Chapter 4 Logarithm Functions

Chapter Outline Exponential Functions The Exponential Function ex Differentiation of Exponential Functions The Natural Logarithm Function The Derivative ln x Properties of the Natural Logarithm Function

Exponential Function Definition Example Exponential Function: A function whose exponent is the independent variable

Properties of Exponential Functions

Simplifying Exponential Expressions EXAMPLE Write each function in the form 2kx or 3kx, for a suitable constant k. SOLUTION (a) We notice that 81 is divisible by 3. And through investigation we recognize that 81 = 34. Therefore, we get (b) We first simplify the denominator and then combine the numerator via the base of the exponents, 2. Therefore, we get

Graphs of Exponential Functions Notice that, no matter what b is (except 1), the graph of y = bx has a y-intercept of 1. Also, if 0 < b < 1, the function is decreasing. If b > 1, then the function is increasing.

Solving Exponential Equations EXAMPLE Solve the following equation for x. SOLUTION This is the given equation. Factor. Simplify. Since 5x and 6 – 3x are being multiplied, set each factor equal to zero. 5x ≠ 0.

The Number e Definition Example e: An irrational number, approximately equal to 2.718281828, such that the function f (x) = bx has a slope of 1, at x = 0, when b = e

The Derivatives of bx and ex

Solving Exponential Equations EXAMPLE Find the equation of the tangent line to the curve at (0, 1). SOLUTION We must first find the derivative function and then find the value of the derivative at (0, 1). Then we can use the point-slope form of a line to find the desired tangent line equation. This is the given function. Differentiate. Use the quotient rule.

Solving Exponential Equations CONTINUED Simplify. Factor. Simplify the numerator. Now we evaluate the derivative at x = 0.

Solving Exponential Equations CONTINUED Now we know a point on the tangent line, (0, 1), and the slope of that line, − 1. We will now use the point-slope form of a line to determine the equation of the desired tangent line. This is the point-slope form of a line. (x1, y1) = (0, 1) and m = −1. Simplify.

Chain Rule for eg(x)

Chain Rule for eg(x) Differentiate. This is the given function. EXAMPLE Differentiate. SOLUTION This is the given function. Use the chain rule. Remove parentheses. Use the chain rule for exponential functions.

Working With Differential Equations Generally speaking, a differential equation is an equation that contains a derivative.

Solving Differential Equations EXAMPLE Determine all solutions of the differential equation SOLUTION The equation has the form y΄ = ky with k = 1/3. Therefore, any solution of the equation has the form where C is a constant.

Solving Differential Equations at Initial Values EXAMPLE Determine all functions y = f (x) such that y΄ = 3y and f (0) = ½. SOLUTION The equation has the form y΄ = ky with k = 3. Therefore, for some constant C. We also require that f (0) = ½. That is, So C = ½ and

Functions of the form ekx

The Natural Logarithm of x Definition Example Natural logarithm of x: Given the graph of y = ex, the reflection of that graph about the line y = x, denoted y = ln x

Properties of the Natural Logarithm

Properties of the Natural Logarithm The point (1, 0) is on the graph of y = ln x [because (0, 1) is on the graph of y = ex]. ln x is defined only for positive values of x. ln x is negative for x between 0 and 1. ln x is positive for x greater than 1. ln x is an increasing function and concave down.

Exponential Expressions EXAMPLE Simplify. SOLUTION Using properties of the exponential function, we have

Solving Exponential Equations EXAMPLE Solve the equation for x. SOLUTION This is the given equation. Remove the parentheses. Combine the exponential expressions. Add. Take the logarithm of both sides. Simplify. Finish solving for x.

Solving Logarithmic Equations EXAMPLE Solve the equation for x. SOLUTION This is the given equation. Divide both sides by 5. Rewrite in exponential form. Divide both sides by 2.

Other Exponential and Logarithmic Functions

Common Logarithms Definition Example Common logarithm: Logarithms to the base 10

Max’s & Min’s of Exponential Equations EXAMPLE The graph of is shown in the figure below. Find the coordinates of the maximum and minimum points.

