Exponential-Logarithmic Equivalence Revision Notes for HSC SSCE Mathematics Advanced

Exponential-Logarithmic Equivalence

Introduction

This note explores the fundamental relationship between exponential and logarithmic functions. Understanding how to convert between these two forms is essential for solving a wide range of equations in mathematics. You'll learn how to transform exponential equations into logarithmic form and vice versa, and discover how the change of base formula enables you to evaluate logarithms using any base.

Learning objectives:

After studying this note, you will be able to:

Convert exponential equations into logarithmic form to solve for unknown exponents
State and apply the change of base formula for logarithms
Use a calculator to evaluate logarithms with any valid base
Solve exponential equations that require the change of base formula

Understanding exponential-logarithmic equivalence

The equivalence between exponential and logarithmic forms is the key to solving equations where the unknown variable appears in the exponent. This relationship allows us to "unlock" the exponent and solve for it directly.

The fundamental relationship

The exponential equation $a^x = b$ can be rewritten in logarithmic form as $x = \log_a(b)$ , where:

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

infoNote

The double arrow ( $\iff$ ) means these two forms are completely equivalent—they express the same mathematical relationship in different ways.

Components:

$a$ is the base (must be 10 or $e$ , or any positive number not equal to 1)
$b$ is the result (must be positive, $b > 0$ )
$x$ is the exponent (the unknown we're solving for)

When to use this equivalence

Use the exponential-logarithmic equivalence when:

You need to solve for a variable that appears as an exponent
You have an exponential equation of the form $a^x = b$
The base is 10, $e$ , or another number that can be converted using logarithms

chatImportant

Remember the restrictions: the base $a$ must satisfy $a > 0$ and $a \neq 1$ , and the result $b$ must be positive ( $b > 0$ ). Without these conditions, the logarithm is undefined.

Solving exponential equations using logarithms

When an unknown appears in the exponent, we can use logarithms to bring it down where we can work with it algebraically.

lightbulbExample

Worked Example: Solving $e^x = 10$

Let's solve for $x$ in the equation $e^x = 10$ , rounding to four decimal places.

Strategy: Since the base is $e$ (Euler's number), we use the natural logarithm (ln) as its inverse to solve for $x$ .

Solution steps:

$e^x = 10$

Converting to logarithmic form using the equivalence:

$x = \ln(10)$

Evaluating using a calculator:

$x \approx 2.3026$

Verification: We can check our answer by calculating $e^{2.3026} \approx 10.0002 \approx 10$ , confirming our solution is correct.

lightbulbExample

Worked Example: Solving $10^x = 30$

Now let's solve for $x$ in the equation $10^x = 30$ , rounding to four decimal places.

Strategy: Since the base is 10, we use the common logarithm ( $\log_{10}$ ) as its inverse to solve for $x$ .

Solution steps:

$10^x = 30$

Converting to logarithmic form using the equivalence:

$x = \log_{10}(30)$

Evaluating using a calculator:

$x \approx 1.4771$

Verification: We can check our answer by calculating $10^{1.4771} \approx 30.0006 \approx 30$ , confirming our solution is correct.

lightbulbExample

Worked Example: Solving logarithmic equations

Sometimes we need to work in reverse, converting a logarithmic equation to exponential form. Let's solve $\log_4(16^x) = 3$ in exact form.

Strategy: Convert the logarithmic equation to exponential form using the equivalence $\log_a(b) = c \iff a^c = b$ . Then express both sides with the same base and equate the exponents.

Solution steps:

$\log_4(16^x) = 3$

Converting to exponential form:

$4^3 = 16^x$

Expressing both sides with base 2:

$(2^2)^3 = (2^4)^x$

Applying the index exponent law $(a^m)^n = a^{mn}$ :

$2^6 = 2^{4x}$

Since the bases are equal, we can equate the exponents:

$6 = 4x$

Solving for $x$ :

$x = \frac{6}{4} = \frac{3}{2}$

Verification:

We can verify: $\log_4\left(16^{\frac{3}{2}}\right)$

Since $16 = 4^2$ , we have $16^{\frac{3}{2}} = (4^2)^{\frac{3}{2}} = 4^3 = 64$

Then $\log_4(64) = 3$ since $4^3 = 64$ , confirming our solution.

infoNote

The equivalence $y = a^x \iff x = \log_a(y)$ enables us to solve equations of the form $a^x = b$ by rewriting with the same base or using logarithms, provided $a > 0$ , $a \neq 1$ , and $b > 0$ .

