We sometimes want to solve equations like $a^{x}=b$, where $a$ and $b$ could be any numbers. For example, we might want to solve $10^{x}=100$. In this case it's very simple. We can quickly see that $10^2=100$, so $x=2$. But what if we had a more complicated equation? Say $10^{x}=200$? $10^2=100$ and $10^3=1000$, so the answer is somewhere in between 2 and 3. But how do we find the exact value of $x$? This is one of the things we will cover in this lesson.

Logarithms are way for us to rewrite exponents. If we have the equation $2^5=32$ we can rewrite this as $\log_232=5$ using logarithms. Similarly, $x^y=z$ can be rewritten as $\log_xz=y$.  Hopefully you can see the pattern here. With exponentials, we input a base and an exponent and get a result. With logarithms, we input a base and the result and get the exponent. Logarithms can thus be said to the "opposite" of exponents. If I have the logarithm $\log_{10}100=x$, I like to think of this as: "What number $x$ do I have to raise $10$ to in order to get $100$?". In this case the answer is $2$ so the $\log_{10}100=2$. Similarly, $\log_ba$ is the value that we have to raise $b$ to in order to get $a$. We can write this mathematically as $b^{\log_ba}=a$. 

In $\log_ba$ we call $b$ the base and $a$ the argument of the logarithm. The logarithm is a type of function, that outputs a different value depending on the base and the argument.

Find the value of

  a) $\log_39$.

  b) $\log_2{\frac{1}{4}}$

There are some special cases that are good to know for calculating logarithms. The first is the value of $\log_bb$. By definition $\log_bb$ is the number that satisfies $b^{\log_bb}=b$. The only value that satisfies this is $1$, so $\log_bb=1$ for any $b$. (Unless $b$ equals $0$ or $1$, in which case there are infinitely many solutions. For this reason, base $0$ and $1$ are never used. For the rest of the lesson, we assume that $b\neq0$ and $b\neq1$.) The next property is the value of $\log_b1$. Since $b^{\log_b1}=1=b^0$ we have $\log_b1=0$.

These two properties, $\log_bb=1$ and $\log_b1=0$, are special cases of the general property $\log_b{b^x}=x$. We can think of the value of $\log_b{b^x}$ as: "What number do I need to raise $b$ to in order to get $b^x$?'' Obviously, the answer is $x$, so $\log_b{b^x}=x$.

From this we see that the logarithm $\log_bx$ is actually the inverse of the exponential $b^x$. This is because if we put $b^x$ as the argument of the logarithm we get back $x$.

Another thing that is good to know is that logarithms with negative bases or arguments are almost always undefined. For example, what is the value $\log_{10}({-10})$? It's the value that satisfies $10^{\log_{10}({-10})}=-10$. However, ten to the power of something will never be a negative number. We can get close to zero by taking for example $10^{-100}=\frac{1}{10^{100}}$, but we'll never reach or go below zero. Therefore $\log_{10}(-10)$ is undefined (unless we allow complex numbers, but we aren't able to handle them yet). It is similar for negative bases. For example, what is the value of $\log_{-4}{5}$? It's the value that satisfies $({-4})^{\log_{-4}{5}}=5$. Although $(-4)^x$ can be a positive number (as in $(-4)^2=16$), it can never be $5$. This is because as soon as the exponent $x$ is not a whole number, then $(-4)^x$ is undefined, and by trial and error there are no whole numbers that satisfy $(-4)^x=5$. Since there are no solutions to $(-4)^x=5$, then $\log_{-4}{5}$ must be undefined.

Some logarithms with negative bases are actually defined. For example, $\log_{-4}16=2$ and $\log_{-2}(-8)=3$. However, it only works in very specific cases. This is not very useful, so in this lesson we will mostly look at logarithms with positive bases and arguments.

Even though the base and the argument must always be positive, the result of a logarithm can still be negative. For example, $\log_{5}{\frac{1}{125}}=-3$ and $\log_{\frac{1}{2}}{2}=-1$.

Let's say that we wanted to solve the equation $10^x=200$. Since $10^2=100$ and $10^3=1000$, the answer must be somewhere in between $2$ and $3$. However, the exact value is hard to guess. We could say that the solution is $x=\log_{10}{200}$, which is technically correct but it doesn't give us a specific value.

This is where we "cheat". We can type $\log200$ into a calculator and it will give us the answer. In this case it's $x=\log200\approx2.301$. This may seem like a very boring way to find solutions to these equations, since we're basically asking the calculator to do it for us, but there is actually no better way. However, the calculator can't handle loads of different bases. It can handle base ten, but for other bases we need a formula to convert them to base ten. But let's first introduce some notation: $\log x$ is assumed to be base ten if the base is not specified. This is just to make it easier to read.

When we want to find logarithms with other bases than ten we need this formula:

$$\log_b{x}=\frac{\log{x}}{\log{b}}$$

Proof: we know that $$ b^{\log_b{x}}=x.$$ Then

$$(10^{\log{b}})^{\log_b{x}}=x$$$$10^{\log{b}\cdot\log_b{x}}=x$$$$10^{\log{b}\cdot\log_b{x}}=10^{\log{x}}.$$

And, since the exponents must be equal,

$$\log{b}\cdot\log_b{x}=\log{x} $$$$\log_b{x}=\frac{\log{x}}{\log{b}}.$$

We can use this to solve equations with other bases than ten.

Find the value of $x$ if $12^x=132$.

