Change of Base & Logarithmic Equations
Your calculator only has $\log_{10}$ and $\ln$ buttons. The change of base formula — $\log_a x = \dfrac{\ln x}{\ln a}$ — converts any logarithm into one you can evaluate. Paired with the log laws you learned in Lesson 7, you'll solve equations that once looked completely untouchable.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
Your calculator only has $\log_{10}$ and $\ln$ buttons. How would you compute $\log_5 30$? Sketch your strategy in one or two lines — you don't need to know the formula yet.
This lesson has two core moves. Master both and you can handle every log and exponential equation the HSC will throw at you.
Move 1 — Convert: use change of base to evaluate any unfamiliar log on your calculator. Move 2 — Solve: rearrange log equations algebraically, then always verify solutions keep arguments positive.
Key facts
- Change of base formula: $\log_a x = \dfrac{\log_b x}{\log_b a}$
- Most common form with $b = e$: $\log_a x = \dfrac{\ln x}{\ln a}$
- Extraneous solutions arise from domain violations
Concepts
- Why any base can be rewritten in terms of another base
- How to combine log laws to solve equations
- Why domain checks are non-negotiable
Skills
- Evaluate any logarithm with a calculator using change of base
- Solve logarithmic equations and check for extraneous solutions
- Apply logarithms to solve exponential equations with different bases
The change of base formula lets us evaluate any logarithm using only $\ln$ or $\log$ on our calculator: $\log_a x = \dfrac{\ln x}{\ln a}$. This is essential for solving equations like $2^x = 5$, where we need $x = \log_2 5 = \dfrac{\ln 5}{\ln 2}$.
When solving logarithmic equations, always check that solutions keep the arguments positive. Squaring or other operations can introduce extraneous solutions that violate the original domain.
Most commonly used with $b = e$ (natural log) or $b = 10$ (common log).
Change of base: $\log_a x = \dfrac{\ln x}{\ln a}$ — argument on top, base on bottom; To solve $a^x = b$: take $\ln$ of both sides, use power law, isolate $x$
Pause — copy the change of base formula $\log_a x = \dfrac{\ln x}{\ln a}$ (argument on top, base on bottom) and the solving strategy for $a^x = b$ (take $\ln$ both sides, use power law, isolate $x$) into your book.
Quick check: Which expression correctly gives $\log_3 7$ using natural logarithms?
Worked examples · 3 in a row, reveal as you go
Evaluate $\log_3 7$ to 3 decimal places.
Solve $\log_2(x+1) + \log_2(x-1) = 3$.
Solve $3^{x+1} = 5^{2x}$.
Did you get this? True or false: when solving a logarithmic equation, any solution that makes a log argument equal to zero must be rejected.
Common errors · the 3 traps that cost marks
Two truths, one lie: Two of these statements are correct. Which one is the lie?
Quick-fire practice · 5 problems
Evaluate $\log_5 12$ to 2 decimal places.
Solve $\log_3(x+2) = 2$.
Solve $\log_2 x + \log_2(x-2) = 3$.
Solve $2^x = 7$ using logarithms.
Solve $5^{x-1} = 3^{x+1}$.
Fill in the blank: Using change of base, $\log_4 x = \dfrac{\ln x}{\ln\!\left(\rule{40px}{0.5px}\right)}$.
Odd one out: Three of these are valid first steps for solving $\log_2(x+3) + \log_2(x-1) = 4$. Which is not a valid first step?
Earlier you were asked how to compute $\log_5 30$ using only $\log_{10}$ or $\ln$. The change-of-base formula gives $\log_5 30 = \dfrac{\ln 30}{\ln 5} \approx \dfrac{3.401}{1.609} \approx 2.11$. When solving log equations, always check the domain at the end — any solution that makes a log argument zero or negative must be rejected as extraneous.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Evaluate $\log_4 15$ to 3 decimal places, showing your working. (2 marks)
Q2. Solve $\log_3(x-1) + \log_3(x+1) = 1$, checking for extraneous solutions. (3 marks)
Q3. Solve $4^{x+1} = 7^x$, giving your answer to 2 decimal places. (4 marks)
Comprehensive answers (click to reveal)
Drill 1: $\dfrac{\ln 12}{\ln 5} \approx 1.54$ · 2: $x + 2 = 3^2 = 9$, so $x = 7$ · 3: $\log_2[x(x-2)] = 3$, $x(x-2) = 8$, $x^2-2x-8=0$, $(x-4)(x+2)=0$, $x=4$ (reject $x=-2$) · 4: $x = \dfrac{\ln 7}{\ln 2} \approx 2.807$ · 5: $(x-1)\ln 5 = (x+1)\ln 3$; $x(\ln 5 - \ln 3) = \ln 3 + \ln 5$; $x = \dfrac{\ln 15}{\ln(5/3)} \approx 5.30$
Q1 (2 marks): $\log_4 15 = \dfrac{\ln 15}{\ln 4}$ [0.5] $\approx \dfrac{2.708}{1.386} \approx 1.953$ [1.5].
Q2 (3 marks): $\log_3[(x-1)(x+1)] = 1$ [0.5]; $(x-1)(x+1) = 3$, so $x^2 - 1 = 3$, $x^2 = 4$, $x = \pm 2$ [1]; domain: $x - 1 > 0$ and $x + 1 > 0$, so $x > 1$. Thus $x = 2$ only [1.5].
Q3 (4 marks): Take logs: $(x+1)\ln 4 = x\ln 7$ [0.5]; $x\ln 4 + \ln 4 = x\ln 7$ [0.5]; $\ln 4 = x(\ln 7 - \ln 4)$ [1]; $x = \dfrac{\ln 4}{\ln 7 - \ln 4} = \dfrac{1.386}{1.946 - 1.386} = \dfrac{1.386}{0.560} \approx 2.48$ [2].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
Enter the arenaClimb platforms by answering change of base and log equation questions. Lighter alternative to the boss.
Mark lesson as complete
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