Logarithms

Part of the Oxford Maths Admissions Test Livestream 2026

Solutions are available here

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TMUA Specification (April 2025, Section 1 § MM5)

$y = a^x$ and its graph, for simple positive values of $a$.
Laws of logarithms:
- $a^b = c \Leftrightarrow b = \log_a c$
- $\log_a x + \log_a y = \log_a (xy)$
- $\displaystyle \log_a x - \log_a y = \log_a \left(\frac{x}{y}\right)$
- $k\log_a x = \log_a (x^k)$ including the special cases:
- $\displaystyle \log_a\left(\frac{1}{x}\right)=-\log_a x$
- $\log_a a = 1$.
Questions requiring knowledge of the change of base formula will not be set.
The solution of equations of the form $a^x = b$, and equations which can be reduced to this form, including those that need prior algebraic manipulation.

Revision

The solution $x$ to the equation $a^x=b$ where $a$ and $b$ are positive numbers (with $a\neq 1$) is called $\log_a (b)$. In this expression, the number $a$ is called the base of the logarithm.
$\log_a (x)$ is a function of $x$ which is defined when $x>0$. Like with $\sin x$, sometimes the brackets are omitted if it's clear what the function is being applied to, so we might write $\log_a x$.
$\log_a x$ is not defined for $x=0$. If $x$ is positive and close to zero, then $\left|\log_a x\right|$ is large.
$\log_a x$ is a one-to-one function on $x>0$; if $\log_a x=\log_a y$ then $x=y$. If $a>1$ then $\log_a x$ is an increasing function.
$\log_a (a^k)=k$ for any value of $k$. To see why, remember that the value of $\log_a(a^k)$ is defined to be the solution $x$ to the equation $a^x=a^k$. That solution is just $k$. This includes the special cases $\log_a 1 = 0$ and $\log_a a =1$.
With fixed $a\neq 1$, you can think of $\log_a x$ as the inverse function for $f(x)=a^x$.
It's also true that $a^{\log_a x}=x$. To see why, let $y=\log_a x$. That would mean that $a^y=x$. Now replace the $y$ in that equation with the expression $\log_a x$.
(Product rule) If $a>0$ and $a\neq 1$ and $x>0$ and $y>0$, then $\log_a (xy)=\log_a(x)+\log_a(y)$.
(Power rule) If $a>0$ and $a\neq 1$ and $x>0$ and $k$ is real, then $\log_a (x^k)=k\log_a x$.
You are not required to know that $\displaystyle \log_a b = \frac{\log_c b}{\log_c a}$ for positive $a$, $b$, $c$ with $a\neq 1$ and $c\neq 1$.

Revision Questions

Simplify $(2^3)^4$ and $(2^4)^3$ and $2^42^3$ and $2^32^4$.
Solve $x^{-2}+4x^{-1}+3=0$.
Simplify $\log_{10} 3+\log_{10} 4$ into a single term.
Write $\log_3(x^2+3x+2)$ as the sum of two terms, each involving a logarithm.
Solve $\log_x (x^2)=x^3$.
Solve $\log_x (2x)=3$ for $x>0$.
Solve $\log_{x+5}(6x+22)=2$.
Let $a=\log_{3} 2$ and $b=\log_{3} 5$. Write the following in terms of $a$ and $b$. \begin{equation*} \log_{3} 1024, \quad \log_{3} 40, \quad \log_{3} \sqrt{\frac{2}{5}}, \quad \log_{3} \frac{1}{10}, \quad \log_{3} 1.024. \end{equation*}
Expand $\left(2^x+2^{-x}\right)\left(2^y-2^{-y}\right)+\left(2^x-2^{-x}\right)\left(2^y+2^{-y}\right)$.
Expand $\left(2^x+2^{-x}\right)\left(2^y+2^{-y}\right)+\left(2^x-2^{-x}\right)\left(2^y-2^{-y}\right)$.
Use logarithms to solve $2^x=3$, then $0.5^x=3$, then $4^x=3$.
Suppose that $a$ and $b$ are positive numbers and $a\neq 1$. Using the definition of $\log_a b$ as the number $x$ that satisfies $a^x=b$, explain why $\log_a b=0$ if and only if $b=1$.
Given $\log_{10}(\log_{10}x)=6$, how many zeros are there at the end of the number $x$?
Use the product rule and the power rule to prove the "quotient rule", which is; if $a>0$ and $a\neq 1$ and $x>0$ and $y>0$, then $\displaystyle \log_a\left(\frac{x}{y}\right)=\log_a(x)-\log_a(y)$.
Solve $2^{x}+2^{-x}=4$.
How many solutions are there to $2^x+2^{-x}=c$? Identify different cases in terms of $c$.
Prove that $\log_a(N+\sqrt{N^2-1})=-\log_a(N-\sqrt{N^2-1})$ for any number $N\geq 1$ and any base $a>1$.
Consider the equation $x^y=y^x$ with $x,y>0$. Use logarithms to turn this into an equation of the form $f(x)=f(y)$.
[Harder] Sketch $f(x)$.
Simplify $a^{k\log_a b}$ for positive numbers $a$, $b$, $k$ with $a\neq 1$ and $b\neq 1$.
Consider the number $x=\left(\log_a b\right) \left(\log_b c\right)$. By simplifying $a^x$, show that $x=\log_a c$.
Deduce that $\displaystyle \log_b c = \frac{\log_a c}{\log_a b}$ for positive numbers $a$, $b$, $c$ with $a\neq 1$ and $b\neq 1$.
Let $a=\log_3 2$ and $b=\log_3 5$. Use the result in the previous question to write $\log_6 (45)$ in terms of $a$ and $b$.
The revision notes say "if $a>1$ then $\log_a x$ is an increasing function". What can you say if $a<1$?

