Pure water has a pH of 7. Lemon juice has a pH of 2. This seemingly small difference means lemon juice is 100,000 times more acidic than pure water. The pH scale is logarithmic — each whole number change represents a tenfold change in hydrogen ion concentration. Logarithms are not just mathematical tools; they are how nature compresses enormous ranges into manageable scales.
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$y = 2^x$ and $y = \log_2(x)$ are inverse functions. If you sketch $y = 2^x$, then reflect it across the line $y = x$, you get the graph of $y = \log_2(x)$.
Before reading on, predict:
Core Content
A logarithm is an exponent. Specifically, $\log_a(x)$ answers the question:
"To what power must I raise $a$ to get $x$?"
This gives the fundamental equivalence:
$$y = \log_a(x) \iff x = a^y$$
The logarithmic function $y = \log_a(x)$ is the inverse function of the exponential function $y = a^x$. This means they "undo" each other:
$$a^{\log_a(x)} = x \quad \text{and} \quad \log_a(a^x) = x$$
The domain of $y = \log_a(x)$ is $x > 0$. This is because the range of $y = a^x$ is $y > 0$, and the domain of an inverse function equals the range of the original function. There is no real power of $a$ that gives zero or a negative number, so $\log_a(0)$ and $\log_a(\text{negative})$ are undefined.
The graph of $y = \log_a(x)$ is the reflection of $y = a^x$ across the line $y = x$. This gives these key features:
Show y = ln(x) in the first quadrant passing through (1, 0), (e≈2.7, 1), (7.4, 2). Draw the vertical asymptote x = 0 as a dashed line. Show y = log_10(x) on the same axes, passing through (1, 0), (10, 1), (100, 2) — shallower than ln(x). On the right side, show y = e^x passing through (0, 1), (1, e), (2, e^2) and y = ln(x) as its reflection across y = x (dashed diagonal). Label all curves and asymptotes.
Evaluate without a calculator: (a) $\log_2(32)$ (b) $\log_3(\frac{1}{9})$ (c) $\ln(e^5)$
The value of each expression.
(a) 2^5 = 32, so log_2(32) = 5.
(b) 3^(-2) = 1/9, so log_3(1/9) = -2.
(c) ln(e^5) = 5·ln(e) = 5·1 = 5.
(a) 5 (b) -2 (c) 5
Evaluate: (a) $\log_5(125)$ (b) $\log_{10}(0.001)$ (c) $\log_4(2)$
Answer:
(a) $5^3 = 125$, so $\log_5(125) = 3$
(b) $10^{-3} = 0.001$, so $\log_{10}(0.001) = -3$
(c) $4^{1/2} = 2$, so $\log_4(2) = \frac{1}{2}$
To evaluate $\log_a(b)$ without a calculator, ask: "What power of $a$ gives $b$?"
| Expression | Question | Answer |
|---|---|---|
| $\log_2(8)$ | $2^{?} = 8$ | 3 |
| $\log_{10}(1000)$ | $10^{?} = 1000$ | 3 |
| $\log_3(\frac{1}{27})$ | $3^{?} = \frac{1}{27}$ | -3 |
| $\ln(1)$ | $e^{?} = 1$ | 0 |
| $\ln(e)$ | $e^{?} = e$ | 1 |
$y = \log_a(x) \iff x = a^y$
Domain: $x > 0$
Range: all real numbers
$\log_a(a) = 1$
$\log_a(1) = 0$
$\ln x = \log_e(x)$
$\log x = \log_{10}(x)$
For $y = \log_2(x)$: the $x$-intercept is $(1, 0)$ because $\log_2(1) = 0$. As $x \to 0^+$, $y \to -\infty$ — the $y$-axis is a vertical asymptote. There is no $y$-intercept because $x = 0$ is not in the domain.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. For $f(x) = \log_2(x)$: (a) State the domain and range. (b) Find $f(8)$ and $f(\frac{1}{4})$. (c) Solve $f(x) = 5$. 3 MARKS
9. The pH of a solution is $\text{pH} = -\log_{10}[\text{H}^+]$. (a) Find the pH when $[\text{H}^+] = 10^{-5}$. (b) Find $[\text{H}^+]$ when pH = 2. (c) If one solution has pH 3 and another has pH 6, how many times more acidic is the first? 3 MARKS
10. Explain why $\log_a(x)$ is only defined for $x > 0$, using both the definition of logarithm as an exponent and the graph of $y = a^x$. Your answer should reference the range of $y = a^x$ and explain why this creates a domain restriction for its inverse. 3 MARKS
1. (a) $\log_2(64) = 6$ (b) $\log_5(1) = 0$ (c) $\log_{10}(0.01) = -2$ (d) $\ln(e^3) = 3$
2. (a) $x = 3^4 = 81$ (b) $x^2 = 16 \Rightarrow x = 4$ (c) $x = e^2 \approx 7.389$
3. $[\text{H}^+] = 10^{-4}$ mol/L
1. $\log_a(0)$ asks "$a$ to what power equals 0?" But $a^y > 0$ for all real $y$. So no such power exists.
2. Difference = 3 pH units. Ratio of acidity = $10^3 = 1000$ times.
3. $y = \log_2(x)$ is increasing; $y = \log_{1/2}(x)$ is decreasing. They are reflections in the $x$-axis because $\log_{1/2}(x) = -\log_2(x)$.
Q8 (3 marks): (a) Domain: $x > 0$; Range: all real numbers [1]. (b) $f(8) = \log_2(8) = 3$; $f(\frac{1}{4}) = \log_2(2^{-2}) = -2$ [1]. (c) $\log_2(x) = 5 \Rightarrow x = 2^5 = 32$ [1].
Q9 (3 marks): (a) pH = $-\log_{10}(10^{-5}) = -(-5) = 5$ [1]. (b) $-\log_{10}[\text{H}^+] = 2 \Rightarrow [\text{H}^+] = 10^{-2}$ [1]. (c) Difference = 3 units, so $10^3 = 1000$ times more acidic [1].
Q10 (3 marks): By definition, $y = \log_a(x)$ means $a^y = x$ [1]. The exponential function $y = a^x$ has range $y > 0$ — it only outputs positive numbers. Since the logarithm is the inverse of the exponential, the domain of the logarithm equals the range of the exponential: $x > 0$ [1]. There is no real number $y$ such that $a^y = 0$ or $a^y = \text{negative}$, so $\log_a(0)$ and $\log_a(\text{negative})$ are undefined. Graphically, the vertical asymptote at $x = 0$ reflects the fact that the horizontal asymptote $y = 0$ of $y = a^x$ becomes a boundary that cannot be crossed [1].
Climb platforms using your knowledge of logarithmic functions, domain, range, and pH calculations. Pool: lesson 5.
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