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A logarithm is the inverse of an exponential: \(\log_a b\) is the power you raise \(a\) to in order to get \(b\). You need to switch between index form and log form, use the natural logarithm \(\ln x\) as the inverse of \(\mathrm{e}^x\), and sketch \(y = \ln x\). These skills are used in almost every later exponential and log question.
\(y = \ln x\) is \(y = \mathrm{e}^x\) reflected in \(y = x\)
Notes
What a logarithm is
\(\log_a b = c\) means \(a^c = b\). The logarithm is the power you raise the base \(a\) to, to get \(b\)
The base \(a\) is positive and not equal to 1, and you can only take the log of a positive number: \(\log_a x\) is defined only for \(x > 0\)
\(\log_a 1 = 0\) because \(a^0 = 1\), and \(\log_a a = 1\) because \(a^1 = a\)
Examples: \(\log_3 81 = 4\) because \(3^4 = 81\); \(\log_2 \dfrac{1}{8} = -3\) because \(2^{-3} = \dfrac{1}{8}\); \(\log_{25} 5 = \dfrac{1}{2}\) because \(25^{1/2} = 5\)
\(\log_a x\) is the inverse of \(a^x\): \(\log_a(a^x) = x\) and \(a^{\log_a x} = x\)
Natural logarithms
\(\ln x\) means \(\log_{\mathrm{e}} x\), the logarithm to base \(\mathrm{e}\). It is the inverse of \(\mathrm{e}^x\)
\(\ln(\mathrm{e}^x) = x\) for all \(x\), and \(\mathrm{e}^{\ln x} = x\) for \(x > 0\). Also \(\ln 1 = 0\) and \(\ln \mathrm{e} = 1\)
To solve \(\mathrm{e}^{(\ldots)} = k\), take ln of both sides. To solve \(\ln(\ldots) = k\), write \((\ldots) = \mathrm{e}^k\)
The graph of \(y = \ln x\)
diagram
\(y = \ln x\) is the reflection of \(y = \mathrm{e}^x\) in the line \(y = x\), so the point \((0, 1)\) becomes \((1, 0)\) and the asymptote \(y = 0\) becomes \(x = 0\)
\(y = \ln x\) is \(y = \mathrm{e}^x\) reflected in \(y = x\): each asymptote becomes the other axis.
Domain \(x > 0\); range all real numbers. The curve always increases, but more and more slowly.
\(y = \log_a x\) (for \(a > 1\)) has the same shape and also passes through \((1, 0)\)
Transformations move the asymptote: \(y = \ln(x - 2)\) has asymptote \(x = 2\) and crosses the x-axis where \(x - 2 = 1\), at \((3, 0)\)
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Write in log form or evaluate without a calculator
1 to 4 marksGrade E
Ask yourself: 'the base to what power gives this number?'
Write the number as a power of the base, e.g. \(8 = 4^{3/2}\) or \(\dfrac{1}{27} = 3^{-3}\)
The power is the value of the logarithm.
Example. (a) Write \(2^5 = 32\) using logarithms. (b) Find the value of \(\log_4 8\). (c) Find the value of \(\log_3 \dfrac{1}{27}\)
Domain \(x > 3\) (because the range of \(f\) is \(f(x) > 3\))
B1
Don’t lose marks
Taking the log of zero or a negative number, or not rejecting a solution that makes the expression inside a log zero or negative.
Writing \(\ln(x + 3) = \ln x + \ln 3\). There is no law like this.
Mixing up the asymptotes: \(y = \ln x\) has a vertical asymptote \(x = 0\); \(y = \mathrm{e}^x\) has a horizontal asymptote \(y = 0\)
Writing \(\mathrm{e}^{\ln x + 2} = x + 2\). In fact \(\mathrm{e}^{\ln x + 2} = \mathrm{e}^{\ln x} \times \mathrm{e}^2 = \mathrm{e}^2 x\)
More tips
Memory tricks
A log asks a question: \(\log_2 32\) asks '2 to what power gives 32?'
ln and e cancel each other out because they are inverses: \(\mathrm{e}^{\ln 5} = 5\) and \(\ln \mathrm{e}^7 = 7\)
For a base bigger than 1, the log of a number between 0 and 1 is negative, and the log of 1 is 0.
The graph of \(y = \ln x\) is \(y = \mathrm{e}^x\) with the coordinates swapped: \((0, 1)\) becomes \((1, 0)\), and the asymptote \(y = 0\) becomes \(x = 0\)
Exam technique
Give exact answers in terms of ln and e when asked; only use decimals at the end if the question wants them.
When you find an inverse, always state its domain: it is the range of the original function.
Label asymptotes with their equations on every sketch.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Convert between index form and log form\(\log_a b = c\) means exactly the same as \(a^c = b\)
Grade E
Evaluate simple logarithms without a calculatorFor example \(\log_2 32 = 5\), \(\log_5 1 = 0\) and \(\log_9 3 = \dfrac{1}{2}\)
Grade D
Use ln and e as inverse functions\(\ln(\mathrm{e}^x) = x\) for all \(x\), and \(\mathrm{e}^{\ln x} = x\) for \(x > 0\)
Grade C
Solve equations using ln and eTake ln of both sides of \(\mathrm{e}^{3x+1} = 10\), or write \(2y - 1 = \mathrm{e}^4\) when \(\ln(2y - 1) = 4\)
Grade C
Sketch \(y=\ln x\)Through \((1, 0)\), increasing, defined only for \(x > 0\), with the y-axis as a vertical asymptote.
