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Functions - Logarithmic functions (HL)

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A logarithmic function y=log⁡a(x)y = \log_a(x) is the inverse of an exponential function y=axy = a^x. Its domain is (0,∞)(0, \infty) and its range is (−∞,∞)(-\infty, \infty). The graph always passes through (1,0)(1, 0) and has a vertical asymptote at x=0x = 0.

Graph of a logarithmic function showing the vertical asymptote at x=0 and the x-intercept at (1,0).
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The base of a logarithm determines the steepness of the curve. Natural logarithms use base ee (2.718...2.718...) and are written as ln⁡(x)\ln(x). Common logarithms use base 1010 and are written as log⁡(x)\log(x).

Graph comparing the natural log function ln(x).
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Logarithmic scales are used to linearize power-law relationships. If y=axny = ax^n, then log⁡(y)=log⁡(a)+nlog⁡(x)\log(y) = \log(a) + n\log(x). Plotting log⁡(y)\log(y) against log⁡(x)\log(x) results in a straight line with gradient nn and intercept log⁡(a)\log(a).

Linearized log-log plot showing a straight line representing a power relationship.
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Transformations of logarithmic functions y=alog⁡b(x−c)+dy = a\log_b(x - c) + d involve vertical stretching/compression (aa), horizontal translation (cc), and vertical translation (dd). The vertical asymptote shifts to x=cx = c.

📐Formulae

log⁡a(m×n)=log⁡a(m)+log⁡a(n)\log_a(m \times n) = \log_a(m) + \log_a(n)

log⁡a(mn)=log⁡a(m)−log⁡a(n)\log_a\left(\frac{m}{n}\right) = \log_a(m) - \log_a(n)

log⁡a(mk)=klog⁡a(m)\log_a(m^k) = k \log_a(m)

log⁡a(x)=log⁡b(x)log⁡b(a)\log_a(x) = \frac{\log_b(x)}{\log_b(a)}

log⁡a(1)=0\log_a(1) = 0

log⁡a(a)=1\log_a(a) = 1

alog⁡a(x)=xa^{\log_a(x)} = x

💡Examples

Problem 1:

Solve the equation 4x−1=154^{x-1} = 15 for xx. Give your answer to 3 significant figures.

Solution:

log⁡(4x−1)=log⁡(15)\log(4^{x-1}) = \log(15) (x−1)log⁡(4)=log⁡(15)(x-1)\log(4) = \log(15) x−1=log⁡(15)log⁡(4)x-1 = \frac{\log(15)}{\log(4)} x=log⁡(15)log⁡(4)+1x = \frac{\log(15)}{\log(4)} + 1 x≈1.95+1=2.95x \approx 1.95 + 1 = 2.95

Explanation:

Take the logarithm of both sides to bring the exponent down using the power rule. Then, isolate xx and use a calculator to evaluate the final value.

Problem 2:

A population of bacteria grows according to the model P(t)=500e0.04tP(t) = 500e^{0.04t}, where tt is time in hours. Find the time it takes for the population to reach 2000.

Solution:

2000=500e0.04t2000 = 500e^{0.04t} 4=e0.04t4 = e^{0.04t} ln⁡(4)=ln⁡(e0.04t)\ln(4) = \ln(e^{0.04t}) ln⁡(4)=0.04t\ln(4) = 0.04t t=ln⁡(4)0.04t = \frac{\ln(4)}{0.04} t≈34.7 hourst \approx 34.7 \text{ hours}

Explanation:

Divide both sides by 500 to isolate the exponential term. Take the natural logarithm (ln⁡\ln) of both sides to cancel the base ee, then solve for tt.

Problem 3:

The relationship between two variables is given by y=3x2.5y = 3x^{2.5}. Show how this can be written as a linear equation.

Solution:

log⁡(y)=log⁡(3x2.5)\log(y) = \log(3x^{2.5}) log⁡(y)=log⁡(3)+log⁡(x2.5)\log(y) = \log(3) + \log(x^{2.5}) log⁡(y)=log⁡(3)+2.5log⁡(x)\log(y) = \log(3) + 2.5\log(x) Y=2.5X+C where Y=log⁡(y),X=log⁡(x),C=log⁡(3)Y = 2.5X + C \text{ where } Y = \log(y), X = \log(x), C = \log(3)

Explanation:

By applying the product and power laws of logarithms, a power function is transformed into a linear form Y=mX+cY = mX + c where the gradient is the original exponent.

Problem 4:

Given the function f(x)=ln⁡(x+2)f(x) = \ln(x + 2), find the coordinates where the graph intersects the yy-axis and state the equation of the vertical asymptote.

Graph of y = ln(x+2) showing the vertical asymptote at x=-2.

Solution:

  1. To find the yy-intercept, set x=0x = 0: f(0)=ln⁡(0+2)=ln⁡(2)≈0.693f(0) = \ln(0 + 2) = \ln(2) \approx 0.693 So the yy-intercept is (0,ln⁡(2))(0, \ln(2)).
  2. The vertical asymptote occurs where the argument of the log is zero: x+2=0  ⟹  x=−2x + 2 = 0 \implies x = -2. The equation of the vertical asymptote is x=−2x = -2.

Explanation:

Logarithmic functions shift horizontally based on the value added to or subtracted from xx inside the function. Here, the graph shifts 2 units to the left.

Problem 5:

Data for the variables PP and QQ suggest a relationship of the form P=kQnP = kQ^n. When log⁡10P\log_{10} P is plotted against log⁡10Q\log_{10} Q, a straight line is formed passing through (0,2)(0, 2) and (2,3)(2, 3). Determine the values of kk and nn.

A straight line on log-log axes passing through (0,2) and (2,3).

Solution:

  1. Express the equation in linear form: log⁡10P=log⁡10(kQn)=nlog⁡10Q+log⁡10k\log_{10} P = \log_{10} (kQ^n) = n\log_{10} Q + \log_{10} k.
  2. This is in the form Y=mX+CY = mX + C where Y=log⁡10PY = \log_{10} P, X=log⁡10QX = \log_{10} Q, m=nm = n, and C=log⁡10kC = \log_{10} k.
  3. Calculate the gradient mm: m=3−22−0=12=0.5m = \frac{3 - 2}{2 - 0} = \frac{1}{2} = 0.5. So n=0.5n = 0.5.
  4. Use the yy-intercept CC: C=2  ⟹  log⁡10k=2C = 2 \implies \log_{10} k = 2. k=102=100k = 10^2 = 100. Final values: k=100k = 100, n=0.5n = 0.5.

Explanation:

By applying log laws, a power model can be transformed into a linear model. The gradient of the log-log plot is the exponent nn and the intercept is log⁡k\log k.