MAT pure2&3 logarithms and exponentials Flashcards
What is an exponential function?
An exponential function is of the form y = a^x. Examples include y = 2^x and y = 3^x.
What is exponential growth and its general form?
Exponential growth occurs when something increases at an ever-increasing rate.
General form: y = c * a^(kt), where c > 0 and k > 0.
What is exponential decay and its general form?
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.
What are logarithms and how are they related to exponentials?
Logarithms are the inverse of exponentials.
If a^x = b, then log_a(b) = x.
Example: Rewrite 3^2 = 9 using logarithms.
log_3(9) = 2
What are the three laws of logarithms?
- log(xy) = log(x) + log(y)
- log(x/y) = log(x) - log(y)
- log(x^n) = n * log(x)
Example: Simplify log(3x^2 / y) using the laws of logarithms.
log(3x^2 / y) = log(3) + 2log(x) - log(y)
What are two special cases of logarithms?
- log_a(a) = 1
- log_a(1) = 0
Example: Evaluate log_2(8).
log_2(8) = 3 because 2^3 = 8.
How do you solve exponential equations using logarithms?
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)
What is an old practical application of logarithms?
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.
Divide 1432627 by 967253 using logarithms.
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
What is the purpose of logarithmic modelling?
To find relationships between variables, especially when data does not produce a straight-line graph in its original form.
How do you model curves involving powers?
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).
Example: How do you verify that y = kx^n fits data and find constants k and n?
- Take logs: log(y) = n * log(x) + log(k).
- Plot log(y) against log(x).
- Gradient of the line gives n.
- Intercept of the line gives log(k); k = 10^(intercept).
How do you model curves involving exponentials?
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).
Example: How do you verify that y = k * a^x fits data and find constants k and a?
- Take logs: log(y) = x * log(a) + log(k).
- Plot log(y) against x.
- Gradient of the line gives log(a); a = 10^(gradient).
- Intercept of the line gives log(k); k = 10^(intercept).
Example: The relationship is y = kx^n. Data points are:
x = {1, 2, 3, 4}, y = {2, 4, 8, 16}. Find k and n.
- Take logs: log(y) = n * log(x) + log(k).
- Plot log(y) = {0.3, 0.6, 0.9, 1.2}, log(x) = {0, 0.3, 0.48, 0.6}.
- Gradient = n ≈ 2.
- Intercept = log(k) ≈ 0.3; k ≈ 10^(0.3) = 2.
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.
- Take logs: log(y) = x * log(a) + log(k).
- Plot log(y) = {1.08, 1.28, 1.48, 1.66}, x = {1.5, 2.0, 2.5, 3.0}.
- Gradient = log(a) ≈ 0.5; a ≈ 10^(0.5) = 3.2.
- Intercept = log(k) ≈ 1.25; k ≈ 10^(1.25) = 18.
How do you determine if a model is appropriate for data?
Plot the transformed variables (e.g., log(y) vs. log(x)). If the plot is approximately a straight line, the model is appropriate.
What should you do if the graph is not a straight line?
The suggested model is either incorrect, or the experimental results are not sufficiently accurate.
What is the general process for applying logarithmic modelling?
- Take logs of the suggested relationship.
- Apply the laws of logarithms to create a straight-line form y = mx + c.
- Plot the transformed variables.
- Use the gradient and intercept to find constants.
What are natural logarithms?
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).
What is the relationship between ln(e) and e^x?
ln(e) = 1 and e^(ln(x)) = x, showing that natural logarithms and exponential functions are inverses of each other.
What is special about the exponential function e^x?
The gradient of the graph of y = e^x is equal to the value of e^x at any point.
Example: Solve 3e^(x-1) = 5.
- Divide both sides by 3: e^(x-1) = 5/3.
- Take ln of both sides: x - 1 = ln(5/3).
- Solve for x: x = ln(5/3) + 1 ≈ 1.51 (3 s.f.).
Example: Solve ln(2x + 1) = 3.
- Exponentiate both sides: 2x + 1 = e^3.
- Subtract 1: 2x = e^3 - 1.
- Solve for x: x = (e^3 - 1)/2 ≈ 9.54 (3 s.f.).
What is the general form of an exponential model?
Exponential growth: y = ce^(kt).
Exponential decay: y = ce^(-kt), where c and k are constants.
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.
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.
What happens to exponential models for large values of t?
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.
What is the derivative of y = e^x?
dy/dx = e^x. If y = e^(kx), then dy/dx = k * e^(kx)
Why is e^x often used for modelling real-life situations?
Its rate of change (dy/dx) is proportional to its value, making it suitable for problems like population growth or radioactive decay.
