Logarithms Solutions

Part of the Oxford Maths Admissions Test Livestream 2026

These are the solutions for the Logarithms worksheet

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$(2^3)^4=2^{12}$. $(2^4)^3=2^{12}$. $2^4 2^3=2^7$. $2^3 2^4=2^7$.
This is a quadratic for $\frac{1}{x}$ with solutions $\frac{1}{x}=-1$ or $\frac{1}{x}=-3$. So $x=-1$ or $x=-\frac{1}{3}$. Alternatively, we could multiply both sides by $x^2$ and solve the quadratic that we get.
This is $\log_{10} 12$.
The quadratic inside the brackets factorises, and this is $\log_3(x+2)+\log_3(x+1)$. Other answers are possible, such as $\log_3 (2(x+2))+\log_3((x+1)/2)$.
The left-hand side is just $2$ so we want $2=x^3$. So $x=\sqrt[3]{2}$.
The left-hand side is $1+\log_x 2$ so we want $\log_x 2=2$. So $x^2=2$ and $x=\sqrt{2}$ (not $-\sqrt{2}$ because we're told that $x>0$).
Take $(x+5)$ to the power of each side to get $6x+22=(x+5)^2$. Expand the square and rearrange for $x^2+4x+3=0$. The solutions are $x=-1$ or $x=-3$. Check these solutions; $\log_4(16)=2$ and $\log_2(4)=2$.
- $\log_{3} 1024=\log_{3}\left(2^{10}\right)=10\log_{3} 2=10a$.
- $\log_{3} 40=\log_{3} 8+\log_{3} 5=3a+b$.
- $\log_{3} \sqrt{\frac{2}{5}}=\frac{1}{2}\log_{3} \left(\frac{2}{5}\right)=\frac{1}{2}\left(a-b\right)$.
- $\log_{3} \left(\frac{1}{10}\right)=-\log_{3} 10=-\log_{3}2-\log_{3} 5=-a-b$.
- $\log_{3} 1.024=\log_{3} 1024+\log_{3} \left(\frac{1}{1000}\right)=10a +3(-a-b)=7a-3b$.
There are other solutions, partly because $b=a\times\log_2 5$ (see the change of base formula below).
$2^{x+y}+2^{y-x}-2^{x-y}-2^{-x-y}+2^{x+y}-2^{y-x}+2^{x-y}-2^{-x-y}$. That's $2^{x+y+1}-2^{-x-y+1}$.
$2^{x+y}+2^{y-x}+2^{x-y}+2^{-x-y}+2^{x+y}-2^{y-x}-2^{x-y}+2^{-x-y}$. That's $2^{x+y+1}+2^{-x-y+1}$.
$2^x=3$ is what it means for $x$ to be $\log_2 3$.
If $0.5^x=3$ then $2^{-x}=3$ so $x=-\log_2 3$. Alternatively, just write down $x=\log_{0.5}3$.
If $4^x=3$ then $2^{2x}=3$ so $x=\frac{1}{2}\log_2 3$. Alternatively, just write down $x=\log_4 3$.
By definition, $\log_a b = 0$ if and only if $a^0 = b$. Since $a$ is positive, we have $a^0=1$ and so $b=1$.
We have $\log_{10}x=10^6$, which is a million. So $x$ is ten to the power of a million. That's got a million zeros at the end.
We have $\displaystyle \log_a \left(\frac{x}{y}\right)=\log_a \left(xy^{-1}\right)=\log_a x +\log_a y^{-1}$ using the product rule, and then $\log_a y^{-1}=- \log_a y$ using the power rule.
Multiply both sides by $2^x$ and rearrange to get $2^{2x}-4\times 2^x+1=0$. This is a quadratic for $2^x$. Solve it for $2^x=2\pm \sqrt{3}$. So $x=\log_2 (2\pm \sqrt{3})$.
Following the previous working, we can see that we'll get two roots for $2^x$ if $c^2-4>0$. But we need these to be positive roots, so we need $c>2$. If $c=2$ there's a repeated root. If $c<2$ there are no roots.
Move both terms onto the left-hand side and use the fact that $(N+\sqrt{N^2-1})(N-\sqrt{N^2-1})=N^2-(N^2-1)=1$; that's the difference of two squares. Remember that $\log_a 1 =0$.
As a result, our solutions to the previous question, $\log_2 (2\pm \sqrt{3})$, are actually $\pm \log_2 (2+\sqrt{3})$, revealing a lovely symmetry. But you could have spotted that from the equation, of course!
I've taken logarithms base 3 on each side to get $\log_{3}(x^y)=\log_{3}(y^x)$. You probably picked a different base for your logarithms, it doesn't make much difference (see the last question below). I can simplify my equation to $y\log_{3} x=x\log_{3} y$ and rearrange to get $\displaystyle \frac{\log_{3} x}{x}=\frac{\log_{3} y}{y}$. You might choose to put this fraction the other way up, or to square both sides or something, so your $f(x)$ might not be the same as mine. Here's a sketch of $\displaystyle y=\frac{\log_{3}(x)}{x}$.
$a^{k\log_a b}$ is the same as $(a^{\log_a b})^k$ which is just $b^k$.
We can use a similar bit of algebra with $k=\log_b c$. \[ a^x=a^{\log_a b \log_b c}=\left(a^{\log_a b}\right)^{\log_b c}=\left(b\right)^{\log_b c}=c. \] So $a^x=c$ and therefore $x=\log_a c$. We have proved that $\left(\log_a b\right) \left(\log_b c\right) = \log_a c$.
We can divide by $\log_a b$ to get \[ \log_b c = \frac{\log_a c}{\log_a b} \]
We can use the result in the previous question to change the base to 3, and then we can rewrite the numerator and denominator separately; \[ \log_6(45)=\frac{\log_3 (45)}{\log_3 6} =\frac{\log_3 5+\log_3 9}{\log_3 2+\log_3 3} = \frac{b+2}{a+1}.\]
If $a<1$ then $\log_a x$ is a decreasing function of $x$. You can deduce this from the corresponding equation $x=a^y$. Since $a^y$ is a decreasing function of $y$ for $a<1$, large values of $y$ correspond to small values of $x$, and vice versa.
You might also note that $\log_a x = - \log_{(1/a)} x$ using the change of base formula.

