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Logarithmic and Exponential Functions (12.1 – 12.5)

12.1 Review of index laws, and definition of a^r for a > 0, where r is rational.

12.2 Definition of logarithms to the base a. Algebraic properties of logarithms and exponents.

12.3 The functions y = a^x and y = log_ax for a > 0 and real x. Change of base.

12.4 The derivatives of y = a^x and y = log_ax. Natural logarithms and exponential function.

12.5 Differentiation and integration of simple composite functions involving exponentials and logarithms.” (syllabus)

For a more detailed description of the requirements for this topic, see the mathematics syllabus on the Board of Studies website click here

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Logarithms (intro)

What is 100 x 1000?

You have come up with the easy way of multiplying by 100 and added two zeros to the 1000 to get 100 000.

Another way is to treat it as indices and we get 10² x 10³ = 10²⁺³ = 10⁵ = 100 000

Now suppose we multiply 8x16

This is really 2³ x 2⁴ = 2⁷ = 128

Multiplying with numbers indices is easy because we simply add the indices to multiply and subtract the indices to divide.

Suppose we could change all numbers to indices and work with indices.

Thankfully we can.

Use your calculator to find 10^0.30103 and 10^0.47712

Hopefully you got close to 2 and 3.

Now add the indices. 10^{0.30103 + 0.47712} = 10^0.77815

Use your calculator to evaluate 10^0.77815

Hopefully you got close to 6.

So we can multiply 2 and 3 by changing them to powers of 10.

You don’t have to use powers of 10. You can use powers of any number.

The number you use is called the base. In the above example we used the base 10.

When we multiplied 8 x 16 we used the base 2.

Calculators have two sets of bases. Numbers to the base of 10, indicated on your calculator by “log” and numbers to the exponential base e, indicated on your calculator by “ln”.

This brings us to our definition of a logarithm.

The logarithm of a number to a base is the power to which the base must be raised to get the number.

e.g. 10² = 100 means that the log of 100 to the base 10 is 2 (The log is the power)

It can be written log₁₀ 100 = 2

Exercise: Evaluate the following logarithms.

(i) log₂ 4 (ii) log₂ 32 (iii) log₃ 27 (iv) log₅ 125 (v) log₁₀ 10

(vi) log₂ 16 (vii) log₄ 16 (viii) log₇ 49 (ix) log₃ 81 (x) log₁₀ 10⁶

Answers: (i) 2, (ii) 5, (iii) 3, (iv) 3 (v) 1, (vi) 4, (vii) 2, (viii) 2, (ix) 4, (x) 6.

Exercise: Use your calculator to find log₁₀ 6 and log₁₀ 5. Add them together.

Find the value of 10 raised to that power.

Answer: log₁₀ 6 = 0.77815 and log₁₀ 5 = 0.69897

log₁₀ 6 + log₁₀ 5 = 0.77815 + 0.69897 = 1.47712

10^1.47712 = 29.9999 30

MORE LOGARITHMS

Log Rules

log_a xy = log_a x + log_a y

log_a = log_a x - log_a y

log_a x^b = b log_a x

log_a 1 = 0

log_a x =

a ^log _a ^x = x

log_a a^x = x

Exercise 1.

Given that log₁₀ 2 = 0.3010, log₁₀ 3 = 0.4771 and log₁₀ 7 = 0.8451, use the log rules to determine the logarithm to the base 10 of all the integers between 1 and 10.

Exercise 2.

Use your answers to Exercise 1 to evaluate the following without using your calculator.

All logs are to the base 10.

(i) log 5 + log 4 – log 2 (ii) log 10 – log 2 – log 5 (iii) log 9 – log 3 + log 2

(iv) (v) 3 log 4 – 2 log 2 (vi) log 8 – log 6 + log 3 (vii) log 7 + log 49 (viii) log 6 + log 2 – log 3 (ix) log 6 – log 12 + log 2 (x)

Exercise 3.

