Year 11 Methods (Unit 1 & 2) Exponential Functions And Logarithms

Using Logarithms To Solve Exponential Equations And Inequalities

Q: How do you solve an exponential equation like a to the power x equals b?

Rewrite it as x equals log base a of b, or take log base ten of both sides to get x times log of a equals log of b, then divide.

Q: What is the change of base rule?

Log base a of b equals log base ten of b divided by log base ten of a. It lets you evaluate any logarithm using the base ten log on a calculator.

Q: Why do you take logarithms of both sides?

Because the log of a to the power x equals x times the log of a. This brings the exponent x down to the front, turning the exponential into a linear equation.

Q: When does the inequality sign flip?

When you divide by a negative number. If the base is less than one then its log is negative, so greater than becomes less than.

Q: Can I use base 10 instead of natural logarithms?

Yes. Year 11 Methods uses base ten or a general base greater than one throughout. Natural logarithms and the number e are Year 12.

Q: Do I need to isolate the power before taking logs?

Yes. Divide off any coefficient so one side is exactly a to the power of the expression. Only then does taking the logarithm free the exponent.

20 practice questions 1 video lesson Theory + worked examples

Learn to use logarithms to solve exponential equations and inequalities for Queensland Year 11 Mathematical Methods (QCAA). When both sides do not share a base, taking the logarithm of each side brings the unknown power down to solve.

You will learn to take logs of both sides, apply the change-of-base rule, and reverse the inequality when the base is under one — useful for any exponential equation, especially in modelling.

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Using Logarithms To Solve Exponential Equations And Inequalities - Video - Using logarithms to solve exponential equations and inequalities Watch

Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), an exponential equation has the unknown in the index, such as \(a^{x}=b\). To solve for \(x\) you take a logarithm of both sides, then use the definition and the change-of-base rule to write \(x\) exactly or as a decimal. The same idea solves exponential inequalities — with a sign reversal when the base is less than \(1\).

An exponential equation has the variable in the exponent, for example \(a^{x}=b\). The definition of a logarithm turns it into a value directly: \(a^{x}=b\) means \(x=\log_a b\).

Because a calculator gives base-10 logarithms, the change-of-base rule rewrites any logarithm using base \(10\): \(\log_a b=\dfrac{\log_{10}b}{\log_{10}a}\). This lets you produce a decimal answer to as many places as required.

An exponential inequality such as \(a^{x}>b\) is solved the same way, but take care: when the base satisfies \(0decreasing, so dividing by \(\log_{10}a\) (a negative number) reverses the inequality sign.

Take a logarithm of both sides. For \(a^{x}=b\), either write \(x=\log_a b\) directly, or take \(\log_{10}\) of both sides and use \(\log_{10}(a^{x})=x\log_{10}a\).

Solving \(2^{x}=5\): the curve meets \(y=5\) at \(x=\log_2 5\approx 2.32\).

An inequality gives a region: \(2^{x}>100\) solves to \(x>\log_2 100\approx 6.64\).

Solving an exponential equation:

\[a^{x}=b \iff x=\log_a b\]

a^{x} = b \Leftrightarrow x = \log_{a} b

Change of base (to evaluate on a calculator):

\[\log_a b=\dfrac{\log_{10}b}{\log_{10}a}\]

\log_{a} b = \frac{\log_{10} b}{\log_{10} a}

Taking logs of a power (the step that frees the exponent):

\[\log_{10}\!\left(a^{x}\right)=x\log_{10}a\]

\log_{10} a^{x} = x \log_{10} a

Sign reversal: if \(0\) into \(<\) (and \(\ge\) into \(\le\)).

How to solve \(a^{x}=b\) (or an inequality)

Isolate the power — divide off any coefficient so one side is exactly \(a^{\text{(expression)}}\).
Take a logarithm of both sides, giving \(\log_a b\); use change of base \(\dfrac{\log_{10}b}{\log_{10}a}\) to get a decimal.
Solve the resulting linear equation for \(x\) (for an inequality, reverse the sign if you divided by a negative logarithm).

Example 1 — Solve a^x = b

Solve \(3^{x}=20\), giving \(x\) correct to two decimal places.

