One Property of Exponential Functions

If b^x = b^y, then x = y. This property is useful for solving exponential equations.

Example: Solve 6³ = 36^x-1

1. Rewrite with the same base: 6³ = (6²)^x-1
2. Apply exponent rules: 6³ = 6^2(x-1)
3. Simplify the exponent: 6³ = 6^2x-2
4. Equate the exponents: 3 = 2x - 2
5. Solve for x: 5 = 2x → x = 5/2

Applications of Exponential Growth Models

Population Growth Problem 1: US Population

Problem: The resident population in the US in 2010 was 309 million people and was growing at the rate of 0.9% per year. Assuming that this growth rate continues, in what year will the population be 488 million people?

Model: A = Pe^rt (Continuous Growth Model)

Given Values:

• A = 488 million (Future Population)
• P = 309 million (Initial Population)
• r = 0.009 (Growth Rate)
• t = ? (Time in years)

Solution Steps:

1. Set up the equation: 488 = 309e^0.009t
2. Divide by P: 488/309 = e^0.009t → 1.579 ≈ e^0.009t
3. Take the natural logarithm of both sides: ln(1.579) = ln(e^0.009t)
4. Simplify: ln(1.579) = 0.009t
5. Solve for t: t = ln(1.579) / 0.009 → t ≈ 50.775 years
6. Calculate the target year: 2010 + 50.775 ≈ 2060

Population Growth Problem 2: MV Population

Problem: In 2000, the population of MV was 2.3 thousand people. In 2010, the population was 3.8 thousand people. If this is modeled by a continuous growth model, and the population was checked at the same time each year:

1. Determine the rate of growth (r):
   • Given: A = 3.8, P = 2.3, t = 10 years (2010 - 2000)
   • Equation: 3.8 = 2.3e^10r
   • Divide: 3.8/2.3 = e^10r → 1.652 ≈ e^10r
   • Take ln: ln(1.652) = 10r
   • Solve for r: r = ln(1.652) / 10 → r ≈ 0.050 (or 5.0%)
2. Write the general equation to model the population:
   • A = 2.3e^0.050t
3. Predict the population in 2040:
   • Time from 2000 to 2040: t = 40 years
   • A = 2.3e^{0.050 * 40} = 2.3e² → A ≈ 16.99 thousand people

Optimization: Maximizing Area with Fencing

Problem: A rancher has 200 feet of fencing to enclose two adjacent rectangular corrals.

Assume the corrals share one side. Let x be the length of the sides perpendicular to the shared fence, and y be the length of the shared fence (and the other parallel sides).

A) Write the area A of the corral as a function of x:

1. Write an equation for the perimeter: The total fencing used is 4 sides of length x and 3 sides of length y (two outer sides and one shared side).
   • 200 = 4x + 3y
2. Solve for y in terms of x:
   • 3y = 200 - 4x
   • y = (200 - 4x) / 3
3. Determine the total area of the adjacent rectangular corrals: If x is the width of each corral and y is the length, the total area is A = (2x)y.
   • Substitute y: A = 2x * [(200 - 4x) / 3]
   • Simplify: A = (400x - 8x²) / 3
4. Answer (Factored Area Function):
   • A = [8x(50 - x)] / 3

B) Write the area function in standard form to find analytically the dimensions that will produce the maximum area.

The area function is A(x) = (-8/3)x² + (400/3)x. This is a downward-opening parabola. The maximum area occurs at the vertex, where x = -b / (2a).

x = -(400/3) / (2 * -8/3) = -(400/3) / (-16/3) = 400/16 = 25.

So, x = 25 feet. Now find y:

y = (200 - 4*25) / 3 = (200 - 100) / 3 = 100 / 3 feet.

The dimensions for maximum area are: each corral is 25 feet wide, and the total length is 100/3 feet. The total enclosure is 50 feet by 100/3 feet.

Solving Logarithmic and Exponential Equations

Example 1: Solving a Natural Logarithm Equation

Equation: 8ln(x) + 1 = 17

1. Isolate the natural logarithm term: 8ln(x) = 16
2. Divide by 8: ln(x) = 2
3. Convert to exponential form (e^ln(x) = x): e² = x
4. Calculate x: x ≈ 7.389

Example 2: Solving a Logarithmic Equation with Product Rule

Equation: log₉(x + 1) + log₉(x - 1) = 2

1. Apply the product rule: log₉[(x + 1)(x - 1)] = 2
2. Simplify the product: log₉(x² - 1) = 2
3. Convert to exponential form: 9² = x² - 1
4. Solve for x: 81 = x² - 1 → 82 = x² → x = ±&sqrt;82
5. Check for extraneous solutions: Logarithms are only defined for positive arguments.
   • If x = 9.055: (9.055 + 1) > 0 and (9.055 - 1) > 0. This is a valid solution.
   • If x = -9.055: (-9.055 + 1) < 0 and (-9.055 - 1) < 0. This makes the original terms undefined. Therefore, x = -9.055 is an extraneous solution.
6. Final Solution: x ≈ 9.055