The change of base formula

When working with logarithms that have bases other than 10 or $e$ , we need a way to evaluate them using a calculator. The change of base formula provides this crucial conversion.

The formula

The change of base rule states that a logarithm with one base can be expressed as the ratio of two logarithms with a different base:

$\log_a(x) = \frac{\log_b(x)}{\log_b(a)}$

where:

$a, b$ are bases with $a, b > 0$ and $a, b \neq 1$
$x$ is a positive number

chatImportant

Why we need it:

The change of base formula is essential because:

Calculators typically only have buttons for $\log_{10}$ (common logarithm) and $\ln$ (natural logarithm)
Exponential equations may have bases other than 10 or $e$
The formula allows us to convert any logarithm to base 10 or base $e$ for calculation
It enables algebraic simplification of expressions with different bases

Proof of the change of base formula

Understanding why this formula works helps you remember and apply it correctly.

Let $y = \log_a(x)$ , so by definition $a^y = x$ .

Let $z = \log_b(a)$ , so by definition $b^z = a$ .

Starting with the exponential form:

$a^y = x$

Substituting $a = b^z$ :

$(b^z)^y = x$

Applying the index exponent law:

$b^{zy} = x$

Taking $\log_b$ of both sides:

$\log_b(b^{zy}) = \log_b(x)$

By the definition of logarithms, $\log_b(b^{zy}) = zy$ :

$zy = \log_b(x)$

Dividing both sides by $z$ :

$y = \frac{\log_b(x)}{z}$

Substituting back $z = \log_b(a)$ :

$y = \frac{\log_b(x)}{\log_b(a)}$

Since $y = \log_a(x)$ :

$\log_a(x) = \frac{\log_b(x)}{\log_b(a)}$

This completes the proof.

infoNote

A helpful memory aid: "New base on bottom" – when changing to a new base $b$ , the logarithm of the original base $a$ goes in the denominator.

lightbulbExample

Worked Example: Evaluating $\log_2(27)$

Let's evaluate $\log_2(27)$ using the change of base formula, rounding to four decimal places.

Strategy: Apply the change of base formula to convert to base 10, which we can evaluate with a calculator.

Solution steps:

Using the change of base formula with $a = 2$ and $b = 10$ :

$\log_2(27) = \frac{\log_{10}(27)}{\log_{10}(2)}$

Evaluating each logarithm separately:

$= \frac{1.43136}{0.30103}$

Calculating the division:

$\approx 4.7549$

Verification: We can check our answer: $2^{4.7549} \approx 27$ , confirming the solution.

lightbulbExample

Worked Example: Solving $2^{x+1} = 11$

Let's solve for $x$ in the equation $2^{x+1} = 11$ , rounding to four decimal places.

Strategy: Since the base is 2 (not 10 or $e$ ), we'll convert to logarithmic form and then use the change of base formula to convert to base 10 for calculation.

Solution steps:

$2^{x+1} = 11$

Converting to logarithmic form:

$x + 1 = \log_2(11)$

Applying the change of base formula:

$x + 1 = \frac{\log_{10}(11)}{\log_{10}(2)}$

Evaluating each logarithm:

$x + 1 \approx 3.459431619$

Subtracting 1 from both sides:

$x \approx 2.459431619$

Rounding to four decimal places:

$x \approx 2.4594$

Verification: We can check: $2^{2.4594+1} = 2^{3.4594} \approx 11.0001 \approx 11$ , confirming our solution.

bookmarkSummary

Key Points to Remember:

Exponential-logarithmic equivalence: $a^x = b \iff x = \log_a(b)$ is the fundamental relationship that allows us to solve for exponents.
Converting forms: When $x$ is in the exponent and hard to isolate, convert the exponential equation to logarithmic form.
Change of base formula: $\log_a(x) = \frac{\log_b(x)}{\log_b(a)}$ converts logarithms to a base suitable for calculator evaluation (typically base 10 or $e$ ).
Verification strategy: Always check your answer by substituting back into the original equation to confirm your solution is correct.
Calculator requirements: Most calculators only have $\log_{10}$ and $\ln$ buttons, so the change of base formula is essential for evaluating logarithms with other bases.

Exponential-Logarithmic Equivalence (HSC SSCE Mathematics Advanced): Revision Notes