Grade B
Sketch transformed log graphsE.g. \(y = \ln(x + 3)\) has asymptote \(x = -3\) and meets the axes at \((-2, 0)\) and \((0, \ln 3)\)
Grade A
Find inverses of exponential and log functionsE.g. for \(f(x) = 3 + \mathrm{e}^{2x}\), \(f^{-1}(x) = \dfrac{1}{2}\ln(x - 3)\) with domain \(x > 3\)
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
(a) Write \(2^{6} = 64\) in logarithmic form.[1]
(b) Find the exact value of \(\log_{25} 5\).[2]
(c) Solve the equation \(\log_{2}(x - 2) = 3\).[2]
Show the answer and mark scheme
(a)Answer: \(\log_{2} 64 = 6\)
B1 for \(\log_{2} 64 = 6\)
Worked solution: \(2^{6} = 64 \iff \log_{2} 64 = 6\) (\(\log_a y = x \iff a^x = y\)).
(b)Answer: \(\frac{1}{2}\)
M1 for writing \(25^{x} = 5\) with both sides as powers of a common base
A1 for \(\frac{1}{2}\)
Worked solution: Let \(x = \log_{25} 5\). Then \(25^{x} = 5\), and writing both as powers of the same number gives \(x = \frac{1}{2}\) (check: \(25^{\tfrac{1}{2}} = 5\)).
(c)Answer: \(x = 10\)
M1 for converting to index form
A1 for \(x = 10\)
Worked solution: \(x - 2 = 2^{3} = 8 \Rightarrow x = 10\).
Question 2Medium6 marks
The function \(\mathrm{g}\) is defined by \[\mathrm{g}(x) = 5\mathrm{e}^{2x} + 1, \quad x \in \mathbb{R}.\]
(a) State the range of \(\mathrm{g}\).[1]
(b) Find \(\mathrm{g}^{-1}(x)\) and state its domain.[3]
(c) Find the exact value of \(x\) for which \(\mathrm{g}(x) = 11\).[2]
Show the answer and mark scheme
(a)Answer: \(\mathrm{g}(x) \gt 1\)
B1 for \(\mathrm{g}(x) \gt 1\)
Worked solution: Since \(\mathrm{e}^{2x} \gt 0\), \(\mathrm{g}(x) \gt 1\).
M1 for making \(\mathrm{e}^{2y}\) the subject of \(x = 5\mathrm{e}^{2y} + 1\)
A1 for \(\mathrm{g}^{-1}(x) = \frac{1}{2}\ln\left(\frac{x - 1}{5}\right)\) oe
B1ft for domain \(x \gt 1\)
Worked solution: \(x = 5\mathrm{e}^{2y} + 1 \Rightarrow \mathrm{e}^{2y} = \frac{x - 1}{5} \Rightarrow y = \frac{1}{2}\ln\left(\frac{x - 1}{5}\right)\). The domain of \(\mathrm{g}^{-1}\) is the range of \(\mathrm{g}\): \(x \gt 1\).
(c)Answer: \(x = \frac{1}{2}\ln 2\)
M1 for \(\mathrm{e}^{2x} = 2\)
A1 for \(x = \frac{1}{2}\ln 2\) oe
Worked solution: \(5\mathrm{e}^{2x} = 10 \Rightarrow \mathrm{e}^{2x} = 2 \Rightarrow x = \frac{1}{2}\ln 2\) (\(\approx 0.347\)).
Question 3Hard9 marks
The diagram shows part of the curve with equation \(y = c + \log_{a}(x - 2)\), where \(a\) and \(c\) are constants and \(a \gt 1\). The line \(x = 2\) (dashed) is an asymptote to the curve. The curve passes through the points \(A(3, -2)\) and \(B(6, -1)\).
[object Object]
(a) Find the value of \(a\) and the value of \(c\).[4]
(b) Find the exact \(x\)-coordinate of the point where the curve crosses the \(x\)-axis.[2]
(c) The curve is the graph of \(y = \mathrm{f}(x)\), \(x \gt 2\). Find \(\mathrm{f}^{-1}(x)\), stating its domain.[3]
Show the answer and mark scheme
(a)Answer: \(a = 4, \ c = -2\)
M1 for substituting both points: \(-2 = c + \log_{a} 1\) and \(-1 = c + \log_{a} 4\)
M1 for subtracting: \(\log_{a} 4 - \log_{a} 1 = 1\) or equivalent
A1 for \(a = 4\)
A1 for \(c = -2\)
Worked solution: Subtracting the equations: \(\log_{a}\frac{4}{1} = 1 \Rightarrow a = 4\). Then \(c = -2 - \log_{4} 1 = -2 - 0 = -2\).
(b)Answer: \(x = 18\)
M1 for \(\log_{4}(x - 2) = 2\)
A1 for \(x = 18\)
Worked solution: \(0 = -2 + \log_{4}(x - 2) \Rightarrow x - 2 = 4^{2} \Rightarrow x = 18\).
Worked solution: \(y = -2 + \log_{4}(x - 2) \Rightarrow x - 2 = 4^{y + 2}\), so \(\mathrm{f}^{-1}(x) = 4^{x + 2} + 2\), with domain the range of \(\mathrm{f}\), i.e. all real \(x\).