There are some rules that can be very useful when you're dealing with logarithms. We have already found one formula:

$$\log_b{a}=\frac{\log{a}}{\log{b}}$$

There are a few more I want to go over in this lesson. The first one is this:

$$\log{a^x}=x\log{a}$$

This is how we prove it. We start with the left hand side and replace $a$ with $10^{\log a}$.

$$\log{a^x}=\log{({(10^{\log{a}})^x})}=\log({10^{x\log{a}}})=x\log{a}$$$$\implies\log{a^x}=x\log{a}$$

Of course, we could have used any positive base instead of ten, so the formula can be generalized to $$\log_b{a^x}=x\log_b{a}.$$.

Solve the equation $5^x=10$

The next logarithm law I want to show you is about the logarithm of a product. We prove it in a similar way as the last law. We start with the expression $\log{ab}$ and replace $a$ with $10^{\log{a}}$ and $b$ with $10^{\log b}$.

$$\log{ab}=\log(10^{\log{a}}\cdot10^{\log{b}})=\log{10^{\log{a}+\log{b}}}=\log{a}+\log{b}$$$$ \implies\log{ab}=\log{a}+\log{b}$$

And that's our formula. Using this and that $\log{a^x}=x\log{a}$, we can prove a similar formula for division.

$$\log{\frac{a}{b}}=\log({a\cdot{b^{-1}}})=\log{a}+\log{b^{-1}}=\log{a}-\log{b}$$$$\implies\log{\frac{a}{b}}=\log{a}-\log{b}$$

That was actually the last logarithm law I wanted to go over. However, since they're all spread out in the text, I'll summarize them here for easier reference. I'll also include some basic properties that are good to know.

$$b^{\log_b{a}}=a$$$$\log_b{b^a}=a$$$$\log_b{b}=1$$$$\log_b1=0$$$$\frac{\log_c{a}}{\log_c{b}}=log_b{a}$$$$\log_b{a^x}=x\log_b{a}$$$$\log_c{ab}=\log_c{a}+\log_c{b}$$$$\log_c{\frac{a}{b}}=\log_c{a}-\log_c{b}$$

Here are some other laws, that you can prove as exercises:

$$(\log_a{b})\cdot(\log_c{d})=(\log_a{d})\cdot(\log_c{b})$$$$\log_ba=\frac{1}{\log_ab}$$$$\log_{a^n}{b^n}=\log_ab$$

While the graph is not very useful in itself, it can often help us better understand the problems we are dealing with. Let's first find some points on the curve $y=\log{x}$. Since $\log1=0$ and $\log10=1$, then $(1,0)$ and $(10,1)$ must lie on the curve. Since $\log0.1=-1$, $\log0.01=-2$, $\log0.001=-3$ and so on, we know that $y$ goes towards minus infinity as $x$ gets closer and closer to zero. For the graph, this means that the curve approaches the $y$-axis asymptotically, that is, for $x$ close to zero the curve is almost parallel to the $y$-axis, and it becomes even more parallel as $x$ gets even closer to zero. For $x=0$, however, $\log x$ is undefined. For negative values of $x$ the logarithm of $x$ is also undefined, so the curve will not exist to the left of the $y$-axis. From this, we can guess that the graph looks something like this:

Although the graph of $y=\log_{b}{x}$ is not necessarily identical to the graph of $y=\log{x}$, it has the same general shape. This is because $\log_{b}{x}=\frac{\log x}{\log{b}}$, which is just a scaled version of $\log x$.

The graph of $y = \log x$ actually has a likeness to the graph of $y=10^x$, and that's not a coincidence. We can rewrite $y=\log{x}$ as $10^y=x$, or $x=10^y$. This is exactly the same as $y=10^x$, except that the variables have switched place. If the $x$-axis and the $y$-axis could also switch place, then $x=10^y$ (with switched axis) would look exactly the same as $y=10^x$ (without switched axis).

When we switch axis we're really reflecting the whole graph in the line $y=x$. That's why we get this symmetry.

Find all solutions to the equation $\log_{\sqrt{2}}{x}=x$, where $\log_{\sqrt2}$ is the base $\sqrt{2}$ logarithm.

Logarithms were first invented as a simpler way to multiply large numbers, before there were calculators. For example, if we wanted to multiply $136.62$ with $428.47$ we could express this as $136.62\cdot428.47=$$10^{\log({136.62\cdot428.47})}=$$10^{\log136.62+\log428.47}$. We would then look up the value of the logarithms $\log{136.62}$ and $\log428.47$ in a logarithm table, add them together and take ten to the power of the sum. This would give the value of $136.62\cdot428.47$, and is much easier than multiplying them directly.

The same method can be used for division. For example, if we wanted to find $1366$ divided by $4284$, we could say $\frac{1366}{4284}=$$ 10^{\log{\frac{1366}{4284}}} =$$ 10^{\log1366-\log4284}$. The value of this expression is easy to find using logarithm tables.

So far we have used base $10$ as our primary base, but the most common base in advanced maths is actually $e$, where $e=1+1+\frac{1}{2!}+\frac{1}{3!}+\dotsb\approx2.72$. The logarithm of base $e$ is called the natural logarithm and is written $\ln$. This might seem a bit random, but it actually turns out to be very useful. Why $e$ is so special is hard to explain without going into derivatives, but the number $e$ and the natural logarithm will be discussed much more in the next lesson. For now, it's just good that you've seen it.