A. Rate of Change · Derivatives of Exponential and Sinusoidal Functions

The Natural Logarithm: ln x as the Inverse of eˣ

A2.7About 65 minutes

Learning goals

I can explain that ln x means logₑ x and that it is the inverse of eˣ.
I can sketch y = eˣ and y = ln x as reflections of each other in the line y = x and state their key features.
I can use the facts e^(ln x) = x (for x > 0) and ln(eˣ) = x to simplify expressions.
I can solve exponential and logarithmic equations involving e and ln, exactly and with a calculator.

Words to know

natural logarithm ln x
The logarithm with base e: ln x = logₑ x. It answers the question "e to what power gives x?"

inverse functions
Two functions that undo each other. Their graphs are reflections in the line y = x.

composition
Applying one function to the output of another, as in ln(eˣ).

vertical asymptote
A vertical line x = a that a graph approaches but never reaches. For y = ln x it is x = 0.

change of base
logₐ x = ln x ÷ ln a, which lets you compute any logarithm with the ln key.

The key that undoes eˣ

In the last lesson you met the natural exponential function eˣ. Questions like "when does the tangent to y = eˣ have slope 5?" lead to equations such as eˣ = 5. To solve them you need a function that undoes eˣ. That function is the natural logarithm, written ln x.

Remember from MHF4U: in Exponential and Logarithmic Functions you learned that y = log_b x means bʸ = x, that y = bˣ and y = log_b x are inverses, and the laws of logarithms. You also saw that your calculator has an ln key and that it means log base e. Now e gets its full role.

Definition

ln x = logₑ x. So y = ln x means eʸ = x.

In words: ln x is the exponent you put on e to get x. For example, ln 20 ≈ 2.996 because e^2.996 ≈ 20.

Two functions that undo each other

Because ln and eˣ are inverses, applying one after the other brings you back to where you started:

ln(eˣ) = x for every real x.
e^(ln x) = x for every x > 0.

The second fact needs x > 0 because ln x only exists for positive x: no power of e is zero or negative.

Function machines: put 3 into eˣ and then into ln and you get 3 back. Put 5 into ln and then into eˣ and you get 5 back. Each function undoes the other.

The graphs are mirror images

Swapping x and y reflects a graph in the line y = x. Every point (a, b) on y = eˣ becomes the point (b, a) on y = ln x. For example, (0, 1) becomes (1, 0), and (1, e) becomes (e, 1).

y = eˣ (solid) and y = ln x (long dashes) are reflections in the line y = x. The dotted segments join matching points, such as (1, e) and (e, 1).

feature	y = eˣ	y = ln x
domain	all real numbers	x > 0
range	y > 0	all real numbers
intercept	y-intercept 1	x-intercept 1
asymptote	horizontal, y = 0	vertical, x = 0
increasing or decreasing?	increasing everywhere	increasing everywhere
key points	(0, 1), (1, e)	(1, 0), (e, 1)

Domain and range swap, and so do the intercepts and the asymptotes.

All the log laws still work

ln is a logarithm, so every law from MHF4U applies, with base e:

ln(xy) = ln x + ln y
ln(x ÷ y) = ln x − ln y
ln(xⁿ) = n ln x
ln 1 = 0 and ln e = 1

The change of base formula lets you compute any logarithm with the ln key: logₐ x = ln x ÷ ln a.

One more identity will be the key to the next lesson. Since a = e^(ln a), we can write any power of a as a power of e:

aˣ = (e^(ln a))ˣ = e^(x ln a).

For example, 2ˣ = e^(0.6931x). Look at that number: ln 2 ≈ 0.6931. You have seen it before. It is the constant k for 2ˣ from the first lesson of this unit. Keep that in mind.

Worked example

Example 1: Simplifying

Evaluate exactly: (a) ln(e⁵) (b) e^(ln 7) (c) e^(2 ln 3) (d) ln(1/e²)

(a) ln and eˣ undo each other: ln(e⁵) = 5.
(b) e^(ln 7) = 7.
(c) Use the power law first: 2 ln 3 = ln(3²) = ln 9. So e^(2 ln 3) = e^(ln 9) = 9.
(d) 1/e² = e⁻², so ln(e⁻²) = −2.

(a) 5 (b) 7 (c) 9 (d) −2

Worked example

Example 2: Solving an exponential equation

Solve 3e^(0.2t) = 15 exactly, then to two decimal places.

