Solution of Exponential Equations Using Logarithms Revision Notes for VCE SSCE Mathematical Methods

Solution of Exponential Equations Using Logarithms

Understanding the problem

When we encounter equations where the unknown variable appears in the exponent, such as $2^x = 11$ or $3^{2x-1} = 28$ , we cannot solve them using basic algebraic techniques. These exponential equations require a different approach using logarithms.

Logarithms provide us with a powerful tool to "bring down" the exponent and turn an exponential equation into one we can solve algebraically. This section shows you how to use the fundamental relationship between exponentials and logarithms to solve these equations.

infoNote

Why do we need logarithms?

In an exponential equation like $2^x = 11$ , we cannot simply isolate $x$ using standard algebraic operations. The variable is "trapped" in the exponent position. Logarithms are specifically designed to extract values from exponents, making them the perfect tool for this type of problem.

The fundamental equivalence

The key principle that allows us to solve exponential equations is the equivalence between exponential and logarithmic forms. These two forms express exactly the same relationship, allowing us to convert freely between them.

chatImportant

The Core Principle

If $a \in \mathbb{R}^+ \setminus \{1\}$ and $x \in \mathbb{R}$ , then:

a^x = b \Leftrightarrow \log_a b = x

This equivalence is the foundation for solving all exponential equations. Memorize this relationship!

For example:

$2^x = 5$ is equivalent to $x = \log_2 5$
$0.3^x = 5$ is equivalent to $x = \log_{0.3} 5$

Understanding that you can convert freely between these two forms is essential for solving exponential equations efficiently.

Worked example: Using logarithm properties

lightbulbExample

Worked Example: Finding k

If $\log_2 6 = k \log_2 3 + 1$ , find the value of $k$ .

Solution:

We need to manipulate the right-hand side to match the left-hand side.

Starting with the equation:

\log_2 6 = k \log_2 3 + 1

Using the power law of logarithms, we can write $k \log_2 3$ as $\log_2(3^k)$ :

\log_2 6 = \log_2(3^k) + \log_2 2

Using the addition law of logarithms:

\log_2 6 = \log_2(2 \times 3^k)

Since the logarithms are equal and have the same base, their arguments must be equal:

6 = 2 \times 3^k

Dividing both sides by 2:

3 = 3^k

Therefore $k = 1$ .

Solving basic exponential equations

For simple exponential equations of the form $a^x = b$ , we can directly convert to logarithmic form to find the solution. This is the most straightforward application of the fundamental equivalence principle.

lightbulbExample

Worked Example: Solving a Basic Exponential Equation

Solve $2^x = 11$ , expressing the answer to two decimal places.

Solution:

Using the equivalence between exponential and logarithmic forms:

2^x = 11 \Leftrightarrow x = \log_2 11

Evaluating this logarithm:

x = 3.45943...

Therefore $x \approx 3.46$ (correct to two decimal places).

This direct conversion method works whenever we have a single exponential term equal to a constant.

Solving complex exponential equations

When the exponential equation involves more complex expressions, we first need to rearrange algebraically before converting to logarithmic form. The key is to isolate the exponential expression on one side of the equation first.

lightbulbExample

Worked Example: Solving a Complex Exponential Equation

Solve $3^{2x-1} = 28$ , expressing the answer to three decimal places.

Solution:

First, we convert the exponential equation to logarithmic form:

3^{2x-1} = 28 \Leftrightarrow 2x - 1 = \log_3 28

Now we can solve algebraically for $x$ :

2x - 1 = \log_3 28

Adding 1 to both sides:

2x = \log_3 28 + 1

Dividing both sides by 2:

x = \frac{1}{2}(\log_3 28 + 1)

Evaluating this expression:

x \approx 2.017

(correct to three decimal places)

Using technology to solve exponential equations

infoNote

Calculator Methods

Calculators can solve exponential equations and provide decimal approximations efficiently. Here are the methods for common calculators:

Using the TI-Nspire:

Use menu > Algebra > Solve and enter the equation
Convert to a decimal answer using ctrl + enter or menu > Number > Convert to Decimal
Round to the required number of decimal places

Using the Casio ClassPad:

In the Main application, enter and highlight the equation
Go to Interactive > Equation/Inequality > solve and tap OK
For a decimal approximation, highlight the answer and use the decimal conversion button

The calculator screenshot shows the complete solution process, including both the exact form using logarithms and the decimal approximation.