Example 3: Solving a Logarithmic Equation with Quotient Rule

Equation: log₃(x - 4)² - log₃(x + 2) = 3

1. Apply the quotient rule: log₃[ (x - 4)² / (x + 2) ] = 3
2. Convert to exponential form: 3³ = (x - 4)² / (x + 2)
3. Simplify: 27 = (x² - 8x + 16) / (x + 2)
4. Multiply by (x + 2): 27(x + 2) = x² - 8x + 16
5. Distribute: 27x + 54 = x² - 8x + 16
6. Rearrange into a quadratic equation: x² - 35x - 38 = 0
7. Solve for x (using quadratic formula or factoring): x ≈ -1.054 and x ≈ 36.054
8. Check for extraneous solutions:
   • If x = 36.054: (36.054 - 4)² > 0 and (36.054 + 2) > 0. This is a valid solution.
   • If x = -1.054: (-1.054 - 4)² > 0 (since it's squared) and (-1.054 + 2) > 0. This is also a valid solution.
9. Final Solutions: x ≈ -1.054 and x ≈ 36.054

General Strategies for Solving Equations

• When 'x' is in the base:
  1. Isolate the function containing the base with 'x'.
  2. Raise both sides to the reciprocal of the power.
  3. Solve for x.
• When 'x' is in the power (exponent):
  1. Isolate the exponential base.
  2. Apply the inverse operation (take the logarithm of both sides, using the base of the exponential if possible, or natural log/common log).
  3. Solve for x.
  4. Use the Change of Base Property if necessary.
• When 'x' is in two powers:
  1. Take the logarithm of both sides (e.g., natural log or common log).
  2. Apply the power rule of logarithms to bring the exponents to the front.
  3. Distribute and rearrange terms to isolate 'x'.
  4. Solve for x.

Half-Life Calculations

Problem: A substance has a half-life of 5715 years. If you start with 10g, how much will remain after a certain time t?

Model: A = Pe^rt

Step 1: Determine the decay rate (r) using the half-life.

• Given: A = 5g (half of initial), P = 10g (initial), t = 5715 years (half-life)
• Equation: 5 = 10e^{r * 5715}
• Divide by 10: 0.5 = e^5715r
• Take natural log: ln(0.5) = 5715r
• Solve for r: r = ln(0.5) / 5715 → r ≈ -0.0001212 (This is a negative rate, indicating decay)

Step 2: Write the general equation for this substance's decay.

• A = 10e^-0.0001212t

Step 3: Plug in the specific time (t) to find the remaining amount (A). (The original document implies this step without providing a specific 't' to plug in.)

Revenue Optimization

Problem: A company sells coats. Initially, they sell 65,000 coats at $400 each. For every $5 drop in price, they sell 1000 more coats.

Let x = the number of $5 price drops.

• New Price: P(x) = 400 - 5x
• New Quantity Sold: Q(x) = 65,000 + 1000x
• Revenue Function: R(x) = P(x) * Q(x) = (400 - 5x)(65,000 + 1000x)

To find the maximum revenue:

The revenue function is a quadratic equation (when expanded, it will be of the form ax² + bx + c) with a negative leading coefficient, meaning its graph is a downward-opening parabola. The maximum revenue occurs at the vertex.

Steps:

1. Expand the revenue function: R(x) = 26,000,000 + 400,000x - 325,000x - 5000x² = -5000x² + 75,000x + 26,000,000
2. Find the x-coordinate of the vertex using x = -b / (2a):
   • x = -75,000 / (2 * -5000) = -75,000 / -10,000 = 7.5
3. This means the optimal number of $5 price drops is 7.5.
4. Calculate the optimal price: 400 - 5(7.5) = 400 - 37.5 = $362.50
5. Calculate the optimal quantity: 65,000 + 1000(7.5) = 65,000 + 7,500 = 72,500 coats
6. Calculate the maximum revenue: R(7.5) = 362.50 * 72,500 = $26,281,250

Complex Numbers and Imaginary Units

Key Concepts:

• If a polynomial equation has one imaginary root (e.g., x = 2i), its conjugate must also be a root (e.g., x = -2i). Imaginary roots always come in conjugate pairs.
• You can factor a polynomial until all imaginary roots are accounted for.

Powers of the Imaginary Unit (i):

• i¹ = i
• i² = -1
• i³ = i² * i = -1 * i = -i
• i⁴ = i² * i² = -1 * -1 = 1
• The cycle of powers of i repeats every four terms.