Lesson 5 ~40 min Logarithms · Path +85 XP

Solving Exponential and Logarithmic Equations

Two kinds of equation, and one idea that solves both. If the unknown is stuck in an index, a logarithm frees it. If it is stuck inside a logarithm, a power frees it. Then check that the answer is legal.

Today's hook: $2^x = 8$ you can solve by inspection. $2^x = 10$ you cannot, because $10$ is not a power of $2$. Yet the answer exists, sits between $3$ and $4$, and can be written down exactly in one symbol. Which one?

0/5QUESTS

Think First

warm-up

Solve $2^x = 32$ in your head. Now think about $2^x = 30$. You know the answer is a little less than $5$. Write down, in words, what you would need to be able to do in order to name that number exactly. Then say what operation would undo the equation $\log_3 x = 4$ instead.

Record your answer in your workbook.

The Big Idea

+5 XP to read

An equation traps the unknown behind an operation, and you free it by applying the inverse of that operation to both sides. Powers and logarithms of the same base are inverses, so each one releases the other. That is the whole method.

$$a^x = b \;\Longrightarrow\; x = \log_a b \qquad\qquad \log_a x = c \;\Longrightarrow\; x = a^c$$

Both arrows are just the definition of a logarithm, used as a tool rather than stated as a fact. The extra care a logarithmic equation needs is the domain check: the solution must leave every argument positive, and a value that fails that test is rejected even though the algebra produced it.

$a^x = b \iff x = \log_a b$

Unknown in the index

Take logarithms of both sides. $a^x = b$ becomes $x = \log_a b$.

Unknown inside a log

Raise the base to both sides. $\log_a x = c$ becomes $x = a^c$.

Always check the domain

Reject any solution that makes an argument zero or negative.

What You'll Master

objectives

Know

That $a^x = b$ has the exact solution $x = \log_a b$
That $\log_a x = c$ has the exact solution $x = a^c$
That every solution of a logarithmic equation must be checked against the domain

Understand

Why taking the logarithm of both sides is a legitimate move on any equation
Why a rejected solution is not an error in the working but a feature of the domain

Can Do

Solve equations with the unknown in the index, exactly and to a given accuracy
Solve equations with the unknown inside a logarithm, including ones needing the laws first
Identify and reject solutions that fall outside the domain

Words You Need

vocabulary

Exponential equationOne with the unknown in the index, such as $3^x = 20$.

Logarithmic equationOne with the unknown inside a logarithm, such as $\log_2(x+1) = 3$.

Inverse operationThe operation that undoes another. Powers and logarithms of the same base are inverses.

Domain checkConfirming a solution leaves every logarithm's argument positive.

Extraneous solutionA value produced by correct algebra that fails the domain check and must be rejected.

Exact formAn answer left as $\log_2 10$ rather than rounded to $3.32$.

When the Unknown Is in the Index

+5 XP to read

If the two sides can be written as powers of the same base, you never need a logarithm at all. Match the bases and equate the indices:

$$2^x = 32 = 2^5 \;\Longrightarrow\; x = 5$$

When that fails, because the right-hand side is not a power of the base, take logarithms of both sides. Doing the same thing to both sides is always legal, and here it releases the unknown:

$$2^x = 10 \;\Longrightarrow\; x = \log_2 10$$

That is the exact answer. If a decimal is wanted, use the change to a calculator base: $\log_2 10 = \dfrac{\log_{10} 10}{\log_{10} 2} = \dfrac{1}{0.3010} \approx 3.32$.

Check it

$2^3 = 8$ and $2^4 = 16$, so a solution of $3.32$ is exactly where it should be.

When the Unknown Is Inside a Logarithm

+5 XP to read

Here the trapping operation is the logarithm, so undo it with a power of the same base.

$$\log_3 x = 4 \;\Longrightarrow\; x = 3^4 = 81$$

If the argument is an expression rather than a bare $x$, undo the logarithm first and solve afterwards:

$$\log_2(x + 5) = 3 \;\Longrightarrow\; x + 5 = 2^3 = 8 \;\Longrightarrow\; x = 3$$

Check: $\log_2(3 + 5) = \log_2 8 = 3$. Correct, and the argument $8$ is positive, so the solution stands.

Equations Needing the Laws First

+5 XP to read

Some equations have several logarithms. Gather them into one, then undo.

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

Product law: $\log_2\big(x(x-2)\big) = 3$. Undo the logarithm: $x(x - 2) = 2^3 = 8$. Expand and solve:

$$x^2 - 2x - 8 = 0 \;\Longrightarrow\; (x - 4)(x + 2) = 0 \;\Longrightarrow\; x = 4 \text{ or } x = -2$$

Now the step that is not optional. Test each against the original equation: $x = 4$ gives arguments $4$ and $2$, both positive, so it is valid. $x = -2$ gives $\log_2(-2)$, which does not exist, so it is rejected.

The answer is $x = 4$ only.

Why a Solution Gets Rejected

+5 XP to read

Rejecting $x = -2$ above is not a sign that something went wrong. It is a consequence of how the working proceeded.

The original equation only makes sense for $x > 2$, because both $x$ and $x - 2$ must be positive. But the moment you write $x(x-2) = 8$ the logarithms are gone and so is that restriction. The quadratic is a wider problem than the one you started with, and it can have solutions the original never permitted.

So the discipline is fixed: state the domain before you solve, and test every candidate against the ORIGINAL equation at the end. A solution rejected with a reason earns marks; one silently dropped does not.

Common Pitfalls

+5 XP to read

Solving $\log_2 x + \log_2(x-2) = 3$ by setting each logarithm to $3$ separately.

Fix: the equation constrains the SUM, not each term. Gather with the product law into a single logarithm before undoing anything.