Use the change of base theorem to find the following: (calculators may be used)

(i) log ₂ 10 (ii) log ₂ 100 (iii) log ₂ 5 (iv) log ₂ 0.5 (v) log ₂ 4

Exercise 1. Answers.

1. 1 = 10⁰ log₁₀ 1 = 0

2. log₁₀ 2 = 0.3010 (given)

3. log₁₀ 3 = 0.4771 (given)

4. log₁₀ 4 = log₁₀ 2² = 2 log₁₀ 2 = 2 x 0.3010 = 0.6020

5. log₁₀ 5 = log₁₀ = log₁₀ 10 – log₁₀ 2 = 1 – 0.3010 = 0.6990

6. log₁₀ 6 = log₁₀ (3 x 2) = log₁₀ 3 + log₁₀2 = 0.3010 + 0.4771 = 0.7781

7. log₁₀7 = 0.8451 (given)

8. log₁₀ 8 = log₁₀ 2³ = 3 log₁₀ 2 = 3 x 0.3010 = 0.9030

9. log₁₀ 9 = log₁₀ 3² = 2 log₁₀ 3 = 2 x 0.4771 = 0.9542

10. 10 = 10¹ log₁₀ 10 = 1

Exercise 2. Answers.

(i) 1 (ii) 0 (iii) 0.7781 (iv) 1.5 (v) 1.2040 (vi) 0.6020 (vii) 2.5353 (viii) 0.6020 (ix) 0 (x) 1.5

Exercise 3. Answers.

(i) 3.3223 (ii) 6.6445 (iii) 2.3223 (iv) -1 (v) 2

EXPONENTIAL FUNCTIONS.

Draw a graph of y = a^x. Choose a value for a, and then plot x & y.

Your graph should have the following properties, regardless of what you chose for a.

It should pass through the point (0, 1)

As x à , y à ,

x à -, y à 0

y cannot be negative or zero i.e. y > 0

Domain; all real x. Range; y > 0

Graph of y = a^x

Draw the graph of y = a^-x

Graph of y = a^-x

Note that the graph is the reflection (mirror image) of y = a^x

Domain; all real x. Range; y > 0

On the same axes, draw graphs of y = 2^x (solid line) and y = 4^x (dotted line)

Graph of y = 2^x ______________

Graph of y = 4^x …………………

EXPONENTIALS

Q.1. Evaluate (i) log₅ 25 (ii) log₃ 27 (iii) log₂ 64

Q.2. Write the following as a logarithm of a single expression.

(i) 2 log a – 4 log b – 3 log c (ii) log k² – log m⁴ + 3 log n²

Q.3. Evaluate (i) (ii)

Q.4. On the same axes, draw the graphs of y = 2^x and y = 10^x.

Q.5. On the same axes, draw the graphs of y = log₁₀ x and y = log₂ x.

What is the domain of these functions?

Q.6. Simplify (i) (3 ^log3⁴) (ii) (8 ^log8⁷)

Q.7. Differentiate (i) y = e^2x (ii) y = e^-2x (iii) y = (y = e^{X2 – 3})

(iv) y = (y = e^{X3 – 4})

Q.8. Integrate the following with respect to x.

(i) e^2x (ii) e^-2x (iii) 3e^{3x – 2} (iv) (–3xe^{3 – X2})

Q.9. Differentiate (i) ln (3x – 2) (ii) ln (x² + x)

Q.10. Integrate the following with respect to x.

(i) (ii) (iii)

Q.11. Find the equation of the normal to the curve y = ln(2x³ – 2) at the point where

x = 2.