Isolate the exponential: divide by 3 to get e^(0.2t) = 5.
Take ln of both sides: ln(e^(0.2t)) = ln 5, so 0.2t = ln 5.
Divide by 0.2: t = 5 ln 5.
Calculator: 5 × 1.6094 ≈ 8.05.

t = 5 ln 5 ≈ 8.05

Worked example

Example 3: Solving a logarithmic equation

Solve ln(x − 2) = 3.

Rewrite in exponential form: e³ = x − 2.
So x = e³ + 2 ≈ 20.09 + 2 = 22.09.
Check the domain: x − 2 = e³ > 0, so the logarithm is defined. ✓

x = e³ + 2 ≈ 22.09

Worked example

Example 4: Writing a power of 10 with base e

Write 10ˣ in the form e^(cx) and give c to four decimal places.

10 = e^(ln 10), so 10ˣ = e^(x ln 10).
c = ln 10 ≈ 2.3026.

10ˣ = e^(2.3026x) (approximately). Compare with the k value for base 10 in the first lesson: 2.303.

Hands-on activity

Fold eˣ into ln x

You need: a sheet of graph paper, a ruler, a pencil, a calculator, and a window or a bright lamp to see through the paper.

Draw x- and y-axes from −3 to 5 with the same scale on both (for example 1 cm per unit). Draw the line y = x lightly.
Plot y = eˣ using x = −2, −1, 0, 0.5, 1, 1.5 (y ≈ 0.14, 0.37, 1, 1.65, 2.72, 4.48). Join the points with a smooth curve and go over it firmly.
Fold the paper exactly along the line y = x, with the curve on the inside. Hold it up to the window and trace the curve through the paper onto the other half.
Unfold. The traced curve is y = ln x. Check three points with your calculator: is ln 1 = 0, ln 2.72 ≈ 1 and ln 4.48 ≈ 1.5?
Measure the slope of each curve where it crosses an axis (at (0, 1) and at (1, 0)) using a ruler laid along the curve. What do you notice?

What to look for: the traced curve should pass through (1, 0), (2.72, 1) and (4.48, 1.5) and hug the negative y-axis without touching it. Both curves cross an axis with slope about 1, because a reflection in y = x turns a slope of 1 into a slope of 1 ÷ 1 = 1.

Just for fun

"ln" comes from the Latin logarithmus naturalis, which is why it is ln and not nl. So every time you press the ln key, you are speaking a little Latin.

Your turn

Evaluate ln(e⁻³).

Evaluate e^(ln 0.4).

Solve eˣ = 12. Give the exact answer and a decimal to three places.

Solve 4e^(2x) = 20.

Solve ln x = −1.

Simplify ln(e²) + ln(1) − ln(e⁴).

State the domain and the asymptote of y = ln x.

More practice: warm-up

What does ln x mean in words?

ln 1 = ?

ln e = ?

The point (2, e²) is on y = eˣ. Which point is on y = ln x?

Use the ln key to find ln 50 to three decimal places.

More practice: on your own

Solve 2000e^(0.03t) = 4000 (a doubling time).

Write 3ˣ as a power of e and give the coefficient to three decimal places.

Solve ln(2x + 1) = 2.

Use change of base to compute log₂ 100.

Solve e^(2x) − 5eˣ + 6 = 0. (Hint: let u = eˣ.)

Challenge

Explain why the equation eˣ = −4 has no solution, using the graph of y = eˣ.

The tangent to y = eˣ at (a, eᵃ) has slope eᵃ. Use the reflection in y = x to find the slope of the tangent to y = ln x at (eᵃ, a), and write it in terms of the x-coordinate of that point.

Solve eˣ + e⁻ˣ = 4 to three decimal places. (Hint: multiply both sides by eˣ.)

Check your understanding

Quiz

ln x is the inverse of which function?

What is ln(e⁷)?

The graph of y = ln x is the reflection of y = eˣ in:

Which is the solution of e^(3x) = 10?

Which statement about y = ln x is true?

e^(3 ln 2) equals:

For parents and teachers

Big idea (A2.7): ln x is the logarithm with base e, the inverse of eˣ. Each undoes the other (ln(eˣ) = x and e^(ln x) = x), and their graphs are mirror images in the line y = x. Writing aˣ = e^(x ln a) sets up the next lesson.

How to help: ask your teen to explain what ln 20 means in words ("the power of e that gives 20") and then check it on the calculator: press ln 20, then raise e to that answer. You should get 20 back.

Common difficulties:

Thinking ln(x + y) = ln x + ln y. The product law is ln(xy) = ln x + ln y.
Taking ln of a negative number or zero (undefined).
Dividing by ln instead of taking ln of both sides ("ln" on its own is not a number).
Forgetting to isolate the exponential first: in 3e^(0.2t) = 15, divide by 3 before taking ln.