TMUA Questions

TMUA 2020 Paper 1 Question 7

The question includes powers of numbers like 2 and 4 and 8 and 16. These are all powers of 2, so we can re-write everything in terms of powers of 2.
The equation $2^{3x}=8^{(y+3)}$ can be re-written as $2^{3x}=2^{3(y+3)}$.
The equation \[4^{(x+1)} = \frac{16^{(y+1)}}{8^{(y+3)}}\] can be re-written as \[2^{2(x+1)}=2^{4(y+1)-3(y+3)}\]
Since $2^x$ is an increasing function of $x$, we can just compare exponents.
So we have the simultaneous equations \begin{align*} 3x &= 3(y+3)\\ 2(x+1) &= 4(y+1)-3(y+3) \end{align*}
The first equation simplifies to $x=y+3$, and that's in a convenient form to substitute into the second equation. This gives $2y+8 = y-5$.
The solution I get has $x=-10$ and $y=-13$, so $x+y=-23$.
The answer is A.

Extension

Is the function \[ \left(\frac{1}{2}\right)^{-x/2}\times 8^{3x} \times \left(2\sqrt{2}\right)^{3x/2}\] an increasing function of $x$, or a decreasing function of $x$?
Find the minimum value of $\displaystyle \mathrm{f}(x)=\frac{2^{\left(x^2\right)}}{\left(2^x\right)^2}.$

TMUA 2021 Paper 1 Question 10

I drew a sketch of the function. The sketch made me think about the fact that the difference between $2$ and $\frac{1}{2}$ is $\frac{3}{2}$, so if we use 3 strips then each will have width $\frac{1}{2}$. I thought about the sign of the function, but I'm not sure if that's going to be relevant.
For the trapezium rule, we need the value of the function at various points. Writing $\mathrm{f}(x)$ for the function $2\log_{10}x$, the trapezium rule calls for \[ \frac{1}{2} \times \frac{1}{2} \times \left( \mathrm{f}\left(\frac{1}{2}\right) + 2\cdot \mathrm{f}\left(1 \right)+ 2\cdot \mathrm{f}\left(\frac{3}{2}\right)+\mathrm{f}\left(2\right) \right) \]
Those values of $\mathrm{f}$ are, in order, $2\log_{10} \frac{1}{2}$, $0$, $2\log_{10} \frac{3}{2}$, and $\log_{10} 4$.
So we have \[ \frac{1}{4}\left(2\log_{10} \frac{1}{2} + 0 + 4\log_{10} \frac{3}{2} + \log_{10} 4\right), \] and the $2\log_{10} \frac{1}{2}$ cancels out the $\log_{10} 4$.
So the answer is just $\log_{10} \frac{3}{2}$.
The answer is B.