Writing $2^x = 10$ as $x = \dfrac{10}{2}$, or dividing both sides by $2$.

Fix: $x$ is an index, not a factor. Division cannot reach it; only a logarithm can.

Giving both roots of the quadratic without checking the domain.

Fix: substitute each candidate into the ORIGINAL equation. If any argument comes out zero or negative, reject that value and say why.

Watch Me Solve It · 3 examples

Watch Me Solve It · Unknown in the index

+15 XP per step

PROBLEM

Solve $5^x = 200$, giving the answer in exact form and correct to two decimal places.

1

Check for a common base

$200 \text{ is not a power of } 5$

$5^3 = 125$ and $5^4 = 625$, so the answer lies between 3 and 4 and is not a whole number.
2

Take logarithms of both sides

$x = \log_5 200$

This is the exact answer; the definition does all the work.
3

Convert for the calculator

$x = \frac{\log_{10} 200}{\log_{10} 5} = \frac{2.3010}{0.6990}$

Calculators carry base 10, so rewrite the base-5 logarithm as a ratio of common logarithms.
4

Evaluate and sanity-check

$x \approx 3.29$

It sits between 3 and 4, as predicted in step 1.

Answer$x = \log_5 200 \approx 3.29$

Watch Me Solve It · Unknown inside a logarithm

+15 XP per step

PROBLEM

Solve $\log_4(3x - 2) = 2$.

1

State the domain first

$3x - 2 > 0 \;\Longrightarrow\; x > \tfrac{2}{3}$

Recording the restriction now makes the final check quick.
2

Undo the logarithm

$3x - 2 = 4^2 = 16$

Raise the base to both sides.
3

Solve the linear equation

$3x = 18 \;\Longrightarrow\; x = 6$
4

Check against the domain and the original

$6 > \tfrac{2}{3} \;\checkmark$

$\log_4(3(6) - 2) = \log_4 16 = 2 \;\checkmark$

Valid on both counts, so the solution stands.

Answer$x = 6$

Watch Me Solve It · Two logarithms and a rejected root

+15 XP per step

PROBLEM

Solve $\log_3 x + \log_3(x + 6) = 3$.

1

State the domain

$x > 0 \text{ and } x + 6 > 0 \;\Longrightarrow\; x > 0$

Both arguments must be positive; the stricter condition wins.
2

Gather with the product law

$\log_3\big(x(x+6)\big) = 3$

Same base, an addition, so the product law applies.
3

Undo and rearrange

$x(x + 6) = 3^3 = 27$

$x^2 + 6x - 27 = 0$
4

Factorise and test both roots

$(x + 9)(x - 3) = 0 \;\Longrightarrow\; x = -9 \text{ or } x = 3$

$x = -9$ fails the domain, since $\log_3(-9)$ does not exist, so it is rejected. $x = 3$ gives $\log_3 3 + \log_3 9 = 1 + 2 = 3$.

Answer$x = 3$; the root $x = -9$ is rejected

Brain Trainer · 4 problems

Brain Trainer · Free the unknown

4 problems

Four quick problems. Work each one, then reveal the answer.

1 Solve $3^x = 81$.

$81 = 3^4$, so match the bases and equate the indices.$x = 4$
2 Solve $\log_5 x = 3$.

Raise $5$ to both sides.$x = 125$
3 Solve $7^x = 50$, exact form.

$50$ is not a power of $7$, so take logarithms.$x = \log_7 50$
4 Solve $\log_2(x - 3) = 4$, and state the domain restriction.

Domain $x > 3$. Undo: $x - 3 = 16$.$x = 19$

Complete in your workbook.

Multiple Choice · 5 questions

MC1

Freeing an index

+10 XP

The exact solution of $4^x = 30$ is:

MC2

Undoing a logarithm

+10 XP

The solution of $\log_6 x = 2$ is:

MC3

Which solution is rejected

+10 XP

Solving $\log_2 x + \log_2(x - 6) = 4$ leads to $x = 8$ and $x = -2$. Which solutions are valid?

MC4

Choosing the method

+10 XP

To solve $9^x = 27$ most efficiently, you should:

MC5

Reading an exact answer

+10 XP

A student solves $2^x = 20$ and writes $x = \log_2 20$. This answer is:

Short Answer · 3 questions

Both kinds of equation

+15 XP

SHORT ANSWER

Solve each equation. Give exact answers unless a decimal is requested.
(a) $3^x = 243$
(b) $\log_7(2x + 1) = 2$
(c) $6^x = 90$, correct to two decimal places

Write your working in your book.

Gather, solve, reject

+15 XP

SHORT ANSWER

Solve $\log_5 x + \log_5(x - 4) = 1$.
Set out the domain, the gathering step, both candidate solutions, and a justified decision on each.

Write your working in your book.

Explain the rejection

+15 XP

SHORT ANSWER

A student solves a logarithmic equation correctly, obtains two values, and is told one must be rejected. They ask why, since every algebraic step was valid.
Explain what happens to the problem during the solving that allows an invalid value to appear.

Write your working in your book.

Stretch Challenge · Doubling time

+25 XP

CHALLENGE

A colony of bacteria doubles every hour. Starting from $500$ cells, the number after $t$ hours is $N = 500 \times 2^{t}$.
(a) Write an equation for the time at which the colony reaches $32\,000$ cells, and solve it exactly.
(b) Solve for the time at which it reaches $20\,000$ cells, correct to one decimal place.
(c) The colony is placed in a dish that can hold $10^{6}$ cells. Show that it fills the dish during the eleventh hour, without evaluating a logarithm.

Quick Review

recap

Index unknown

$a^x = b \Rightarrow x = \log_a b$

Log unknown

$\log_a x = c \Rightarrow x = a^c$

Several logs

Gather into one, then undo

Never skip

Test every candidate in the original

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