SL 2.10 Solve exponential equations using logarithms | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Functions

IB Mathematics: Analysis and Approaches SL — Study topic SL 2.10

An exponential equation has an unknown in an exponent, such as

4^x=30

. Ordinary inverse operations do not isolate

x

directly because it is an exponent. Logarithms provide a way to bring the exponent down so that it can be found. This lesson reviews the needed algebra, develops that method, and connects exact steps to calculator and graph checks.

What you will learn

Recognise when an unknown is in an exponent and explain why logarithms can help solve for it.
Use logarithms and the change-of-base relationship to solve exponential equations.
Check solutions algebraically, numerically, and with a graph when appropriate.
State relevant restrictions and give numerical answers to a suitable accuracy.

1. Prior knowledge: powers and logarithms

A logarithm answers the question: “To what power must the base be raised to give this number?” Thus,

\log_b a=c

means that

b^c=a

. Here

b

is the base and

a

is the number whose logarithm is taken. For real logarithms, require

b>0

,

b\ne1

, and

a>0

.

For example,

\log_2 8=3

because

2^3=8

. In an exponential equation such as

2^x=8

, the logarithm with base

2

gives

x=\log_2 8=3

. When the base is not convenient, a calculator’s common logarithm,

\log

, or natural logarithm,

\ln

, can be used. Both are logarithms; the choice does not change the solution.

The key rule is that taking a logarithm of a positive power brings its exponent down as a multiplier. The change-of-base relationship lets us evaluate a logarithm in any base using calculator keys for

\log

or

\ln

. Use parentheses carefully when entering a quotient or a complete exponent.

Before taking logarithms, use ordinary algebra to isolate the exponential expression. For instance, in

6\cdot 3^x=54

, divide both sides by

6

first. The resulting power equals a positive number, so its logarithm is defined.

\log_b(a^x)=x\log_b a,\qquad \log_b a=\frac{\ln a}{\ln b}

A logarithm is the inverse operation to raising a positive base to a power.
The number inside a real logarithm must be positive.
Isolate the exponential expression before taking logarithms.

2. The solving method and its representations

For an equation of the form

a^x=c

, with

a>0

,

a\ne1

, and

c>0

, take the logarithm of both sides. The exponent becomes a factor, and dividing by the logarithm of the base gives an exact expression for

x

. The same method works when the exponent is a linear expression, such as

2x-1

: after taking logarithms, solve the resulting linear equation.

If the equation has a multiplier, first divide by it. For example, to solve

k a^{mx+n}=d

, where the quantities permit a real logarithm, first obtain

a^{mx+n}=d/k

. This requires

d/k>0

. Then take logarithms and solve for

x

. Keeping each algebra step visible helps avoid losing a factor or sign.

There are several useful ways to interpret the result. Algebraically, logarithms isolate the exponent. Numerically, a calculator evaluates the logarithmic quotient. Graphically, the solution is the intersection of the graph of the left-hand side and the horizontal graph of the right-hand side. In context, the solution may represent time or another measured quantity, so the final answer should include appropriate units and respect any stated domain.

A graphing calculator can check a solution by plotting both sides as separate functions or by plotting their difference and locating its zero. Use a sensible viewing window and treat the graph as a check, not as the explanation: it suggests the solution, while the logarithm steps justify it. If the model only allows non-negative time, disregard any intersection outside that domain.

a^x=c\;\Longrightarrow\;x=\frac{\ln c}{\ln a}

Take logarithms only after isolating a positive exponential expression.
Solve the resulting linear equation for the unknown in the exponent.
A graph or numerical substitution checks a result but does not replace the algebraic method.

3. Calculator accuracy and solution checks

A calculator may display a decimal approximation, but the logarithmic expression is often the most useful exact form. Keep that exact form until the last step, then round to the precision requested. If no precision is specified, give a reasonable number of decimal places and identify it as an approximation.

Check a numerical answer by substituting it into the original equation. The two sides should agree to the displayed accuracy. For a contextual model, check that the value satisfies the stated domain and report units. A graphing calculator can provide an independent visual check: the two graphs should meet near the calculated value.

When the exponential base is between

0

and

1

, the same logarithm method applies. The calculator quotient still gives the solution; do not assume that every exponential graph increases. The restrictions on the base and on the logarithm argument still matter.

0<a<1 or a>1, c>0

Retain exact logarithmic forms until the final rounding step.
Substitute an approximation into the original equation to check it.
State units and domain restrictions in contextual answers.

4. Exam-style decisions

A clear solution shows the rearrangement, the logarithm step, the equation for the unknown, and the final value. This makes the method assessable even when a calculator is used. Avoid rounding intermediate values because small rounding errors can affect the final answer.

Before accepting a result, ask whether the logarithm was applied to a positive quantity, whether the unknown was isolated correctly, and whether the answer fits the question. For a graph-based check, confirm that the plotted functions represent the original two sides and that the displayed intersection lies in the relevant domain.

Show enough algebra to make the logarithm method clear.
Round only at the end and match the requested accuracy.
Check both the original equation and any contextual restrictions.

Worked example

A power with a linear exponent

Solve

3^{2x-1}=17

. Give the answer to three significant figures.

Take logarithms
Both sides are positive, so take natural logarithms. The power rule brings the full exponent down as a multiplier.
$\ln\left(3^{2x-1}\right)=\ln 17$
Bring down the exponent
Use the logarithm power rule, then divide by $\ln 3$ to isolate the linear expression.
$(2x-1)\ln 3=\ln 17$
Solve and round
Rearrange for $x$ and evaluate only at the end. The logarithmic expression is exact; the decimal is rounded to three significant figures.
$x=\frac{1}{2}\left(1+\frac{\ln 17}{\ln 3}\right)\approx1.79$

Answer:

x\approx1.79

to three significant figures.

