Part of the Oxford MAT Livestream

MAT syllabus

Laws of logarithms and exponentials. Solution of the equation $a^x = b$.


  • $a^ma^n=a^{m+n}$ for any positive real number $a$ and any real numbers $m$ and $n$.
  • $(a^m)^n=a^{mn}$ for any positive real number $a$ and any real numbers $m$ and $n$.
  • $\displaystyle a^{-n}=\frac{1}{a^n}$ for any real positive number $a$ and any real number $n$.
  • $a^0=1$ for any non-zero real number $a$.
  • The solution to $a^x=b$ where $a$ and $b$ are positive numbers (with $a\neq 1$) is $\log_a (b)$. 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 $\log_a x$.
  • $\log_a x$ doesn't repeat any values; if $\log_a x=\log_a y$ then $x=y$.
  • Note the special case $\log_a a =1$ because $\log_a a$ is the solution $x$ to the equation $a^x=a$, and that solution is 1.
  • In fact, $\log_a (a^x)=x$.
  • In that sense, the logarithm function is the inverse function for $y=a^x$.
  • $a^{\log_a x}=x$.
  • $\log_a (xy)=\log_a(x)+\log_a(y)$.
  • $\log_a (x^k)=k\log_a x$ including $\displaystyle\log_a \frac{1}{x}=-\log_a x$.
  • There's a mathematical constant called $e$, which is just a number (it's about 2.7).
  • $e^x$ is called the exponential function.
  • The laws of indices and laws of logarithms above hold when the base $a$ is equal to $e$.
  • $\log_e x$ is sometimes written as $\ln x$, and the function is sometimes called the natural logarithm.


  1. Simplify $(2^3)^4$ and $(2^4)^3$ and $2^42^3$ and $2^32^4$.
  2. Solve $x^{-2}+4x^{-1}+3=0$.
  3. Solve $\log_x (x^2)=x^3$.
  4. Solve $\log_{x+5}(6x+22)=2$.
  5. Simplify $\log_{10} 3+\log_{10} 4$ into a single term.
  6. Let $a=\ln 2$ and $b=\ln 5$. Write the following in terms of $a$ and $b$.\[\ln 1024, \quad \ln 40, \quad \ln \sqrt{\frac{2}{5}}, \quad \ln \frac{1}{10}, \quad\ln 1.024.\]
  7. Expand $\left(e^x+e^{-x}\right)\left(e^y-e^{-y}\right)+\left(e^x-e^{-x}\right)\left(e^y+e^{-y}\right)$.
  8. Expand $\left(e^x+e^{-x}\right)\left(e^y+e^{-y}\right)+\left(e^x-e^{-x}\right)\left(e^y-e^{-y}\right)$.
  9. Solve $2^x=3$. Solve $0.5^x=3$. Solve $4^x=3$.
  10. For which values of $x$ (if any) does $1^x=1$? For which values of $x$ (if any) does $1^x=3$?
  11. For which values of $b$ (if any) does $0^b=0$? For which values of $a$ (if any) does $a^0=0$?

MAT questions

MAT 2015 Q1H

How many distinct solutions does the following equation have?

\log_{x^2 + 2}(4 - 5x^2 - 6x^3) = 2
(a) None,

(b) 1,

(c) 2,

(d) 3,

(e) Infinitely many.


Hint: that's a scary logarithm! How can we get rid of it?


MAT 2017 Q1I

The equation
\log _b\left(\left(b^x\right)^x\right) + \log_a\left(\frac{c^x}{b^x}\right) + \log_a\left(\frac{1}{b}\right)\log_a(c) = 0 

has a repeated root when

(a)  $b^2=4ac$,

(b) $\displaystyle b=\frac{1}{a}$,

(c) $\displaystyle c=\frac{b}{a}$,

(d) $\displaystyle c=\frac{1}{b}$,

(e) $a=b=c$.


Hint: there's a lot of simplifying to do here before this turns into a polynomial.

MAT 2013 Q1F

Three positive numbers $a$, $b$, $c$, satisfy
\log_b a=2, \qquad \log_b\left(c-3\right)=3,\qquad \log_a\left(c+5\right)=2.

This information

(a) specifies $a$ uniquely.

(b) is satisfied by exactly two values of $a$

(c) is satisfied by infinitely many values of $a$

(d) is contadictory.


Hint: this is secretly a system of equations for $a$, $b$, and $c$. The option "specifies $a$ uniquely" would be true if there is exactly one value of $a$ that works in these equations, and perhaps either one or more solutions for $b$ and $c$. The option "is contradictory" would be true if there are no solutions for one or more of the variables.


MAT 2013 Q1J

For a  real number $x$ we denote by $[x]$ the largest integer less than or equal to $x$.

Let $n$ be a natural number. The integral
\int_0^n \left[2^x\right]\,\mathrm{d}x

(a) $\log_2\left(\left(2^n-1\right)!\right)$,

(b) $n2^n-\log_2\left(\left(2^n\right)!\right)$,

(c) $n2^n$,

(d) $\log_2\left(\left(2^n\right)!\right)$,

where $k!=1\times2\times3\times\dots\times k$ for a positive integer $k$.


Hint: split the integral up into different regions where $2^x$ takes values in between different whole numbers.


The following material is included for your interest only, and not for MAT preparation.

This isn't on the MAT syllabus, but $\ln x$ plays a special role in calculus. It's the indefinite integral of $x^{-1}$. Let's explore that. First, here's a quick reminder that we can't integrate $x^{-1}$ with our normal rule for integrating $x^n$, which would give $\displaystyle\frac{x^{n+1}}{n+1}$, because we can't divide by $n+1$ if $n=-1$. But the area under the graph $y=x^{-1}$ from, say, $x=1$ to $x=2$ is just some real number! It's perhaps surprising that it's $\ln 2$.

To get an idea of the link between integrating $x^{-1}$ and $\ln x$, let's write down the problem we're trying to solve;

Find a function $y(x)$ such that $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{x}.$

Here's a trick - we can flip both sides of this equation the other way up to get


This sort of manipulation is definitely not on the MAT syllabus, and it's really not obvious that the inverse of $\displaystyle\frac{\mathrm{d}y}{\mathrm{d}x}$ should be $\displaystyle\frac{\mathrm{d}x}{\mathrm{d}y}$ because the derivative there is not really a fraction, it's more like a notation for a limit, but trust me, this operation does actually work.

Now if we squint at this new equation, it's telling us something about the derivative of $x$ in terms of $y$ (if it helps, switch the $x$ in this equation for something that looks like a fancy $y$ and switch the $y$ for something that looks like a fancy $x$). It says that when we differentiate $x$ with respect to $y$, we get $x$ back. That's exactly what $e^y$ does! So we can integrate and write $x=e^y$. Rearranging, this gives $y=\ln x$. Magic!

(Technical note: we could in fact have chosen $x=A e^y$ for any constant $A$, and then we'd have got $y=\ln x+ c$ for some constant $c$. This is exactly what we should have expected from the original problem; don't forget the constant of integration!)

Exponentials can also be used to define two new functions; 
f(x)=\frac{1}{2}\left(e^x+e^{-x}\right)\quad\text{and}\quad g(x)=\frac{1}{2}\left(e^x-e^{-x}\right).
These have lots of nice properties which you can check, such as $f'(x)=g(x)$ and $g'(x)=f(x)$ and $f(x)^2-g(x)^2=1$. They're called hyperbolic trigonometric functions, and they're usually written as $f(x)=\cosh x$ and $g(x)=\sinh x$. You can probably guess what $\tanh x$ is.

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