Q.12. Differentiate (i) y = 4^x (ii) y = 6^x (iii) y = log₅ x

ANSWERS:

Q.1. (i) 2 (ii) 3 (iii) 6 Q.2. (i) log (a² / b⁴c³) (ii) log (k²n⁶ / m⁴)

Q.3. (i) 2/3 (ii) 2

Q.6. (i) 4 (ii) 7 Q.7. (i) 2e^2x (ii) –2e^-2x (iii) 2xe^{(x2 – 3)} (iv) 3x² e ^{(x3 – 4)}

Q.8. (i) ½ e^2x + c (ii) – ½ e^-2x + c (iii) e^{(3x – 2)} + c (iv)+ c i.e.(e^{(3 – x2)}) + c

Q.9. (i) (ii)

Q.10. (i) ln (x² + 4) (ii) ln (2x² + 3x) (iii) ln (2x³ + 4x)

Q.12. (i) ln4. 4^x (ii) ln6. 6^x (iii) 1 / (x ln5)

GRADIENT OF y = a^x

Gradient at the point (0, 1)

Since all curves of the form y = a^x pass through the point (0,1), it is useful to be able to determine the gradient of such a curve at (0,1)

Suppose P is the point (0,1) and Q is a point on the curve near P.

Complete the triangle PQR so that QR is the rise and PR is the run.

The gradient is =

Since R is 1 unit from the x-axis, QR = a^Dx -1 = Dy

= =

And is the gradient at (0,1)

GRADIENT OF EXPONENTIALS

Consider the graph of the function y = a^x Let P (x, y) and Q(x + Dx, y + Dy) be two points on the graph.

Since y = a^x and y + Dy = a ^{x + Dx}

Then y + Dy - y = a ^{x + Dx} – a^x

and Dy = a ^{x + Dx} – a^x and by the index laws

Dy = a ^x. a ^Dx – a^x = a^x(a^Dx -1)

=

=

=

= a^x

But is the gradient , m, at the point (0, 1)

So = a^x m = m a^x where m is the gradient at (0,1)

Exercise: Using the formula; = and taking Dx as 0.000001, find the gradients of (i) y = 2^x, (ii) y = 3^x and (iii) y = 2.5^x, at the point (0,1).

Answers: (i) 0.693 (ii) 1.099 (iii) 0.916

WHAT IS “e”?

By considering the equation y = a^x and using the formula = , it is possible to find a value for a such that is very close to 1. This value turns out to be about 2.7183 and is known as “e” (it is irrational so does not have an exact decimal value).

y = e^x is called the exponential function and has the property that = e^x.

The syntax of calculators varies but to find “e” press inv ln 1 = or 1 inv ln =.

“inv” is replaced by “shift” on some calculators.

DIFFERENTIATING EXPONENTIAL FUNCTIONS

If y = e^x then = e^x. If y = e^f(x) then = f ’(x) e^f(x).

Example: Differentiate e^3x+4 with respect to x.

Answer: Apply the chain rule;

Let y = e^u where u = 3x + 4

= e^u = e^3x+4. 3 = 3e^3x+4

Exercise 1: Differentiate the following with respect to x:

(i) e^5x (ii) e^7x (iii) e^-3x (iv) e^5x-6 (v) e^5x – 6 (vi) e^6x-3 (vii) e^7-2x (viii) e^3-5x

(ix) 5x² – e^5x (x) 5x² – 10x + 4 + e^4x

Exercise 2: Differentiate the following with respect to x:

(i) 6e^5x (ii) 12e^-6x (iii) (iv) (v) (vi) (vii) (viii)

(ix) (x) e^x + e^-x

Exercise 3: What is the point of intersection of the curves y = e^x and y = e^2x? What is the gradient of each of these curves at the point of intersection?

Answers:

Exercise 1: (i) 5e^5x (ii) 7e^7x (iii) -3e^-3x (iv) 5e^5x-6 (v) 5e^5x (vi) 6e^6x-3 (vii) -2e^7-2x

(viii) -5e^3-5x (ix) 10x – 5e^-5x (x) 10x – 10 + 4e^4x

Exercise 2: (i) 30e^5x (ii) -72e^-6x (iii) (iv) (10x – 6) (v) 6x² (vi) 20x³ (vii) ½ x (viii)– (ix) (x) e^x - e^-x

Exercise 3: intersection (0,1) (all exponentials intersect at (0,1)