Solving exponential inequalities

When solving exponential inequalities, we must pay careful attention to the base of the exponential function. The direction of the inequality may change depending on whether the base is greater than or less than 1.

chatImportant

Critical Principle for Inequalities

The behavior of logarithmic functions depends on the base:

For $a > 1$ : the function $y = \log_a x$ is strictly increasing, so inequality directions remain the same
For $0 < a < 1$ : the function $y = \log_a x$ is strictly decreasing, so inequality directions reverse

This is crucial! Missing this step is one of the most common mistakes in solving exponential inequalities.

lightbulbExample

Worked Example: Solving an Exponential Inequality

Solve the inequality $0.7^x \geq 0.3$ .

Solution:

Method 1: Taking logarithms of both sides

We can take $\log_{10}$ of both sides:

\log_{10}(0.7^x) \geq \log_{10} 0.3

Using the power law:

x \log_{10} 0.7 \geq \log_{10} 0.3

Now we divide both sides by $\log_{10} 0.7$ . Since $\log_{10} 0.7 < 0$ (because $0.7 < 1$ ), we must reverse the inequality direction:

x \leq \frac{\log_{10} 0.3}{\log_{10} 0.7}

Method 2: Direct conversion using properties

Since $0 < 0.7 < 1$ , the function $y = 0.7^x$ is strictly decreasing. Therefore, the inequality $0.7^x \geq 0.3$ holds for:

x \leq \log_{0.7} 0.3

Both methods give the same result.

Summary of inequality rules

Understanding when to reverse inequality directions is essential for success with exponential inequalities.

bookmarkSummary

Key Rules for Exponential Inequalities

For equations: The equivalence works in both directions regardless of the base

$2^x = 5 \Leftrightarrow x = \log_2 5$
$0.3^x = 5 \Leftrightarrow x = \log_{0.3} 5$

For inequalities with base $a > 1$ : The direction stays the same

$2^x \geq 5 \Leftrightarrow x \geq \log_2 5$

For inequalities with base $0 < a < 1$ : The direction reverses

$0.3^x \geq 5 \Leftrightarrow x \leq \log_{0.3} 5$

Why does reversal occur? Exponential functions with bases between 0 and 1 are decreasing functions, so larger input values produce smaller output values.

Exam tips

infoNote

Common Mistakes to Avoid and Tips for Success

Always check whether you need to reverse the inequality when the base is between 0 and 1
When using a calculator, verify that your answer makes sense by substituting back into the original equation
Express your final answer to the required number of decimal places
Remember that $\log_a b$ only exists when $a > 0$ , $a \neq 1$ , and $b > 0$
Show clear working steps, especially when rearranging before converting to logarithmic form

Remember!

bookmarkSummary

Key Points to Remember

Core equivalence: The statements $a^x = b$ and $\log_a b = x$ are equivalent. This is your primary tool for solving exponential equations.
Basic method: For simple equations like $2^x = 11$ , directly convert to logarithmic form: $x = \log_2 11$ .
Complex equations: For equations like $3^{2x-1} = 28$ , first rearrange to isolate the exponential term, then convert to logarithmic form.
Inequality direction matters: When solving inequalities with base $a > 1$ , the inequality direction stays the same. With base $0 < a < 1$ , the direction reverses because the function is decreasing.
Use technology wisely: Calculators can quickly provide decimal approximations, but you should understand the underlying logarithmic conversion process.

Solution of Exponential Equations Using Logarithms (VCE SSCE Mathematical Methods): Revision Notes