MAT pure2&3 logarithms and exponentials

Question 1

What is an exponential function?

Answer 1

An exponential function is of the form y = a^x. Examples include y = 2^x and y = 3^x.

Question 2

What is exponential growth and its general form?

Answer 2

Exponential growth occurs when something increases at an ever-increasing rate.
General form: y = c * a^(kt), where c > 0 and k > 0.

Question 3

What is exponential decay and its general form?

Answer 3

Exponential decay occurs when something decreases at a rate that slows over time.
General form: y = c * a^(-kt), where c > 0 and k > 0.

Question 4

What are logarithms and how are they related to exponentials?

Answer 4

Logarithms are the inverse of exponentials.
If a^x = b, then log_a(b) = x.

Question 5

Example: Rewrite 3^2 = 9 using logarithms.

Answer 5

log_3(9) = 2

Question 6

What are the three laws of logarithms?

Answer 6

1. log(xy) = log(x) + log(y)
2. log(x/y) = log(x) - log(y)
3. log(x^n) = n * log(x)

Question 7

Example: Simplify log(3x^2 / y) using the laws of logarithms.

Answer 7

log(3x^2 / y) = log(3) + 2log(x) - log(y)

Question 8

What are two special cases of logarithms?

Answer 8

1. log_a(a) = 1
2. log_a(1) = 0

Question 9

Example: Evaluate log_2(8).

Answer 9

log_2(8) = 3 because 2^3 = 8.

Question 10

How do you solve exponential equations using logarithms?

Answer 10

Take the logarithm of both sides and apply the laws of logarithms.
Example: Solve 2^x = 10.
log(2^x) = log(10)
x * log(2) = log(10)
x = log(10) / log(2)

Question 11

What is an old practical application of logarithms?

Answer 11

Before calculators, logarithms were used for complex calculations like division:
log(a/b) = log(a) - log(b)
Inverse logarithms were used to find the result.

Question 12

Divide 1432627 by 967253 using logarithms.

Answer 12

log(1432627) ≈ 6.1562
log(967253) ≈ 5.9855
log(1432627 / 967253) = 6.1562 - 5.9855 = 0.1707
Inverse log(0.1707) ≈ 1.481
Result: 1432627 / 967253 ≈ 1.481

Question 13

What is the purpose of logarithmic modelling?

Answer 13

To find relationships between variables, especially when data does not produce a straight-line graph in its original form.

Question 14

How do you model curves involving powers?

Answer 14

Use the form y = kx^n. Taking logs of both sides:
log(y) = n * log(x) + log(k)
Plot log(y) against log(x) to get a straight line with gradient n and intercept log(k).

Question 15

Example: How do you verify that y = kx^n fits data and find constants k and n?

Answer 15

1. Take logs: log(y) = n * log(x) + log(k).
2. Plot log(y) against log(x).
3. Gradient of the line gives n.
4. Intercept of the line gives log(k); k = 10^(intercept).

Question 16

How do you model curves involving exponentials?

Answer 16

Use the form y = k * a^x. Taking logs of both sides:
log(y) = x * log(a) + log(k)
Plot log(y) against x to get a straight line with gradient log(a) and intercept log(k).

Question 17

Example: How do you verify that y = k * a^x fits data and find constants k and a?

Answer 17

1. Take logs: log(y) = x * log(a) + log(k).
2. Plot log(y) against x.
3. Gradient of the line gives log(a); a = 10^(gradient).
4. Intercept of the line gives log(k); k = 10^(intercept).

Question 18

Example: The relationship is y = kx^n. Data points are:
x = {1, 2, 3, 4}, y = {2, 4, 8, 16}. Find k and n.

Answer 18

1. Take logs: log(y) = n * log(x) + log(k).
2. Plot log(y) = {0.3, 0.6, 0.9, 1.2}, log(x) = {0, 0.3, 0.48, 0.6}.
3. Gradient = n ≈ 2.
4. Intercept = log(k) ≈ 0.3; k ≈ 10^(0.3) = 2.

Question 19

Example: The relationship is y = k * a^x. Data points are:
x = {1.5, 2.0, 2.5, 3.0}, y = {12, 19, 30, 46}. Find k and a.

Answer 19

1. Take logs: log(y) = x * log(a) + log(k).
2. Plot log(y) = {1.08, 1.28, 1.48, 1.66}, x = {1.5, 2.0, 2.5, 3.0}.
3. Gradient = log(a) ≈ 0.5; a ≈ 10^(0.5) = 3.2.
4. Intercept = log(k) ≈ 1.25; k ≈ 10^(1.25) = 18.

Question 20

How do you determine if a model is appropriate for data?

Answer 20

Plot the transformed variables (e.g., log(y) vs. log(x)). If the plot is approximately a straight line, the model is appropriate.

Question 21

What should you do if the graph is not a straight line?

Answer 21

The suggested model is either incorrect, or the experimental results are not sufficiently accurate.

Question 22

What is the general process for applying logarithmic modelling?

Answer 22

1. Take logs of the suggested relationship.
2. Apply the laws of logarithms to create a straight-line form y = mx + c.
3. Plot the transformed variables.
4. Use the gradient and intercept to find constants.

Question 23

What are natural logarithms?

Answer 23

Natural logarithms are logarithms to base e, where e ≈ 2.718. They obey the same rules as other logarithms and are denoted as ln(x).

Question 24

What is the relationship between ln(e) and e^x?

Answer 24

ln(e) = 1 and e^(ln(x)) = x, showing that natural logarithms and exponential functions are inverses of each other.

Question 25

What is special about the exponential function e^x?

Answer 25

The gradient of the graph of y = e^x is equal to the value of e^x at any point.

Question 26

Example: Solve 3e^(x-1) = 5.

Answer 26

1. Divide both sides by 3: e^(x-1) = 5/3.
2. Take ln of both sides: x - 1 = ln(5/3).
3. Solve for x: x = ln(5/3) + 1 ≈ 1.51 (3 s.f.).

Question 27

Example: Solve ln(2x + 1) = 3.

Answer 27

1. Exponentiate both sides: 2x + 1 = e^3.
2. Subtract 1: 2x = e^3 - 1.
3. Solve for x: x = (e^3 - 1)/2 ≈ 9.54 (3 s.f.).

Question 28

What is the general form of an exponential model?

Answer 28

Exponential growth: y = ce^(kt).
Exponential decay: y = ce^(-kt), where c and k are constants.

Question 29

Example: The temperature of a coffee cup is given by T = 20 + 60e^(-0.1t). Find:
a) Initial temperature.
b) Temperature after 5 minutes
c) Time for T = 25.

Answer 29

a) Initial temperature: T(0) = 20 + 60 = 80°C.
b) T(5) = 20 + 60e^(-0.5) ≈ 56.4°C.
c) Solve 25 = 20 + 60e^(-0.1t):
e^(-0.1t) = 5/60,
-0.1t = ln(5/60),
t ≈ 24.8 minutes.

Question 30

What happens to exponential models for large values of t?

Answer 30

Exponential growth models predict continuous increase, while exponential decay models approach a limiting value, but neither perfectly fits real-life scenarios for very large t.

Question 31

What is the derivative of y = e^x?

Answer 31

dy/dx = e^x. If y = e^(kx), then dy/dx = k * e^(kx)

Question 32

Why is e^x often used for modelling real-life situations?

Answer 32

Its rate of change (dy/dx) is proportional to its value, making it suitable for problems like population growth or radioactive decay.