Gradient of y = e^x is e^x. At (0,1) gradient = e⁰ = 1

Gradient of y = e^2x is 2e^2x. At (0,1) gradient = 2e⁰ = 2

INTEGRATING EXPONENTIAL FUNCTIONS

Since integration is the reverse of differentiation, the following properties apply:

= e^x + c

= e^f(x) + c

Example: Find

Answer. f(x) = 5x – 4 so f ’(x) = 5

= =

=

Example: Find

Answer. f(x) = 2x + 4 so f’(x) = 2

= =

= ½ (e⁶ – e⁴)

Exercise 1: Integrate the following with respect to x:

(i) e^2x (ii) e^-3x (iii) e^{4x - 2} (iv) e^{3 - 2x} (v) e^{7 - 5x} (vi) (vii)

(viii) (ix) (x)

Exercise 2: Evaluate the following integrals between the values shown:

(i) (ii) (iii) (iv) (v) (vi) (vii)

(viii) (ix) (x)

Exercise 3.

(i) Find the equation of the tangent to the curve y = e^x+1 at the point where x = 1.

(ii) Find the area bounded by the curve y = e^x+1 and the x-axis between the values x = 1 and x = 2.

Answers:

Exercise 1: (i) ½ e^2x + c (ii) + c (iii) + c (iv) + c

(v) + c (vi) + c (vii) + c (viii) + c

(ix) + c (x) + c

Exercise 2: (i) ½ (e²-1) (ii) (iii) (iv) (v)

or (vi) (vii) (viii) 2e⁴ (ix) (x) 0

Exercise 3: (i) y = e⁴x + e² (ii) e³ – e²

DERIVATIVES OF LOG FUNCTIONS

While differentiating most log functions is difficult and beyond the scope of the HSC course, differentiating functions involving log_e is fairly straight forward.

Log_e can also be written as ln and since this is the notation used on calculators it is the notation that will be used here.

If y = ln f(x) then =

Example 1: Differentiate y = ln (x²+3)

Answer: = =

An important result is ln x =

Example 2: Differentiate y = ln[(x + 2) (2x – 3)] by (i) expanding and

(ii) by the rule log ab = log a + log b

Answer : (i) y = ln[(x + 2) (2x – 3)] = ln(2x² + x – 6)

= =

(ii) y = ln[(x + 2) (2x – 3)] = ln (x + 2) + ln (2x – 3)

=

= =

=

Exercise 1: Differentiate the following with respect to x:

(i) ln (x + 4) (ii) ln (2x – 5) (iii) ln (3 – 7x) (iv) ln (8 – 5x) (v) ln (x² -9)

(vi) ln (18 – 7x²) (vii) ln (1 – x) (viii) ln (5x² + 17) (ix) ln (7x³ – 27)

(x) ln (13 + 2x + 7x² – 5x³)

Exercise 2: Differentiate the following with respect to x.

(i) ln (x - 3)³ (ii) ln (x² – 8)² (iii) ln (7 – x³)⁴ (iv) ln (8 – 5x²)⁵ (v) ln

(vi) ln- (vii) ln (viii) ln [(5x + 4) (3 – 7x) (ix) ln (x) ln

Answers: Exercise 1: (i) (ii) (iii) (iv) (v)

(vi) (vii) (viii) (ix) (x)

Exercise 2: (i) = (ii) = (iii) =

(iv) (v) (vi) (vii) (viii) - (ix)- (x) -

DERIVATIVES OF y = a^x

Suppose a = e^m

Then from the definition of logarithms, m = log_ea

(Remember: The logarithm of a number to a base is the power to which the base must be raised to get the number. Log_ba = c means a = b^c)

m = log_ea = ln a a = e^m so a = e^{ln a} *This is important

y = a^x

(a^x) = e^{ln a x} (remember that ln a is a constant)

So e^{ln a x} = ln a e^{ln a x} = ln a. a^x

Example 1. Find if y = 10^x

Answer 1. (a^x) = ln a. a^x

10^x = ln 10. 10^x

Example 2. Find if y = 4^2x

Answer 2. Let 2x = u then y = 4^u

Then by the chain rule; = .

