# Logarithms and powers

Part of the Oxford MAT Livestream

### MAT syllabus

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

### Revision

- $a^ma^n=a^{m+n}$ 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.

### Warm-up

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

\begin{equation*}

\log_{x^2 + 2}(4 - 5x^2 - 6x^3) = 2

\end{equation*}

(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

\begin{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

\end{equation*}

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

\begin{equation*}

\log_b a=2, \qquad \log_b\left(c-3\right)=3,\qquad \log_a\left(c+5\right)=2.

\end{equation*}

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

\]

equals

(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.

Extension

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

\begin{equation*}

\frac{\mathrm{d}x}{\mathrm{d}y}=x

\end{equation*}

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;

\begin{equation*}

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).

\end{equation*}

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.