Extension

Is this an overestimate or an underestimate?
If I were using this as the start of an interview question, then next I might try to find out whether you know enough about integration and logarithms to work out the exact value of that integral. If not then I might use the interview to teach you that \[\int \ln x \,\mathrm{d}x = x\ln x - x + c,\] where $\ln x$ means $\log_e x$ with $e$ a particular irrational number that's about 2.7, and where $c$ is the constant of integration. Or perhaps we might work that out together.
Use the trapezium rule with 101 strips to estimate \[ \int_0^{101} \log_{10}\! \left(3^{1-x}\right)\,\mathrm{d}x.\]

TMUA 2022 Paper 1 Question 11

I wrote out a few terms and the last term, like this \[ \log_{10} \left(3^0\right) + \log_{10} \left(3^1\right) + \log_{10} \left(3^2\right) + ... + \log_{10} \left(3^{-99}\right), \] where I had to think really hard about whether the last term would involve $3^{-99}$ or $3^{-100}$.
Each term looks like it can be simplified with a law of logarithms to bring down the exponent (from the options, I can see that we want everything to be in terms of $\log_{10} 3$).
I now have \[ -\left(0+1+2+...+ 99\right)\log_{10} 3 \] where I've pulled the minus sign out the front, the $\log_{10}3$ out the end, and I've kept the $0$ at the front of the sum so that it's still clear that I've got 100 terms.
In the brackets I can see the sum of the numbers from 1 to 99. That's $\frac{1}{2}\times99\times 100$, using the rule for the sum of the terms of an arithmetic progression, or the formula for triangle numbers $\binom{n}{2}$, whichever you prefer.
That's $-4950 \log_{10}3$.
The answer is A.

Extension

Check your understanding: suppose that $a_n$ is a geometric progression with $a_1=a$ and $a_{n+1}=r a_n$ for $n\geq 1$, where $a$ and $r$ are positive constants. Find an expression for \[ \sum_{n=1}^{100} \log_{10} a_n.\]
This sum is related to the last bullet point in the Extension section of the previous question. Noting that the function $\log_{10}\! \left(3^{1-x}\right)$ is linear, can you comment on whether the trapezium rule gives an overestimate or an underestimate?
MAT 2016 Q1A asked for the product of the first 15 terms of the geometric progression with first term $a_1=1$ and subsequent terms defined by $a_{n+1}=l a_n$ for $n\geq 1$, in terms of $l$. That question also involved careful manipulation of powers.

TMUA 2021 Paper 2 Question 14

I would quite like to solve these equations, if I can.
I see that $x$ only appears as part of $2^x$, and $y$ only appears as part of $\log_2 y$. I'd like to solve for $2^x$ and $\log_2 y$, treating those as the variables, if I can.
In my rough work I've underlined $2^x$ and written "problem!" because I can see that a solution for $2^x$ might not correspond to a solution for $x$.
Taking the difference between the equations in the question will eliminate the bit that depends on $y$, whatever we're calling it. I'm left with $(p-1)2^x =1$
I've rearranged this for $p=1+2^{-x}$. I know that $2^{-x}$ is positive, so I really want $p>1$.
Then $x=-\log_2 \left(p-1\right)$, which would be fine.
And $y$ would be... something else. If I'm honest, I didn't write down what $y$ would be, because it seems pretty clear that the equation $2^x + \log_2 y = 1$ could be rearranged for $\log_2 y$ in terms of this solution we've just found for $x$, and then $y$ would be 2 to the power of whatever we get. That's all well-defined, and the question doesn't ask me to find $y$.
If I'm being less lazy, then the solution is \[ x=-\log_2 \left(p-1\right), \qquad y= 2^{\left(\frac{p-2}{p-1}\right)}. \]
Anyway, the solution exists if and only if $p>1$.
The answer is C.

Extension

Check your understanding: what is the complete range of $p$ for which the simultaneous equations \begin{align*} p \cos x + \sin y &= 2\\ \cos x + \sin y &= 1 \end{align*} have at least one real solution $(x,y)$?
What is the complete range of $p$ for which the simultaneous equations \begin{align*} p \cdot\sqrt{x} + y^2 &= 2\\ \sqrt{x} + y^2 &= 1\\ \end{align*} have at least one real solution $(x,y)$?
What is the complete range of $p$ for which the simultaneous equations \begin{align*} p 2^x + \log_2 y &= 2\\ 2^x + \log_2 y &= 2 \end{align*} have at least one real solution $(x,y)$?