= .

= ln u. 4^u .

= ln 2x. 4^2x .

= ln 2x. 4^2x. 2

= 2 ln 2x.4^2x

Exercise 1: Find the derivatives of the following with respect to x:

(i) y = 3^x (ii) y = 4^x (iii) y = -3^x (iv) y = 2^2x (v) y = 3^4x (vi) y = 8^x (vii) y = 2^3x (viii) y = -5^4x (ix) y = 6^4x (x) y = 4^3x+4

Exercise 2: Find the derivatives of the following with respect to x:

(i) 3^8x+3 (ii) 4^2x-5 (iii) 6^7-3x (iv) 5^3-2x (v) 5^-x (vi) -5^x (vii) (-5)^x (viii) 4^-2x (ix) 2^-4x (x) 8^5-2x

Answers:

Exercise 1. (i) ln3.3^x (ii) ln4.4^x (iii) –ln3.3^x (iv) 2ln2.2^2x (v) 4ln3.3^4x

(vi) ln8.8^x = 3ln2.2^3x (vii) 3ln2.2^3x (viii) -4ln5.5^4x (ix) 4ln6.6^4x (x) 3ln4.4^3x+4

Exercise 2. (i) 8ln3.3^8x+3 (ii) 2ln4.4^2x-5 = 4ln2.2^4x-10 (iii) -3ln6.6^7-3x (iv) -2ln5.5^3-2x (v) –ln5.5^-x (vi) –ln5.5^x (vii) Does not exist since ln(-5) does not exist.

(viii) -2ln4.4^-2x = -4ln2.2^-4x (ix) -4ln2.2^-4x (x) -2ln8.8^5x-2 = -6ln2.2^15x-6

DERIVATIVE OF y = log_ax

y = log_ax

log_ax = and since ln a is a constant

= ln x =

log_ax =

Example 1. Find log₁₀ x

Answer 1. =

Example 2. Find log₂ (4x+3)

Answer 2. Let 4x + 3 = u

= .

= ln₂ u.

=

=

Exercise: Differentiate the following with respect to x:

(i) y = log₄ x (ii) y = log₁₀ 100x (iii) y = log₁₀ (2x + 3) (iv) y = log₁₀ (7x – 2)

(v) y = log₂ 8x (vi) y = log₂ (4x + 6) (vii) y = log₃ (8x – 4) (viii) y = log₁₀ x²

(ix) y = log₁₀ (8x² – 3x) (x) y = log_e (8x² – 3x) (xi) y = log₈ 24

(xii) y = log₆ (3x² + 4) (xiii) y = log₁₀ x³ (xiv) y = ln (7x² – 8x)

Answers.

Exercise 1. (i) (ii) (iii) (iv) (v)

(vi) (vii) (viii) (ix) (x)

(xi) 0 (xii) (xiii) (xiv)

Graphs of Log Functions

1. On the same axes draw graphs of y = ln3x and y = ln4x.

Find the gradient function of y = ln3x and y = ln4x and hence show that their gradients are equal for all values of x.

Answer.

Gradient function of y = ln f(x) =

Gradient function of y = ln3x = =

Gradient function of y = ln4x = =

Since gradient function of y = ln3x = Gradient function of y = ln4x = the gradients are the same for all values of x.

Draw graphs of the following:

(i) y = 2^x (ii) y = 3^x (iii) y = 2^-x (iv) y = 3^-x (v) y = 1^x

Example 2. Find if y = 3^2x+2

Answer 2. Let 2x + 2 = u then y = 3^u

Then by the chain rule; = .

=.

= ln u. 3^u .

= ln 2x + 2. 3^2x+2 .

= ln 2x+2. 3^2x+2. 2

= 2 ln 2x+2.3^2x+2