Max’s & Min’s of Exponential Equations CONTINUED At the maximum and minimum points, the graph will have a slope of zero. Therefore, we must determine for what values of x the first derivative is zero. This is the given function. Differentiate using the product rule. Finish differentiating. Factor. Set the derivative equal to 0. Set each factor equal to 0. Simplify.

Max’s & Min’s of Exponential Equations CONTINUED Therefore, the slope of the function is 0 when x = 1 or x = −1. By looking at the graph, we can see that the relative maximum will occur when x = − 1 and that the relative minimum will occur when x = 1. Now we need only determine the corresponding y-coordinates. Therefore, the relative maximum is at (− 1, 0.472) and the relative minimum is at (1, − 1).

Derivative Rules for Natural Logarithms

Differentiating Logarithmic Expressions EXAMPLE Differentiate. SOLUTION This is the given expression. Differentiate. Use the power rule. Differentiate ln[g(x)]. Finish.

Differentiating Logarithmic Expressions EXAMPLE The function has a relative extreme point for x > 0. Find the coordinates of the point. Is it a relative maximum point? SOLUTION This is the given function. Use the quotient rule to differentiate. Simplify. Set the derivative equal to 0.

Differentiating Logarithmic Expressions CONTINUED The derivative will equal 0 when the numerator equals 0 and the denominator does not equal 0. Set the numerator equal to 0. Write in exponential form. To determine whether the function has a relative maximum at x = 1, let’s use the second derivative. This is the first derivative. Differentiate.

Differentiating Logarithmic Expressions CONTINUED Simplify. Factor and cancel. Evaluate the second derivative at x = 1. Since the value of the second derivative is negative at x = 1, the function is concave down at x = 1. Therefore, the function does indeed have a relative maximum at x = 1. To find the y-coordinate of this point So, the relative maximum occurs at (1, 1).

Properties of the Natural Logarithm Function

Simplifying Logarithmic Expressions EXAMPLE Write as a single logarithm. SOLUTION This is the given expression. Use LIV (this must be done first). Use LIII. Use LI. Simplify.

Differentiating Logarithmic Expressions EXAMPLE Differentiate. SOLUTION This is the given expression. Rewrite using LIII. Rewrite using LI. Rewrite using LIV. Differentiate.

Differentiating Logarithmic Expressions CONTINUED Distribute. Finish differentiating. Simplify.

Logarithmic Differentiation Definition Example Logarithmic Differentiation: Given a function y = f (x), take the natural logarithm of both sides of the equation, use logarithmic rules to break up the right side of the equation into any number of factors, differentiate each factor, and finally solving for the desired derivative. Example will follow.

Logarithmic Differentiation EXAMPLE Use logarithmic differentiation to differentiate the function. SOLUTION This is the given function. Take the natural logarithm of both sides of the equation. Use LIII. Use LI.

Logarithmic Differentiation CONTINUED Use LIV. Differentiate. Solve for f ΄(x). Substitute for f (x).

Chapter 5 Applications of the Exponential and Natural Logarithm Functions

Chapter Outline Exponential Growth and Decay Applications of the Natural Logarithm Functions

Exponential Growth Definition Example Exponential Growth: A quantity, such that, at every instant the rate of increase of the quantity is proportional to the amount of the quantity at that instant

Exponential Decay Definition Example Exponential Decay: A quantity, such that, at every instant the rate of decrease of the quantity is proportional to the amount of the quantity at that instant

Exponential Growth & Decay Model

Exponential Growth & Decay in Application EXAMPLE A 50 mg dose of quinine is given to a patient to prevent malaria. The quinine remaining in the body t hours after taking the dose satisfies the differential equation Q′ (t) = −0.06Q (t). Find a formula for the amount Q of quinine in the body t hours after the dose is given. How much quinine in the body after 24 hours? How long will it take for the quinine in the body to reach 10 mg? What is the half-life of the quinine in the body? SOLUTION

