Review of Logarithmic Functions Revision Notes for HSC SSCE Mathematics Advanced

Review of Logarithmic Functions

This section reviews logarithmic functions with base $e$ , focusing on properties, transformations, and solving techniques. Two new topics are introduced: dilations of logarithmic functions and exponential and logarithmic equations reducible to quadratics.

The function $y = \log_e x$

Definition and inverse relationship

A logarithmic function is the inverse of an exponential function. Since exponential functions are one-to-one, they have inverse functions.

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The Key Relationship:

$y = \log_e x$ means $x = e^y$

This can be understood with the phrase: "The log is the index, when the number is written as a power of the base."

For example:

$3 = \log_2 8$ because $8 = 2^3$
$y = \log_e x$ means $x = e^y$

The inverse relationship algebraically

Since $y = \log_e x$ is the inverse function of $y = e^x$ , composing these two functions in either order gives the identity function:

$\log_e e^x = x \text{ for all real } x$

$e^{\log_e x} = x \text{ for all } x > 0$

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These identities are fundamental for simplifying expressions and solving equations. You will use them frequently when working with logarithmic and exponential expressions.

Graphical properties

When both functions are graphed together, they appear as reflections of each other in the diagonal line $y = x$ .

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Key Features of the Graphs:

Domain and range:

For $y = e^x$ : domain is $(-\infty, \infty)$ and range is $(0, \infty)$
For $y = \log_e x$ : domain is $(0, \infty)$ and range is $(-\infty, \infty)$

Notice that the domain and range are swapped between the two functions.

Asymptotes:

$y = e^x$ has a horizontal asymptote at $y = 0$
$y = \log_e x$ has a vertical asymptote at $x = 0$

Intercepts:

$y = e^x$ crosses the $y$ -axis at $(0, 1)$
$y = \log_e x$ crosses the $x$ -axis at $(1, 0)$
Both curves have gradient $1$ at their intercepts

Rate of change:

Both functions are increasing throughout their domains
$y = e^x$ increases at an increasing rate (concave up)
$y = \log_e x$ increases at a decreasing rate (concave down)

One-to-one property: Both functions are one-to-one, meaning each output value corresponds to exactly one input value.

Notation and calculator usage

Standard notation

The function can be written in two equivalent ways:

$y = \log_e x$ (emphasises that the base is $e$ )
$y = \ln x$ (pronounced "logs naperian" or "logs natural")

Both notations mean exactly the same thing.

Calculator usage

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Calculator Warning:

Be careful when using your calculator!

The $\ln$ button means $\log_e x$ (natural logarithm)
The $\log$ button means $\log_{10} x$ (common logarithm)

In mathematics, when you see " $\log x$ " without a base specified, it usually means $\log_e x$ . However, this is not true on calculators or in other fields.

Transformations of $y = \log_e x$

Logarithmic functions can be transformed using reflections, shifts, and dilations, just like other functions.

Example: Reflections and shifts

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Worked Example: Reflection in the $y$ -axis

a) For $y = \log_e(-x)$ :

This reflects the graph across the $y$ -axis
Domain becomes $x < 0$
$x$ -intercept is at $(-1, 0)$
Vertical asymptote remains at $x = 0$

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Worked Example: Vertical shift down

b) For $y = \log_e x - 2$ :

This shifts the entire graph down by $2$ units
Domain remains $x > 0$
$x$ -intercept moves to $(e^2, 0)$
Vertical asymptote remains at $x = 0$

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Worked Example: Horizontal shift left

c) For $y = \log_e(x + 3)$ :

This shifts the entire graph left by $3$ units
Domain becomes $x > -3$
$x$ -intercept is at $(-2, 0)$
Vertical asymptote moves to $x = -3$

Dilations

Some transformations can be achieved using either dilations or shifts, due to logarithm properties.

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Worked Example: Horizontal dilation

a) For $y = \log_e(3x)$ :

To graph this function, apply a horizontal dilation to $y = \log_e x$ with factor $\frac{1}{3}$ (the graph is compressed horizontally).

Key features:

The point $(1, 0)$ moves to $\left(\frac{1}{3}, 0\right)$
The point $(e, 1)$ moves to $\left(\frac{e}{3}, 1\right)$

Alternatively, using logarithm laws: $y = \log_e(3x) = \log_e 3 + \log_e x$ , which is a vertical shift up by $\log_e 3$ .

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Worked Example: Vertical dilation

b) For $y = 3\log_e x$ :

To graph this function, apply a vertical dilation to $y = \log_e x$ with factor $3$ (the graph is stretched vertically).

Key features:

The $x$ -intercept remains at $(1, 0)$
The point $(e, 1)$ moves to $(e, 3)$

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The vertical shift down by $2$ can also be written as $y = \log_e x - 2 = \log_e x - \log_e e^2 = \log_e(e^{-2}x)$ , which represents a horizontal dilation with factor $e^2 \approx 7.39$ .

Using the inverse identities

The inverse identities allow us to simplify many expressions involving logarithms and exponentials.

Simplification examples

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Worked Example: Using $\log_e e^x = x$

$\log_e e^6 = 6$
$\log_e e = \log_e e^1 = 1$
$\log_e \frac{1}{e} = \log_e e^{-1} = -1$
$\log_e \frac{1}{\sqrt{e}} = \log_e e^{-\frac{1}{2}} = -\frac{1}{2}$

Using $e^{\log_e x} = x$ :

$e^{\log_e 10} = 10$
$e^{\log_e 0.1} = 0.1$

These identities are particularly useful for solving equations.

