01 · Exact values
Find ln(e⁴), ln(1) and ln(e⁻²).
Hint
ln(eᵏ) = k.
Worked solution
4, 0 and −2.
Understand · explore · practise
Natural logarithms (ln)
Understand ln as logarithm to base e, use inverse identities, sketch natural logarithm transformations and solve simple equations exactly.
Before you startExponential graphs and the meaning of a logarithm
01 / What does ln mean?
It is the logarithm whose base is e.
ln(x) = loge(x)
ln(x) = y ⇔ eʸ = x, with x > 0
ln(1) = 0
ln(e) = 1
ln(e²) = 2
ln(1/e) = −1
A natural logarithm can be negative. For example, ln(1/e) is −1 because e⁻¹ = 1/e. It is the input, not the result, that must be positive.
02 / The inverse graphs
Swap the input and output coordinates.
If (t,eᵗ) lies on y = eˣ, then (eᵗ,t) lies on y = ln(x). Swapping coordinates reflects a graph in the line y = x.
Move the point to see the pair. The graph uses equal coordinate scales so the reflection is shown faithfully. The exponential point can have negative x, but the logarithm point always has positive x.
Move an inverse pairMove at your pace
At t = 0, the exponential point is (0,1). Its reflected logarithm point is (1,0), so ln(1) is approximately 0. The displayed exponential value is rounded; the exact pair is (t,e^t) and (e^t,t). The gold line is y = x and the coordinate scales are equal.
Watch exponential points reflect into natural-logarithm points
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Sketch y = ln(x)
The positive-input restriction gives a vertical boundary.
Domain: x > 0
Range: all real y
x-intercept: (1,0)
Vertical asymptote: x = 0
The graph increases throughout its domain. It tends to −∞ as x approaches zero from the right, and grows without bound as x increases, though increasingly slowly. There is no y-intercept because x = 0 is not allowed.
The domain comes from the inverse relationship with the positive-valued exponential function. The presence of an asymptote on a sketch is not, by itself, a general rule for deciding a function’s domain.
04 / Use the inverse identities
Check which expression is inside which function.
ln(eˣ) = x, for every real x
eln x = x, for x > 0
ln(e3x−1) = 3x − 1
eln(2x−5) = 2x − 5, with x > 5/2
The second simplification retains the original logarithm’s domain. It does not extend the expression to every real x just because 2x − 5 can be written there.
A sum does not cancel term by term
ln(e² + e³) = ln[e²(1 + e)]
= 2 + ln(1 + e)
It is not 2 + 3. Factor the sum before using the product law.
05 / Transform a natural-logarithm graph
Make the logarithm’s input positive first.
For y = 2 + ln(3 − x), the condition 3 − x > 0 gives x < 3. The graph is decreasing, with vertical asymptote x = 3.
y-intercept: (0, 2 + ln 3)
At y = 0:
ln(3 − x) = −2
x = 3 − e⁻²
Thus its x-intercept is (3 − e⁻²,0). Its range is all real numbers. The outside +2 shifts the graph vertically and does not change its domain.
06 / A scale and shift inside ln
Equivalent forms must keep the same domain.
For y = ln(2x − 4), the domain is x > 2:
ln(2x − 4) = ln[2(x − 2)]
= ln 2 + ln(x − 2), for x > 2
The vertical asymptote is x = 2. The x-intercept solves 2x − 4 = 1, giving x = 5/2. There is no y-intercept because zero is outside the domain.
07 / Solve simple inverse equations
Undo the outer function before solving the linear part.
e3x−2 = 7
3x − 2 = ln 7
x = (ln 7 + 2)/3
ln(4x − 1) = 2
4x − 1 = e²
x = (e² + 1)/4
The second answer satisfies x > 1/4. Keep e and ln in an exact answer; a decimal approximation may be useful only after that.
For example, ln(x) = −3 gives x = e⁻³, which is positive. Rejecting a negative logarithm value would incorrectly lose this solution.
08 / Your turn
Apply the inverse operation and retain the domain.
Give exact answers unless a decimal is requested.
02 · Undo an exponential
Solve e2x+1 = 9.
Hint
Take ln of both positive sides.
Worked solution
x = (ln 9 − 1)/2.
03 · Undo a logarithm
Solve ln(3x + 2) = 4.
Hint
Set 3x + 2 = e⁴.
Worked solution
x = (e⁴ − 2)/3, satisfying x > −2/3.
04 · Negative output
Solve ln(x) = −2.
Hint
The input can lie between zero and one.
Worked solution
x = e⁻².
05 · A shifted graph
Give the domain and asymptote of y = ln(x + 5).
Hint
x + 5 must be positive.
Worked solution
Domain x > −5; vertical asymptote x = −5.
06 · Reflection
The point (−2,e⁻²) lies on y = eˣ. Give the corresponding point on y = ln(x).
Hint
Swap the coordinates.
Worked solution
(e⁻²,−2).
07 · An intercept
Find the x-intercept of y = 1 + ln(4 − x).
Hint
At y = 0, the logarithm equals −1.
Worked solution
(4 − e⁻¹,0), within the domain x < 4.
08 · Preserve the domain
Simplify eln(x−2) and state its domain.
Hint
The original logarithm needs a positive input.
Worked solution
x − 2, defined here only for x > 2.
09 · A sum inside ln
Simplify ln(e³ + e⁴) as far as the laws allow.
Hint
Factor out e³.
Worked solution
3 + ln(1 + e).
10 · A scaled input
Find the domain, asymptote and x-intercept of y = ln(5x − 10).
Hint
Use input > 0 for the domain and input = 1 for the intercept.
Worked solution
Domain x > 2; asymptote x = 2; x-intercept (11/5,0).
09 / Recap
ln and e undo one another on the right domains.
- ln is logarithm to base e.
- The input to ln is positive; its output can be any real number.
- The graphs of exp and ln reflect in y = x.
- Use the inside expression to find a transformed logarithm’s domain.
- Keep exact e and ln expressions when requested.
- Inverse simplification never removes an original domain restriction.
Section 1 of 9 · What does ln mean?