Exponential Growth & Decay in Application EXAMPLE The world’s population was 5.51 billion on January 1, 1993 and 5.88 billion on January 1, 1998. Assume that at any time the population grows at a rate proportional to the population at that time. In what year will the world’s population reach 7 billion? SOLUTION The year 1993 will be the year t = 0 and the population at time t = 0 is 5.51 (measured in billions). Therefore, the year 1998 will be year t = 5 and the population at time t = 5 is 5.88 (measured in billions). Since the population grows at a rate proportional to the size of the population, we can use the exponential growth model P(t) = P0ekt to describe the population of the world. Since P0 is the initial quantity, P0 = 5.51. Therefore, our formula becomes

Exponential Growth & Decay in Application CONTINUED Now we use the other given information (5.88 billion in 1998) to determine k. When t = 5, the population is 5.88 billion people. Rewrite in logarithmic form. Solve for k. Therefore, our formula to model this situation is Now we can determine when the world’s population will be 7 billion. Replace P(t) with 7. Divide. Rewrite in logarithmic form. Solve for t. Therefore, the world’s population will be 7 billion people about 18.36 years after our initial year, 1993. That is, around the year 2011.

Exponential Growth & Decay in Application EXAMPLE Radium-226 is used in cancer radiotherapy, as a neutron source for some research purposes, and as a constituent of luminescent paints. Let P(t) be the number of grams of radium-226 in a sample remaining after t years, and suppose that P(t) satisfies the differential equation (a) Find the formula for P(t). (b) What was the initial amount? (c) What is the decay constant? (d) Approximately how much of the radium will remain after 943 years? (e) How fast is the sample disintegrating when just one gram remains? Use the differential equation.

Exponential Growth & Decay in Application CONTINUED (f) What is the weight of the sample when it is disintegrating at the rate of 0.004 grams per year? (g) The radioactive material has a half-life of about 1612 years. How much will remain after 1612 years? 3224 years? SOLUTION (a) Since the function y = Cekt satisfies the differential equation y΄ = ky, the function P(t) = Cekt = Ce-0.00043t (where k = -0.00043). Since for the function y = Cekt, C is always the initial quantity (at time t = 0), C = 12 (since P(0) = 12). Therefore, our function is

Exponential Growth & Decay in Application CONTINUED (b) We were given P(0) = 12. Therefore the initial amount is 12 grams. (c) Since our exponential decay function is , the decay constant, being the coefficient of t, is − 0.00043. (d) To determine approximately how much of the radium will remain after 943 years, we will evaluate the function at t = 943. This is the decay function. Evaluate the function at t = 943. Simplify. Therefore, after 943 years, there will be approximately 8 grams remaining.

Exponential Growth & Decay in Application CONTINUED (e) To determine how fast the sample is disintegrating when just one gram remains, we must first recognize that this is a situation concerning the rate of change of a quantity, namely the rate at which the radium is disintegrating. This of course involves the derivative function. This function was given to us and is . Now we will determine the value of the derivative function at P(t) = 1 (when one gram remains). This is the derivative function. Replace P(t) with 1. So, when there is just one gram remaining, the radium is disintegrating at a rate of 0.00043 grams/year. (f) To determine the weight of the sample when it is disintegrating at the rate of 0.004 grams per year, we must determine P(t) when P΄(t) = −0.004.

Exponential Growth & Decay in Application CONTINUED This is the derivative function. Replace P΄(t) with − 0.004. Solve for P(t). So, the weight of the sample when it is disintegrating at the rate of 0.004 grams per year, is 9.3 grams. (g) To determine how much of the radium will remain after 1612 years, that is one half-life, we will simply recognize that after one half-life, half of the original amount of radium will be disintegrated. That is, 12/2 = 6 grams will be disintegrated and therefore 6 grams will remain. After 3224 years, two half-lives, half of what was remaining at the end of the first 1612 years (6 grams) will remain. That is, 6/2 = 3 grams. These results can be verified using the formula for P(t).