TMUA Questions

TMUA 2020 Paper 1 Question 7

Given that $$2^{3x} = 8^{(y+3)}$$ and $$4^{(x+1)} = \frac{16^{(y+1)}}{8^{(y+3)}}$$ what is the value of $x + y$?

(A) $-23$
(B) $-22$
(C) $-15$
(D) $-14$
(E) $-11$
(F) $-10$

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TMUA 2021 Paper 1 Question 10

Use the trapezium rule with 3 strips to estimate \[ \int_{\frac{1}{2}}^{2} 2\log_{10} x \,\mathrm{d}x \]

(A) $\log_{10} \dfrac{\sqrt{6}}{2}$
(B) $\log_{10} \dfrac{3}{2}$
(C) $\log_{10} \dfrac{9}{4}$
(D) $\log_{10} 3$
(E) $\log_{10} \dfrac{81}{16}$
(F) $\log_{10} \dfrac{\sqrt{23}}{2}$

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TMUA 2022 Paper 1 Question 11

Evaluate \[ \sum_{n=1}^{100} \log_{10}\!\left(3^{1-n}\right) \]

(A) $-4950\log_{10}3$
(B) $4950\log_{10}3$
(C) $-5050\log_{10}3$
(D) $5050\log_{10}3$
(E) $1 - 4950\log_{10}3$
(F) $1 + 4950\log_{10}3$
(G) $1 - 5050\log_{10}3$
(H) $1 + 5050\log_{10}3$

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TMUA 2021 Paper 2 Question 14

Consider the following simultaneous equations, where $p$ is a real number: \begin{align*} p 2^x + \log_2 y &= 2\\ 2^x + \log_2 y &= 1 \end{align*} What is the complete range of $p$ for which these simultaneous equations have a real solution $(x, y)$?

(A) $p < 1$
(B) $p \ne 1$
(C) $p > 1$
(D) $p < 1$ or $p > 2$
(E) $p \ne 1$ and $p < 2$
(F) $p > 1$ and $p < 2$
(G) $p > 2$
(H) All real values of $p$

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TMUA 2021 Paper 2 Question 17

Consider the following functions defined for $x > 1$: \[ \mathrm{f}(x) = \log_2\!\left(\log_2 \sqrt{x}\right) \] \[ \mathrm{g}(x) = \log_2\!\left(\sqrt{\log_2 x}\right) \] Which one of the following is true for all values of $x > 1$?

