Your calculator has buttons for $\ln$ and $\log_{10}$, but not for $\log_2$, $\log_5$, or $\log_7$. How do scientists calculate $\log_2(1000)$ when studying binary search algorithms? How do biologists solve $P = P_0 e^{kt}$ for $t$ when studying population growth? The change of base formula bridges any logarithm to the ones your calculator knows. Combined with logarithmic equations, it gives us the full toolkit for solving real-world exponential problems.
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Your calculator can compute $\ln$ and $\log_{10}$. Can you find $\log_3(20)$ using only these? Make a prediction about what formula might help.
Core Content
Let $y = \log_a(x)$. By definition, $a^y = x$. Take natural log of both sides:
ln(a^y) = ln(x) → y ln(a) = ln(x) → y = ln(x) / ln(a)
Therefore:
log_a(x) = ln(x) / ln(a)
This works for any valid base $b$:
log_a(x) = log_b(x) / log_b(a)
Most common use: $\log_a(x) = \frac{\ln(x)}{\ln(a)}$ or $\frac{\log_{10}(x)}{\log_{10}(a)}$. Both give the same answer.
Example: $\log_3(20) = \frac{\ln(20)}{\ln(3)} \approx \frac{2.996}{1.099} \approx 2.727$.
There are two main types of logarithmic equations:
Convert to exponential form: $x = a^c$. Then check that $x > 0$.
Example: log_2(x) = 5 → x = 2^5 = 32. Check: 32 > 0 ✓
Since log is one-to-one: $M = N$. Then check that $M > 0$ and $N > 0$.
Example: log_3(x+2) = log_3(2x-1) → x+2 = 2x-1 → x = 3. Check: x+2=5>0, 2x-1=5>0 ✓
Sometimes you need to use log laws to combine terms before solving.
Example: log_2(x) + log_2(x-2) = 3 → log_2(x(x-2)) = 3 → x(x-2) = 2^3 = 8 → x^2 - 2x - 8 = 0 → (x-4)(x+2) = 0 → x = 4 or x = -2. Check: x=4: log_2(4)+log_2(2)=2+1=3 ✓. x=-2: log_2(-2) undefined ✗. Final answer: x = 4.
Critical: Always check solutions. Logarithms are only defined for positive arguments.
Solve: $\log_5(x) + \log_5(x - 4) = 2$.
All valid solutions for $x$.
log_5(x) + log_5(x-4) = 2
log_5(x(x-4)) = 2
x(x-4) = 5^2 = 25
x^2 - 4x - 25 = 0
x = (4 ± sqrt(16+100))/2 = (4 ± sqrt(116))/2 = 2 ± sqrt(29)
x ≈ 2 + 5.385 = 7.385 or x ≈ 2 - 5.385 = -3.385
Check x > 4 (need x > 0 and x-4 > 0)
x = 2 + sqrt(29) ≈ 7.385 > 4 ✓
x = 2 - sqrt(29) ≈ -3.385 < 0 ✗
$x = 2 + \sqrt{29} \approx 7.39$.
Solve: $\log_3(x - 1) = 2 + \log_3(2)$.
Answer:
$\log_3(x-1) - \log_3(2) = 2 \Rightarrow \log_3(\frac{x-1}{2}) = 2 \Rightarrow \frac{x-1}{2} = 9 \Rightarrow x-1 = 18 \Rightarrow x = 19$. Check: $x-1 = 18 > 0$ ✓.
When we combine logarithmic terms using log laws, we sometimes create equations that have extra solutions. This happens because:
Rule: After solving, substitute back into the original equation. Reject any solution that makes any log argument non-positive.
$\log_a(x) = \frac{\ln x}{\ln a}$
$\log_a(x) = c \Rightarrow x = a^c$
$\log_a(M) = \log_a(N) \Rightarrow M = N$
All log arguments must be positive
$\log_3(20)$ can be found using $\log_3(20) = \frac{\ln(20)}{\ln(3)} \approx 2.727$. The change of base formula works because if $3^y = 20$, taking $\ln$ of both sides gives $y \ln(3) = \ln(20)$, so $y = \frac{\ln(20)}{\ln(3)}$. This is not just a calculator trick — it is a direct consequence of the definition of logarithms.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Use change of base to evaluate $\log_7(50)$ to 3 decimal places. Show your working. 3 MARKS
9. Solve $\log_2(x - 1) + \log_2(x + 1) = 3$. Show all working and check for extraneous solutions. 3 MARKS
10. A binary search algorithm halves the search space with each comparison. Show that the maximum number of comparisons needed to search $N$ items is $\lceil \log_2(N) \rceil$. Calculate how many comparisons are needed for 1 million items. Explain why this is dramatically faster than a linear search, which needs up to $N$ comparisons. 3 MARKS
1. $\log_5(100) = \frac{\ln(100)}{\ln(5)} \approx \frac{4.605}{1.609} \approx 2.861$.
2. $x = 2^7 = 128$.
3. $x + 3 = 4^2 = 16 \Rightarrow x = 13$. Check: $13 + 3 = 16 > 0$ ✓.
4. $\log_3(x(x-2)) = 2 \Rightarrow x(x-2) = 9 \Rightarrow x^2 - 2x - 9 = 0 \Rightarrow x = 1 + \sqrt{10} \approx 4.16$ (reject $1 - \sqrt{10}$ since $x > 2$).
1. $\log_3(x)$ requires $x > 0$; $\log_3(x-4)$ requires $x > 4$. Combined: $x > 4$.
2. $2^{20} = 1{,}048{,}576$ items. Maximum ~1.05 million items.
3. $\ln(x(x-3)) = \ln(10) \Rightarrow x(x-3) = 10 \Rightarrow x^2 - 3x - 10 = 0 \Rightarrow (x-5)(x+2) = 0$. $x = 5$: check $\ln(5) + \ln(2) = \ln(10)$ ✓. $x = -2$: $\ln(-2)$ undefined ✗. Final: $x = 5$.
Q8 (3 marks): $\log_7(50) = \frac{\ln(50)}{\ln(7)}$ [1] $= \frac{3.912}{1.946}$ [0.5] $\approx 2.010$ [1.5].
Q9 (3 marks): $\log_2((x-1)(x+1)) = 3$ [0.5]. $(x-1)(x+1) = 8$ [0.5]. $x^2 - 1 = 8$, so $x^2 = 9$, giving $x = \pm 3$ [0.5]. Check $x = 3$: $\log_2(2) + \log_2(4) = 1 + 2 = 3$ ✓ [0.5]. Check $x = -3$: $\log_2(-4)$ undefined ✗ [0.5]. Valid solution: $x = 3$ [0.5].
Q10 (3 marks): After $n$ comparisons, remaining items = $N/2^n$ [0.5]. For success, need $N/2^n \leq 1$, so $2^n \geq N$, thus $n \geq \log_2(N)$ [0.5]. For $N = 1{,}000{,}000$: $n = \log_2(10^6) = \frac{\ln(10^6)}{\ln(2)} = \frac{6\ln(10)}{\ln(2)} \approx 19.93$, so $\lceil 19.93 \rceil = 20$ comparisons [0.5]. Linear search needs up to 1,000,000 comparisons in the worst case [0.5]. Binary search is faster because each comparison eliminates half the remaining items — the number of steps grows logarithmically with $N$ rather than linearly [0.5]. For large $N$, $\log_2(N)$ is vastly smaller than $N$ (e.g., 20 vs 1,000,000) [0.5].
Climb platforms using change of base, log equation solving, and domain checking. Pool: lesson 8.
Tick when you've finished all activities and checked your answers.