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hscscience Maths Adv · Y11
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Module 4 · L8 of 15 ~40 min ⚡ +95 XP available

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.

Today's hook, Currency exchange: if you know the AUD/USD rate and the AUD/EUR rate, you can calculate the USD/EUR rate. Change of base works the same way, if you can evaluate $\ln x$ and $\ln a$ on your calculator, you can find $\log_a x$ for any base $a$, simply by dividing.
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Orient and recall

Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
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Recall, your gut answer first
+5 XP warm-up

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.

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The two moves

Work through the core explanation before applying it.

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The two moves
+5 XP to read

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.

Rewrite any log in terms of natural logs (or log₁₀) using the change-of-base formula.
$$\log_a x = \dfrac{\ln x}{\ln a} = \dfrac{\log_{10} x}{\log_{10} a}$$
Convert then solve
Use change of base to evaluate unfamiliar logs, then solve equations algebraically.
Always check the domain
After solving, verify all log arguments are positive. Reject any solution that makes an argument zero or negative.
Equal log rule
$\log_a x = \log_a y$ implies $x = y$ when both arguments are positive and the base is the same.
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What you'll master

Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.

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What you'll master
Know

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
Understand

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
Can do

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
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Key terms
Change of base formula$\log_a x = \dfrac{\log_b x}{\log_b a}$ for any positive $b \neq 1$. Most used with $b = e$.
Extraneous solutionA solution obtained algebraically that fails the original domain, it makes a log argument zero or negative.
Domain of a logarithmThe argument of a logarithm must be strictly positive: $\log_a x$ is defined only for $x > 0$.
Exponential equationAn equation where the unknown appears in an exponent, e.g. $2^x = 5$. Take logs of both sides to solve.
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Change of base and solving log equations

Work through the core explanation before applying it.

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Change of base and solving log equations
core concept

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 $x = \log_2 5 = \dfrac{\ln 5}{\ln 2}$.

Proving it. Let $y = \log_a x$. By the definition of a logarithm that is exactly the statement $a^y = x$.

Take $\log_b$ of both sides, where $b$ is any base you can actually evaluate: $\log_b (a^y) = \log_b x$.

The power law from Lesson 7 brings the index to the front: $y \log_b a = \log_b x$.

Divide by $\log_b a$, which is never zero because $a \neq 1$: $y = \dfrac{\log_b x}{\log_b a}$. Since $y$ was defined as $\log_a x$, that is the change of base rule, and choosing $b = e$ gives the $\ln$ form above.

Why $a \neq 1$ matters. If $a$ were $1$ then $\log_b a = 0$ and the division would be undefined, which is the same reason $1$ is excluded as a logarithm base in the first place. The proof needs that exclusion; it does not merely inherit it.

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.

$$\log_a x = \frac{\log_b x}{\log_b a} = \frac{\ln x}{\ln a}$$

Most commonly used with $b = e$ (natural log) or $b = 10$ (common log).

Domain check is a mark, always do it. After solving a logarithmic equation, substitute your answers back into the original equation and confirm every log argument is positive. Any value that makes an argument zero or negative must be rejected. This is the most common source of lost marks in this topic.

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

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Work examples end to end

Follow the reasoning through complete worked solutions.

PROBLEM 1 · CHANGE OF BASE

Evaluate $\log_3 7$ to 3 decimal places.

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$\log_3 7 = \dfrac{\ln 7}{\ln 3}$
Change of base to natural logarithm.
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PROBLEM 2 · LOG EQUATION WITH DOMAIN CHECK

Solve $\log_2(x+1) + \log_2(x-1) = 3$.

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$\log_2[(x+1)(x-1)] = 3$
Product law: sum of logs = log of product.
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PROBLEM 3 · EXPONENTIAL EQUATION DIFFERENT BASES

Solve $3^{x+1} = 5^{2x}$.

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Take logarithms of both sides: $\ln(3^{x+1}) = \ln(5^{2x})$
Use logs to bring down the exponents.
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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

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Dodge the traps, then apply

Meet the mistakes that cost marks, then do it yourself.

Trap 01
Forgetting to check the domain after solving
After solving, always verify that all log arguments are positive. Reject any solution that makes an argument zero or negative. This is the most common source of lost marks in logarithmic equations.
Trap 02
Dropping the logs and adding the arguments
$\log_a x + \log_a y = \log_a z$ does not mean $x + y = z$. It means $\log_a(xy) = \log_a z$, so $xy = z$. Use the product law first, then convert to exponential form.
Trap 03
Inverting the change of base fraction
$\log_a x = \dfrac{\ln x}{\ln a}$, not $\dfrac{\ln a}{\ln x}$. The argument $x$ goes on top; the base $a$ goes on the bottom. Swapping them gives the reciprocal, which is a completely different value.

Two truths, one lie: Two of these statements are correct. Which one is the lie?

Quick-fire practice · 5 problems

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Drill it, then lock it in

Run the quick drill and copy the summary into your book.

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Evaluate $\log_5 12$ to 2 decimal places.

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Solve $\log_3(x+2) = 2$.

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Solve $\log_2 x + \log_2(x-2) = 3$.

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Solve $2^x = 7$ using logarithms.

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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?

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Revisit your thinking

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.

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Multiple choice

Answer the drill bank and rate your confidence.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer

Write full responses, then check them against the model answers.

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Short answer
ApplyBand 42 marks

Q1. Evaluate $\log_4 15$ to 3 decimal places, showing your working. (2 marks)

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ApplyBand 43 marks

Q2. Solve $\log_3(x-1) + \log_3(x+1) = 1$, checking for extraneous solutions. (3 marks)

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ApplyBand 54 marks

Q3. Solve $4^{x+1} = 7^x$, giving your answer to 2 decimal places. (4 marks)

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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].

Module overview · Maths Advanced · Checkpoint 2