Using Logarithms to Solve Exponential Equations and Inequalities Revision Notes for VCE SSCE Mathematical Methods

Using Logarithms to Solve Exponential Equations and Inequalities

Understanding the fundamental relationship

When working with exponential equations and inequalities, logarithms provide a powerful tool for finding solutions. The key principle is the equivalence between exponential and logarithmic forms.

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For any positive number $a$ where $a \neq 1$ , the following statements mean exactly the same thing:

$a^x = b$ (exponential form)
$\log_a b = x$ (logarithmic form)

This equivalence works because logarithms are defined as the inverse operation of exponentiation. When you see an exponential equation, you can immediately rewrite it as a logarithmic equation, and vice versa.

This relationship is the foundation for solving all exponential equations and inequalities.

Solving exponential equations

Method 1: Converting to logarithmic form

The most direct way to solve an exponential equation is to convert it to logarithmic form using the equivalence principle above.

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Worked Example: Simple exponential equation

Find the value of $x$ in the equation $2^x = 11$ .

Solution:

Converting to logarithmic form gives us:

x = \log_2 11

We can calculate this using a calculator:

x = \log_2 11 \approx 3.45943

This method is straightforward and works well when you have access to a calculator that can compute logarithms with different bases.

Method 2: Using the change of base formula

An alternative approach uses common logarithms (base 10) along with a useful formula called the change of base formula. This method is particularly helpful when your calculator only has buttons for $\log_{10}$ or natural logarithms.

Using the same equation $2^x = 11$ , we apply the logarithm base 10 to both sides:

\log_{10}(2^x) = \log_{10} 11

Using the power law of logarithms, we bring down the exponent:

x \log_{10} 2 = \log_{10} 11

Solving for $x$ :

x = \frac{\log_{10} 11}{\log_{10} 2}

This demonstrates that:

\log_2 11 = \frac{\log_{10} 11}{\log_{10} 2}

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The change of base formula

In general, for any positive numbers $a$ , $b$ , and $c$ where $a \neq 1$ and $b \neq 1$ :

\log_a c = \frac{\log_b c}{\log_b a}

This formula allows you to convert any logarithm to a base you can work with (usually base 10 or base $e$ ). The numerator is the logarithm of the number you're interested in, and the denominator is the logarithm of the original base.

Worked example: Equations with expressions in the exponent

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Worked Example: Exponential equation with expression in exponent

Find the value of $x$ in the equation $3^{2x-1} = 28$ .

Solution:

Converting to logarithmic form:

2x - 1 = \log_3 28

2x = 1 + \log_3 28

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

x \approx 2.017 \text{ (to three decimal places)}

When the exponent contains an expression like $2x - 1$ , we first convert to logarithmic form, then solve the resulting linear equation for $x$ .

Solving exponential inequalities

Exponential inequalities follow similar principles to equations, but we must be careful about the direction of the inequality sign.

The crucial rule about inequality direction

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Critical Rule: When to reverse inequality signs

When solving inequalities involving exponentials, the direction of the inequality sign depends on the base:

If the base is greater than 1, the inequality direction stays the same
If the base is between 0 and 1, the inequality direction reverses

Why this happens:

When the base is greater than 1, larger exponents give larger values
When the base is between 0 and 1, larger exponents give smaller values

Worked example: Inequality with base less than 1

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Worked Example: Inequality with fractional base

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

Method 1: Using common logarithms

Apply the logarithm base 10 to both sides:

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

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

Now we divide both sides by $\log_{10} 0.7$ . Because $\log_{10} 0.7$ is negative (since $0.7 < 1$ ), the inequality sign flips:

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

x \leq 3.376 \text{ (to three decimal places)}

Method 2: Direct conversion

We can also solve this by recognizing that $0 < 0.7 < 1$ . As $x$ decreases, $y = 0.7^x$ increases (the function is decreasing). Therefore, the inequality holds for $x \leq \log_{0.7} 0.3$ .

Both methods give the same answer, but Method 2 is more efficient once you understand the behaviour of exponential functions with bases less than 1.

Summary of inequality rules

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Pattern for exponential inequalities:

When base $> 1$ :

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

(inequality direction unchanged)

When $0 <$ base $< 1$ :

(0.3)^x \geq 5 \Leftrightarrow x \leq \log_{0.3} 5

(inequality direction reversed)

Application to exponential graphs

Now that we can use logarithms, we can find exact $x$ -intercepts for exponential graphs. Previously, we could only estimate these intercepts or describe them as "somewhere between" certain values.

Finding intercepts using logarithms

When sketching exponential graphs, we typically need to find:

The equation of any horizontal asymptote
The $y$ -intercept (by substituting $x = 0$ )
The $x$ -intercept (by setting $f(x) = 0$ and solving)

The $x$ -intercept now requires logarithms to solve exactly.

Worked example: Complete graph sketch

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Worked Example: Sketching an exponential graph

Sketch the graph of $f(x) = 2 \times 10^x - 4$ , giving the equation of the asymptote and the axis intercepts.

Solution:

Finding the asymptote:

As $x \to -\infty$ , we have $10^x \to 0^+$ , so $y \to -4^+$ .

The equation of the horizontal asymptote is $y = -4$ .

Finding the $y$ -intercept:

When $x = 0$ :

f(0) = 2 \times 10^0 - 4 = 2 - 4 = -2

The $y$ -intercept is at $-2$ .

Finding the $x$ -intercept:

Set $f(x) = 0$ :

2 \times 10^x - 4 = 0

2 \times 10^x = 4

10^x = 2

x = \log_{10} 2

x \approx 0.3010 \text{ (to four decimal places)}

The graph crosses the $x$ -axis at exactly $\log_{10} 2$ , crosses the $y$ -axis at $-2$ , and has a horizontal asymptote at $y = -4$ .

Key takeaways

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Remember these essential points:

Equivalence principle: $a^x = b \Leftrightarrow \log_a b = x$ - use this to convert between exponential and logarithmic forms
Change of base formula: $\log_a c = \frac{\log_b c}{\log_b a}$ - use this to calculate logarithms with any base using base 10 or base $e$
Inequality direction: When solving inequalities, flip the inequality sign if the base is between 0 and 1, or if you divide by a negative number
Finding $x$ -intercepts: Logarithms allow us to find exact $x$ -intercepts of exponential graphs by solving equations like $a \times b^x + c = 0$
Always check your work: Substitute your answer back into the original equation to verify it's correct

Using Logarithms to Solve Exponential Equations and Inequalities (VCE SSCE Mathematical Methods): Revision Notes