Solution

Write in logarithm form, then change to base \(10\):

\(x\)	\(=\)	\(\log_3 20\)
	\(=\)	\(\dfrac{\log_{10}20}{\log_{10}3}\)

Evaluate on a calculator:

\(x\)	\(\approx\)	\(\dfrac{1.30103}{0.47712}\)
	\(\approx\)	\(2.7268\)

\(x\approx 2.73\).

x \approx 2.73

Example 2 — Exact answer by change of base

Solve \(5^{x}=2\), giving the exact value and then a decimal to two places.

Solution

Take \(\log_{10}\) of both sides and bring the index down:

\(\log_{10}(5^{x})\)	\(=\)	\(\log_{10}2\)
\(x\log_{10}5\)	\(=\)	\(\log_{10}2\)

Divide by \(\log_{10}5\) — this is the change-of-base form:

\(x\)	\(=\)	\(\dfrac{\log_{10}2}{\log_{10}5}=\log_5 2\)
\(x\)	\(\approx\)	\(0.43\)

Exact: \(x=\log_5 2=\dfrac{\log_{10}2}{\log_{10}5}\); \(\ x\approx 0.43\).

x = \log_{5} 2

Example 3 — Exponent that is an expression

Solve \(5^{\,2x-1}=100\), correct to two decimal places.

Solution

Write in logarithm form (the whole exponent equals the log):

\(2x-1\)	\(=\)	\(\log_5 100\)
	\(=\)	\(\dfrac{\log_{10}100}{\log_{10}5}\)

Evaluate the logarithm:

\(2x-1\)	\(\approx\)	\(\dfrac{2}{0.69897}\)
\(2x-1\)	\(\approx\)	\(2.8614\)

Solve the linear equation for \(x\):

\(2x\)	\(\approx\)	\(3.8614\)
\(x\)	\(\approx\)	\(1.9307\)

\(x\approx 1.93\).

x \approx 1.93

Example 4 — Inequality with a base less than 1

Solve \(0.8^{x}<0.5\), giving \(x\) to two decimal places.

Solution

Take \(\log_{10}\) of both sides and bring the index down:

\(\log_{10}(0.8^{x})\)	\(<\)	\(\log_{10}0.5\)
\(x\log_{10}0.8\)	\(<\)	\(\log_{10}0.5\)

Now \(\log_{10}0.8\approx-0.09691\) is negative, so dividing by it reverses the inequality.

Divide by \(\log_{10}0.8\) and flip the sign:

\(x\)	\(>\)	\(\dfrac{\log_{10}0.5}{\log_{10}0.8}\)
\(x\)	\(>\)	\(\dfrac{-0.30103}{-0.09691}\)
\(x\)	\(>\)	\(3.1063\)

\(x>3.11\) (to 2 d.p.).

x > 3.11

Common pitfalls

Not isolating the power first. In \(4\cdot 2^{x}=100\) you must divide by \(4\) to get \(2^{x}=25\) before taking logarithms — you cannot take the log of \(4\cdot2^{x}\) term by term.

Forgetting to reverse the inequality. When the base is less than \(1\) (or you divide by any negative logarithm), the \(>\) must become \(<\).

Getting change of base upside down. \(\log_a b=\dfrac{\log_{10}b}{\log_{10}a}\): the argument goes on top, the base on the bottom.

Frequently asked questions

How do you solve an exponential equation like a to the power x equals b?

Rewrite it as \(x=\log_a b\), or take \(\log_{10}\) of both sides to get \(x\log_{10}a=\log_{10}b\), then divide.

What is the change of base rule?

\(\log_a b=\dfrac{\log_{10}b}{\log_{10}a}\). It lets you evaluate any logarithm using the base-10 log on a calculator.

Why do you take logarithms of both sides?

Because \(\log_{10}(a^{x})=x\log_{10}a\) brings the exponent \(x\) down to the front, turning the exponential into a linear equation.

When does the inequality sign flip?

When you divide by a negative number. If the base is less than \(1\) then \(\log_{10}a<0\), so \(>\) becomes \(<\).

Can I use base 10 instead of natural logarithms?

Yes — Year 11 Methods uses base-10 (or a general base \(a>1\)) logarithms throughout; natural logarithms and \(e\) are Year 12.

Do I need to isolate the power before taking logs?

Yes. Divide off any coefficient so one side is exactly \(a^{\text{expression}}\); only then does taking the logarithm free the exponent.

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More in Exponential Functions And Logarithms

Using Logarithms To Solve Exponential Equations And Inequalities

📖 Theory

How to solve \(a^{x}=b\) (or an inequality)

Common pitfalls

Frequently asked questions

Theory