Pearson Edexcel · A-Level · Mathematics
Course version: 9MA0 · Version from Mar 2026
Know and use the definition of log_a(x) as the inverse of a^x, where a is…
Full specification wording
Exponentials and logarithms: 6.3 Know and use the definition of log_a(x) as the inverse of a^x, where a is positive and x ≥ 0. Know and use the function ln x and its graph. Know and use ln x as the inverse function of e^x.
The logarithm log_a(x) answers the question: to what power must a be raised to give x?
Checked against the Pearson Edexcel 9MA0 specification (opens in a new tab) on .
- Point 3 of 7
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
The logarithm log_a(x) answers the question: to what power must a be raised to give x? So y = log_a(x) exactly when a^y = x, making log_a the inverse of a^x. Because a^x is always positive, the domain of log_a(x) is x > 0, and its graph is the reflection of y = a^x in the line y = x, passing through (1, 0) with the y-axis as a vertical asymptote. The natural logarithm ln x is log_e(x), the inverse of e^x, so ln(e^x) = x and e^(ln x) = x for x > 0. Its graph passes through (1, 0) and (e, 1) and rises ever more slowly. These inverse relationships let you solve equations by taking logs or exponentiating both sides.
What earns the marks
- Defines log_a(x) as the power to which a must be raised to obtain x, and rewrites between forms a^y = x and y = log_a(x).
- Explains that log_a is the inverse function of a^x, so the two graphs are reflections in y = x.
- States the domain x > 0 and identifies the vertical asymptote at x = 0, with intercept (1, 0).
- Defines ln x as log_e(x) and uses the inverse identities ln(e^x) = x and e^(ln x) = x for x > 0.
- Sketches y = ln x with correct shape, intercept and asymptote, and relates it to the graph of y = e^x.
- Uses the inverse relationship to solve equations, for example converting e^(2x) = 7 into 2x = ln 7.
Common mistakes and how to put them right
Writing log_a(x) for negative or zero x.
Put it right by stating the domain x > 0 and rejecting invalid inputs.
Treating ln x as a power function such as x^n.
Put it right by sketching the logarithmic curve with its asymptote and slow growth.
Confusing log_a(x) with (log a) × x or with a^x.
Put it right by translating to the exponent form a^y = x.
Applying ln to a sum term by term, for example replacing ln(x + 3) with ln x + ln 3.
Put it right by noting no such law exists for sums.
Exam tips
- Convert between exponential and logarithmic form before attempting to solve, as this often reveals the required operation.
- Mark the intercept (1, 0) and the vertical asymptote whenever you sketch y = ln x or y = log_a(x).
- Use ln to undo e and e to undo ln, applying the same operation to both sides of an equation.
- Check that any solution lies in the domain x > 0 before stating it as final.
Try the lesson
The start of the Exponentials and logarithms: 6.3 Know and use the definition of log_a(x) as the inverse of a^x, where a is positive and x ≥ 0. Know and use the function ln x and its graph. Know and use ln x as the inverse function of e^x. lesson, made from checked teaching notes for this specification point.
Have a go first
Find the exact value of e^(3 ln 2).
Listen
The logarithm log_a(x) answers the question: to what power must a be raised to give x? So y = log_a(x) exactly when a^y = x, making log_a the inverse of a^x. Because a^x is always positive, the domain of log_a(x) is x > 0, and its graph is the reflection of y = a^x in the line y = x, passing through (1, 0) with the y-axis as a vertical asymptote. The natural logarithm ln x is log_e(x), the inverse of e^x, so ln(e^x) = x and e^(ln x) = x for x > 0. Its graph passes through (1, 0) and (e, 1) and rises ever more slowly. These inverse relationships let you solve equations by taking logs or exponentiating both sides. A strong answer earns marks for these points. Defines log_a(x) as the power to which a must be raised to obtain x, and rewrites between forms a^y = x and y = log_a(x). Explains that log_a is the inverse function of a^x, so the two graphs are reflections in y = x. States the domain x > 0 and identifies the vertical asymptote at x = 0, with intercept (1, 0). Defines ln x as log_e(x) and uses the inverse identities ln(e^x) = x and e^(ln x) = x for x > 0. A common slip: Writing log_a(x) for negative or zero x. Instead, stating the domain x > 0 and rejecting invalid inputs. Exam tip: Convert between exponential and logarithmic form before attempting to solve, as this often reveals the required operation.
Key words
Say what each word means before you turn the card.
Inverse functions
Try answering aloud or in your head before turning the card.
Logarithmic graphs and asymptotes
Try answering aloud or in your head before turning the card.
The natural logarithm and base e
Try answering aloud or in your head before turning the card.
Solving exponential equations
Try answering aloud or in your head before turning the card.
See every definition
- Inverse functions
- log_a is the inverse function of a^x, so the two graphs are reflections in y = x.
- Logarithmic graphs and asymptotes
- The graph of log_a(x) passes through (1, 0) with the y-axis as a vertical asymptote, and the domain is x > 0.
- The natural logarithm and base e
- ln x is log_e(x), the inverse of e^x, so ln(e^x) = x and e^(ln x) = x for x > 0.
- Solving exponential equations
- The inverse relationship lets you solve equations by taking logs or exponentiating both sides, for example converting e^(2x) = 7 into 2x = ln 7.
There is more to this lesson
The full lesson goes on to check your understanding and help you earn the marks. Create a free account to carry on, with your progress saved.
A week of Premium from your first lesson · Then the free plan continues · No card, no charge
Test your understanding of Exponentials and logarithms: 6.3 Know and use the definition of log_a(x) as the inverse of a^x, where a is positive and x ≥ 0. Know and use the function ln x and its graph. Know and use ln x as the inverse function of e^x.
Active recall beats re-reading. Take an instant quiz now to test what you’ve learned and lock it into long-term memory.
A week of Premium from your first lesson · Then the free plan continues · No card, no charge