Check: Substituting the unrounded value into

3^{2x-1}

gives

17

. A graph of

y=3^{2x-1}

and

y=17

intersects near

x=1.79

.

Worked example

Isolate the power first

Solve

5\cdot2^{x+1}=37

. Give the answer to three significant figures.

Isolate the exponential expression
Divide both sides by $5$ before taking logarithms. The resulting value is positive, as required for a real logarithm.
$2^{x+1}=\frac{37}{5}$
Take logarithms
Take natural logarithms on both sides and use the power rule to bring down $x+1$ .
$(x+1)\ln 2=\ln\left(\frac{37}{5}\right)$
Solve and round
Divide by $\ln 2$ , then subtract $1$ . The calculator value is rounded only after the rearrangement.
$x=\frac{\ln(37/5)}{\ln 2}-1\approx1.89$

Answer:

x\approx1.89

to three significant figures.

Check: Using the unrounded value makes

5\cdot2^{x+1}

approximately

37

. The graph of the left side and the horizontal line

y=37

should meet at this value.

Worked example

An exponential model in context

A quantity is modelled by

A=1200(1.06)^t

, where

A

is measured in units and

t

is time in years. Find when the model reaches

2000

units. Give the time to two decimal places.

Set the model equal to the target
The target is $2000$ units. Divide by $1200$ to isolate the exponential expression. The model uses non-negative time, so the solution must have $t\geq0$ .
$(1.06)^t=\frac{2000}{1200}=\frac{5}{3}$
Use logarithms
Take natural logarithms and use the power rule. Since $1.06>0$ and $5/3>0$ , both logarithms are defined.
$t\ln(1.06)=\ln\left(\frac{5}{3}\right)$
Calculate with units
Divide by $\ln(1.06)$ and round the result to two decimal places. The positive result satisfies the model's time domain.
$t=\frac{\ln(5/3)}{\ln(1.06)}\approx8.77$

Answer: The model reaches

2000

units after approximately

8.77

years.

Check: Substituting the unrounded time into

1200(1.06)^t

gives approximately

2000

. A graph of this model and the horizontal line

A=2000

shows an intersection near

t=8.77

.

Common mistakes and how to avoid them

Taking a logarithm before isolating the exponential expression in an equation with a multiplier.

Correction: Use inverse operations such as division first, then take logarithms of the isolated positive power.

Writing

\log(a^x)

as

\log(a)x

without the logarithm-of-a-base relationship.

Correction: Use

\log(a^x)=x\log(a)

, keeping the base and argument clear.

Rounding logarithmic values during the algebra.

Correction: Keep the exact quotient of logarithms until the final calculation, then round to the stated accuracy.

Accepting a calculator or graph result without checking it in the original equation.

Correction: Substitute the result into the original equation and confirm that both sides agree to the stated accuracy.

Ignoring a contextual restriction such as non-negative time.

Correction: Compare the calculated solution with the domain described in the question and include units.

Lesson summary

For an isolated equation $a^x=c$ , take logarithms and solve for the exponent.
If the exponent contains an expression such as $mx+n$ , use the logarithm power rule, then solve the resulting linear equation.
For equations with a multiplier, isolate the power first and check that the logarithm argument is positive.
Use calculator or graphing technology to evaluate and check, while showing the mathematical reasoning.
Retain exact forms where useful, round at the end, and state units or domain restrictions when relevant.

Check your understanding

Question 1

Solve

4^x=11

. Which expression gives the exact solution?

$x=\frac{\ln 4}{\ln 11}$
$x=\frac{\ln 11}{\ln 4}$
$x=\ln\left(\frac{11}{4}\right)$
$x=\frac{11}{4}$

Show answer and explanation

x=\frac{\ln 11}{\ln 4}

Taking logarithms gives

x\ln 4=\ln 11

, so divide by

\ln 4

.

Question 2

Solve

7\cdot3^x=63

. What is

x

?

$x=1$
$x=2$
$x=3$
$x=9$

Show answer and explanation

x=2

Divide by

7

to get

3^x=9=3^2

, so

x=2

.

Question 3

For which value of

c

is the real logarithm step in solving

2^x=c

valid?

$c=-2$
$c=0$
$c=\frac{1}{2}$
$c=1$

Show answer and explanation

c=\frac{1}{2}

A real logarithm requires a positive argument. Of these choices, only

1/2

is positive.

Key terms

Exponential equation: An equation in which the unknown appears in an exponent, such as $5^x=20$ .
Logarithm: The exponent that a stated base must have to produce a given positive number.
Common logarithm: A logarithm with base $10$ , usually written $\log$ .
Natural logarithm: A logarithm with base $e$ , written $\ln$ ; a calculator can use it to evaluate logarithmic quotients.
Change of base: A relationship that rewrites a logarithm using a different base, allowing calculator evaluation with $\log$ or $\ln$ .

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SL 2.10 · Solve exponential equations using logarithms

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

IB Mathematics: Analysis and Approaches SL — Study topic SL 2.10

What you will learn

1. Prior knowledge: powers and logarithms

2. The solving method and its representations

3. Calculator accuracy and solution checks

4. Exam-style decisions

A power with a linear exponent

Isolate the power first

An exponential model in context

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review