TMUA 2021 Paper 2 Question 17

The options are quite confusing to me. They each seem to involve one case where $\mathrm{f}(x)<\mathrm{g}(x)$ and one case where $\mathrm{g}(x)<\mathrm{f}(x)$. So perhaps I should start by working out which values of $x$ give each of those behaviours.
I rewrote $\mathrm{f}(x)=\log_2 \left(\frac{1}{2}\log_2 x\right)$ to make the two functions look more similar.
Both functions end with an application of the function $\log_2 x$. I know that $\log_2 x$ is an increasing function of $x$, so \[ \log_2 \left(\frac{1}{2}\log_2 x\right) < \log_2 \left(\sqrt{\log_2 x}\right) \quad\text{if and only if}\quad \frac{1}{2}\log_2 x < \sqrt{\log_2 x}. \]
To get to grips with that, I'm going to write $u=\log_2 x$. So now I'm looking at $\frac{1}{2}u < \sqrt{u}$.
I know that $\frac{1}{2}u<\sqrt{u}$ if and only if $0 < u < 4$. I drew a little sketch to remind myself which way around that inequality goes. $\sqrt{u}$ starts fast but $\frac{1}{2}u$ eventually catches up.
So $\mathrm{f}(x)<\mathrm{g}(x)$ if and only if $\log_2 x <4$, and $\mathrm{f}(x)>\mathrm{g}(x)$ if and only if $\log_2 x>4$. I'm not including the fact that $u>0$ because this is guaranteed by the range of $x$ in the question.
For each of the possible answers, we need to know about the values in each case. It's not immediately clear whether I should look at $\mathrm{f}(x)$ or $\mathrm{g}(x)$, so I'll look at both.
If $\log_2 x<4$ then what can I say about $\mathrm{f}(x)$? I could say $\mathrm{f}(x)<\log_2\left(\frac{1}{2}\times 4\right) = 1$.
If I had to say something about $\mathrm{g}(x)$ it would be $\mathrm{g}(x)<\log_2\left(\sqrt{4}\right) =1$.
In the other case, if $\log_2 x >4$, then all the inequality signs flip, and so $\mathrm{g}(x)>1$ and $\mathrm{f}(x)>1$.
I've missed the case where $\log_2 x = 4$ and both $\mathrm{f}(x)$ and $\mathrm{g}(x)$ are equal to 1.
The answer is F.

Extension

If I were using this as the start of an interview question, then next I might introduce a third function: \[ \mathrm{h}(x)=\sqrt{\log_2\left(\log_2 x\right)}. \] Where is this function defined? For what value or values of $x$ does $\mathrm{h}(x)=\mathrm{f}(x)$? For what value or values of $x$ does $\mathrm{h}(x)=\mathrm{g}(x)$?
[If your answers involve massive powers of 2, leave them as massive powers of 2.]
These functions involve the three permutations of two "$\log_2$"s and one "$\sqrt{\phantom{x}}$". You could invent your own extension, using the six possible permutations of two of each function!

TMUA 2022 Paper 2 Question 15

I think that the following approach is unintended, but it's what I did, so I'll start by showing you how it goes.
I know that the logarithms in the question, $\log_x y$ and $\log_y z$ and $\log_z x$, are related via the identity $\left(\log_x y\right) \left(\log_y z\right) \left(\log_z x\right)=1$
We're told that the first two are equal to $z$ and $x$ respectively.
Therefore the third must be $\dfrac{1}{xz}$ to make the identity work. That's one of the options.
I think that the intended route is more like this: start by writing out what the given information tells us in terms of exponentials.
We have $y=x^z$ and $z=y^x$.
So, substituting the first into the second to eliminate $y$, we have $z= x^{zx}$.
I want to say something about $\log_z x$, so I want a statement where some power of $z$ is equal to $x$.
I can take each side to the power of $\dfrac{1}{xz}$ to get $z^{(1/xz)}=x$.
Therefore $\displaystyle \log_z x = \frac{1}{xz}$.
Either way, the answer is F.

Extension

Use the change of base formula to prove that \[ \log_x y = \left(\log_y x\right)^{-1}\] for all positive $x$ and $y$, with neither equal to 1.
Use the change of base formula to prove that \[\log_{\frac{1}{b}}c = \log_b \frac{1}{c}\] for all positive $b$ and $c$, with $b$ not equal to 1.
Use the change of base formula to prove that \[\left(\log_x y\right) \left(\log_y z\right) \left(\log_z x\right)=1\] for all positive $x$, $y$, and $z$, with none of them equal to 1.
Use the change of base formula to prove that \[\left(\log_a b\right) \left(\log_b c\right) \left(\log_c d\right) \left(\log_d a\right)=1\] for all positive $a$, $b$, $c$, and $d$, with none of them equal to 1.
If I were using this as the start of an interview question, then next I might ask you to continue the pattern that I've set up with the previous two bullet points.