(A) $0 \le \mathrm{f}(x) \le \mathrm{g}(x)$ or $\mathrm{g}(x) \le \mathrm{f}(x) \le 0$
(B) $0 \le \mathrm{g}(x) \le \mathrm{f}(x)$ or $\mathrm{f}(x) \le \mathrm{g}(x) \le 0$
(C) $\tfrac{1}{2} \le \mathrm{f}(x) \le \mathrm{g}(x)$ or $\mathrm{g}(x) \le \mathrm{f}(x) \le \tfrac{1}{2}$
(D) $\tfrac{1}{2} \le \mathrm{g}(x) \le \mathrm{f}(x)$ or $\mathrm{f}(x) \le \mathrm{g}(x) \le \tfrac{1}{2}$
(E) $1 \le \mathrm{f}(x) \le \mathrm{g}(x)$ or $\mathrm{g}(x) \le \mathrm{f}(x) \le 1$
(F) $1 \le \mathrm{g}(x) \le \mathrm{f}(x)$ or $\mathrm{f}(x) \le \mathrm{g}(x) \le 1$

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TMUA 2022 Paper 2 Question 15

The real numbers $x$, $y$ and $z$ are all greater than 1, and satisfy the equations \[ \log_x y = z \quad \text{and} \quad \log_y z = x \] Which one of the following equations for $\log_z x$ must be true?

(A) $\log_z x = y$
(B) $\log_z x = \dfrac{1}{y}$
(C) $\log_z x = xy$
(D) $\log_z x = \dfrac{1}{xy}$
(E) $\log_z x = xz$
(F) $\log_z x = \dfrac{1}{xz}$
(G) $\log_z x = yz$
(H) $\log_z x = \dfrac{1}{yz}$

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Hints

TMUA 2020 Paper 1 Question 7

The question just asks for $x+y$, but we should probably try to solve for both $x$ and $y$.
The variable $x$ only appears in exponents. The variable $y$ only appears in exponents.
How are the numbers 2, 4, 8, and 16 related?
Try not to evaluate $8^3$ as 512, unless you have to.

TMUA 2021 Paper 1 Question 10

If you haven't met the trapezium rule yet, come back to this question after you have.
Don't forget the factor of $\frac{1}{2}$, or the factor of $2$, or the other factor of $\frac{1}{2}$.
During your work, you can choose whether you prefer to write things like $4\log_{10} 3$ or $\log_{10} 81$. I personally prefer the first one.

TMUA 2022 Paper 1 Question 11

Write out the first couple of terms of the sum, and the last term.
Careful with the last term!
Use laws of logarithms on each term.
Remember that you can recognise certain sorts of sequences (in particular, arithmetic progressions and geometric progressions).

TMUA 2021 Paper 2 Question 14

Try to solve these equations for $x$ and $y$, and see what happens.
The variable $x$ only appears in terms like $2^x$, and the variable $y$ only appears in terms like $\log_2 y$. If you write $a=2^x$ and $b=\log_2 y$, can you solve the equations for $a$ and $b$?
By making that substitution, we've obscured the fact that one of the variables appears in an exponential and one appears in a logarithm. Don't forget that! Given that we actually want to solve for $x$ and $y$, not $a$ and $b$, what might go wrong?

TMUA 2021 Paper 2 Question 17

Don't try to understand the options! Compare $\mathrm{f}(x)$ and $\mathrm{g}(x)$. For which values of $x$ is $\mathrm{f}(x)\leq \mathrm{g}(x)$?
You know that $\log_2 x$ is an increasing function of $x$, so you can just compare the arguments of that function if you want to know which will give the larger output.
If it helps, write $u$ for $\log_2 x$ to make things look simpler.
The question wants bounds on $\mathrm{f}(x)$ and $\mathrm{g}(x)$ in various cases. If you work out values of $x$ for which $\mathrm{f}(x)\leq \mathrm{g}(x)$, then that will give you a starting point to use when you try to find inequalities for the values of $\mathrm{f}(x)$ and $\mathrm{g}(x)$.

TMUA 2022 Paper 2 Question 15

Write out what the given equations mean in terms of exponentials.
Don't think too much about what it means for $x$ to appear both as a base in the first equation, and then also as a value in the second statement. Aim for equations that you can work with!
If you can eliminate $y$ from your equations then you've shown that there is some relationship between $x$ and $z$. That might let you rule out six of the eight options.
You're aiming for a fact about $\log_z x$, so you're looking for an expression with a power of $z$ that involves $x$.