A whisper at 30 decibels. A conversation at 60 decibels. A jet engine at 120 decibels. Each 10 dB increase sounds roughly twice as loud, but represents a tenfold increase in sound intensity. The decibel scale is logarithmic — without it, we'd need numbers from 1 to 1,000,000,000,000 to describe the range of sounds humans can hear.
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The Richter scale for earthquakes is logarithmic (base 10). A magnitude 6 earthquake is:
Make a prediction before reading on.
Core Content
Since $y = \log_a(x)$ is the inverse of $y = a^x$, every feature of the exponential graph has a corresponding feature in the log
| Exponential $y = a^x$ | Logarithmic $y = \log_a(x)$ |
|---|---|
| Domain: all real numbers | Range: all real numbers |
| Range: $y > 0$ | Domain: $x > 0$ |
| $y$-intercept: $(0, 1)$ | $x$-intercept: $(1, 0)$ |
| Horizontal asymptote: $y = 0$ | Vertical asymptote: $x = 0$ |
| Point $(1, a)$ | Point $(a, 1)$ |
Show y = e^x in the first quadrant passing through (0,1), (1,e≈2.7), (2,7.4). Draw y = x as a dashed diagonal line. Show y = ln(x) as the reflection of y = e^x across y = x, passing through (1,0), (e≈2.7, 1), (7.4, 2). Label the vertical asymptote x = 0 for ln(x) and the horizontal asymptote y = 0 for e^x. Mark corresponding points and draw small perpendicular connectors showing the reflection.
For logarithmic functions ($a > 1$):
This is the opposite of exponential functions, where larger bases give steeper graphs. It makes sense: if $a^x$ grows faster, then $\log_a(x)$ must grow slower to be its inverse.
All log graphs pass through $(1, 0)$. For $x > 1$, the graph with the smaller base is higher. For $0 < x < 1$, the graph with the larger base is higher.
Sketch $y = \log_2(x - 3) + 1$, showing all key features.
The vertical asymptote, $x$-intercept, and one other point.
Start with y = log_2(x): asymptote x=0, passes through (1,0) and (2,1).
Shift right 3: y = log_2(x-3).
Asymptote: x = 3.
Shift up 1: y = log_2(x-3) + 1.
New x-intercept: log_2(x-3) = -1 => x-3 = 0.5 => x = 3.5.
Point: when x = 5, y = log_2(2) + 1 = 1 + 1 = 2, so (5, 2).
Asymptote: $x = 3$.
$x$-intercept: $(3.5, 0)$.
Point: $(5, 2)$.
Sketch $y = \ln(x + 2) - 1$, labelling the asymptote, $x$-intercept, and the point where $x = -1$.
Answer:
Asymptote: $x = -2$. $x$-intercept: $\ln(x+2) = 1 \Rightarrow x+2 = e \Rightarrow x = e - 2 \approx 0.718$. At $x = -1$: $y = \ln(1) - 1 = 0 - 1 = -1$, so $(-1, -1)$.
All standard transformations apply to logarithmic functions:
The vertical asymptote is particularly important: for $y = \log_a(x - h) + k$, the asymptote is the vertical line $x = h$. This is where the argument of the logarithm becomes zero.
$y = \log_a(x)$ reflects $y = a^x$ across $y = x$
$x$-intercept: $(1, 0)$
Asymptote: $x = 0$
$(a, 1)$ since $\log_a(a) = 1$
$y = \log_a(x-h)+k$: asymptote $x = h$
A magnitude 6 earthquake is $10^{6-4} = 10^2 = 100$ times as strong as a magnitude 4 earthquake. The Richter scale is base 10 logarithmic, so each whole number increase represents a tenfold increase in amplitude and roughly a 31.6-fold increase in energy.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Sketch $y = \log_2(x - 3) + 1$, labelling the vertical asymptote, $x$-intercept, and the point where $x = 5$. 3 MARKS
9. On the same set of axes, sketch $y = \ln(x)$ and $y = \log_{10}(x)$. Label both curves, the asymptote, and explain which curve is steeper for $x > 1$. 3 MARKS
10. The sound intensity level in decibels is given by $L = 10 \log_{10}(\frac{I}{I_0})$. A rock concert measures 110 dB and normal conversation measures 60 dB. Calculate how many times more intense the rock concert is than normal conversation. Explain why the decibel scale uses a logarithm. 3 MARKS
1. Asymptote $x = 0$, passes through $(1, 0)$ and $(3, 1)$.
2. Asymptote $x = 2$. $x$-intercept: $\log_2(x-2) = -1 \Rightarrow x-2 = 0.5 \Rightarrow x = 2.5$.
3. Shift left 1, down 2. Asymptote moves to $x = -1$.
1. Since $10 > e$, $\log_{10}(x)$ grows slower than $\ln(x)$. For $x > 1$, smaller bases give steeper log graphs.
2. $y = \ln(-x)$ is the reflection of $y = \ln(x)$ in the $y$-axis. Domain: $x < 0$.
3. Difference = 30 dB. Ratio of intensities = $10^{30/10} = 10^3 = 1000$ times.
Q8 (3 marks): Asymptote: $x = 3$ [1]. $x$-intercept: $\log_2(x-3) = -1 \Rightarrow x-3 = 0.5 \Rightarrow x = 3.5$, so $(3.5, 0)$ [1]. At $x = 5$: $y = \log_2(2) + 1 = 1 + 1 = 2$, so $(5, 2)$ [1].
Q9 (3 marks): Both pass through $(1, 0)$ [0.5]. $y = \ln(x)$ passes through $(e, 1) \approx (2.72, 1)$ [0.5]. $y = \log_{10}(x)$ passes through $(10, 1)$ [0.5]. For $x > 1$, $y = \ln(x)$ is steeper because $e < 10$ — smaller bases produce steeper logarithmic graphs [1]. Both have asymptote $x = 0$ [0.5].
Q10 (3 marks): $110 = 10 \log_{10}(\frac{I_c}{I_0}) \Rightarrow \frac{I_c}{I_0} = 10^{11}$ [0.5]. $60 = 10 \log_{10}(\frac{I_n}{I_0}) \Rightarrow \frac{I_n}{I_0} = 10^6$ [0.5]. Ratio: $\frac{I_c}{I_n} = \frac{10^{11}}{10^6} = 10^5 = 100{,}000$ times more intense [1]. The decibel scale uses logarithms because the range of human hearing spans an enormous range of intensities (from $10^{-12}$ W/m² to $10^0$ W/m² — a factor of $10^{12}$) [0.5]. A logarithmic scale compresses this vast range into a manageable 0–120 dB scale that matches human perception, where equal steps sound roughly equally different [0.5].
Climb platforms using your knowledge of logarithmic graphs, transformations, and inverse relationships. Pool: lesson 6.
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