Converting between exponential and logarithmic forms

Being able to convert between exponential and logarithmic statements is essential.

The key relationship

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Remember: " $y = \log_e x$ means $x = e^y$ "

The base of the exponential becomes the base of the logarithm, and the logarithm is the power (index).

Conversion examples

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Worked Example: Converting between forms

From exponential to logarithmic form:

If $x = e^3$ , then $\log_e x = 3$
If $e^x = 10$ , then $x = \log_e 10$

From logarithmic to exponential form:

If $\log_e x = -1$ , then $x = e^{-1} = \frac{1}{e}$
If $x = \log_e 10$ , then $e^x = 10$
If $e^x = \frac{1}{2}$ , then $x = \log_e \frac{1}{2}$

The change-of-base formula

Sometimes we need to work with logarithms in different bases. The change-of-base formula allows us to convert any logarithm to base $e$ .

Formula

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Change-of-Base Formula:

For any positive base $b \neq 1$ :

$\log_b x = \frac{\log_e x}{\log_e b}$

This can be remembered as: "The log of the number over the log of the base."

Using the formula

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Worked Example: Solving $2^x = 100$

Solve $2^x = 100$ correct to three decimal places.

Step 1: Recognise that $2^6 = 64$ and $2^7 = 128$ , so the answer is between $6$ and $7$ .

Step 2: Convert to logarithmic form: $x = \log_2 100$

Step 3: Use the change-of-base formula: $x = \frac{\log_e 100}{\log_e 2} \approx 6.644$

Alternative method: Take logarithms base $e$ of both sides of the original equation:

$2^x = 100$ $\log_e(2^x) = \log_e 100$ $x \log_e 2 = \log_e 100$ $x = \frac{\log_e 100}{\log_e 2} \approx 6.644$

Solving equations reducible to quadratics

Some exponential and logarithmic equations can be transformed into quadratic equations using substitution.

Exponential equations

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Worked Example: Solving $4^x - 7 \times 2^x + 12 = 0$

Step 1: Notice that $4^x = (2^x)^2$ . The equation becomes: $(2^x)^2 - 7 \times 2^x + 12 = 0$

Step 2: Let $u = 2^x$ : $u^2 - 7u + 12 = 0$

Step 3: Factorise: $(u - 4)(u - 3) = 0$ $u = 4 \text{ or } u = 3$

Step 4: Return to $x$ : $2^x = 4 \text{ or } 2^x = 3$ $x = 2 \text{ or } x = \log_2 3$

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Worked Example: Solving $3e^{2x} - 11e^x - 4 = 0$

Step 1: Notice that $e^{2x} = (e^x)^2$ . The equation becomes: $3(e^x)^2 - 11e^x - 4 = 0$

Step 2: Let $u = e^x$ : $3u^2 - 11u - 4 = 0$

Step 3: Factorise using sum = $-11$ , product = $-12$ :

3u^2 - 12u + u - 4 = 0 3u(u - 4) + (u - 4) = 0 (3u + 1)(u - 4) = 0

$u = -\frac{1}{3} \text{ or } u = 4$

Step 4: Return to $x$ : $e^x = -\frac{1}{3} \text{ or } e^x = 4$

Since $e^x$ is always positive, $e^x = -\frac{1}{3}$ has no solution.

Therefore: $e^x = 4$ , so $x = \log_e 4$

Logarithmic equations

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Worked Example: Solving $\log_e x - \frac{9}{\log_e x} = 0$

Method 1 (with substitution):

Step 1: Let $u = \log_e x$ : $u - \frac{9}{u} = 0$

Step 2: Multiply both sides by $u$ : $u^2 - 9 = 0$

Step 3: Factorise: $(u - 3)(u + 3) = 0$ $u = 3 \text{ or } u = -3$

Step 4: Return to $x$ : $\log_e x = 3 \text{ or } \log_e x = -3$ $x = e^3 \text{ or } x = e^{-3}$

Method 2 (without substitution):

Step 1: Multiply both sides by $\log_e x$ : $(\log_e x)^2 - 9 = 0$

Step 2: Solve for $\log_e x$ : $(\log_e x)^2 = 9$ $\log_e x = 3 \text{ or } \log_e x = -3$

Step 3: Convert to exponential form: $x = e^3 \text{ or } x = e^{-3}$

Remember!

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Key Points to Remember:

The function $y = \log_e x$ is the inverse of $y = e^x$ , meaning their graphs are reflections of each other in the line $y = x$ .
The key inverse identities are $\log_e e^x = x$ (for all real $x$ ) and $e^{\log_e x} = x$ (for all $x > 0$ ).
On your calculator, $\ln$ means $\log_e$ (natural logarithm), while $\log$ means $\log_{10}$ (common logarithm).
To convert between bases, use the change-of-base formula: $\log_b x = \frac{\log_e x}{\log_e b}$ .
Complex exponential and logarithmic equations can often be solved by recognising quadratic patterns and using substitution methods.

Review of Logarithmic Functions (HSC SSCE Mathematics Advanced): Revision Notes