Logarithms

Logarithms are weird.

Logarithms are really, really weird.

They have lots of properties that are similar to both functions and variables, but they aren’t either. They’re lying to you. They’re Exponents!

Logarithms basics

At its core, a logarithm is the inverse of the following exponential function:

$$\begin{array}{c}f(x)=a^x\\ y=a^x\\ x=a^y\end{array}$$

The graph of which is:

Note

• blue is $10^x$
• red is $x=10^y$, aka $\log x$

The above is reformatted as this:

$$\begin{array}{rl}{a}^y& =x\\ & \downarrow \\ {\log }_{a}x& =y\end{array}$$

For reasons I do not know why, of the time it’s assumed that the $a$ is 10 and is left out, so you usually see ${\log }_{10}x$ as $\log x$. Also for reasons I do not know why atm, is that ${\log }_{e}x$ is used so much that it gets its own dedicated shorthand, $\ln x$

Logarithm interactions

When adding, subtracting, dividing, or even exponentially increasing logarithms, you can’t simply do it. Remember, these are exponents in disguise!

Because of that, they behave like exponents when they interact with each other!

$$\begin{array}{rl}\log (w\cdot n)& =\log w+\log n\\ \log (\frac{w}{n})& =\log w-\log n\\ \log w^n& =n\cdot \log w\end{array}$$

But the really weird bit about logarithms is how if you want to evaluate a logarithm that isn’t base 10, you can use the change of base formula and you get the answer, no extra steps needed!!

$${\log }_{a}x=\frac{{\log }_{b}x}{{\log }_{b}a}$$

The crazy thing is, b can be anything, not just base 10 and it’ll work! For that reason it’s often that people just use $\ln$ for $\log$ and vice versa.

Solving Logarithmic (and exponential!) equations

There are a few different ways you can solve logarithm problems when the x variable is in the exponent or inside the logarithmic argument.

However, remember that if the original problem was a $log$ then x cannot be $x<=0$

Special Cases

Since logs are exponents, there are a few special cases when it comes to solving them.

A log of 1 equals zero

If you take the log of 1, it always equals zero no matter the base. This is because any number raised to 0 equals 1!

$$\begin{array}{rl}{\log }_{4}1& =0\\ 4^0& =1\end{array}$$

A log that matches the base equals 1

This is because the only exponent that results in the same number, is 1, virtually no exponent!

$$\begin{array}{rl}\log 10& =1\\ 10^1& =10\\ & \downarrow \\ 10& =10\end{array}$$

Same goes for $\ln$ and $e$

$$\begin{array}{rl}\ln e& =1\\ e^1& =e\\ & \downarrow \\ e& =e\end{array}$$

Set bases to be the same

If both sides of the equation are able to be rewritten so the bases are the same, you can set them to the same

$$\begin{array}{rl}{2}^{4x}& =8\\ 2^{4x}& =2^3\end{array}$$

Then you can simply ignore the bases and solve as normal!

$$\begin{array}{rl}{\cancel{2}}^{4x}& ={\cancel{2}}^3\\ 4x& =3\end{array}$$

Rewrite as exponential (or logarithmic)

Sometimes its easiest to just convert an exponent to a logarithm!

$$\begin{array}{rl}{4}^{3x}& =7\\ & \downarrow \\ {\log }_{4}7& =3x\\ & \downarrow \\ \frac{\ln 7}{\ln 4}& =3x\end{array}$$

Or an logarithm to an exponent!

$$\begin{array}{rl}\log (2x+1)& =3\\ & \downarrow \\ 10^3& =2x+1\end{array}$$

Warning

If the problem was originally an exponent and you converted it into a logarithm, it’s totally ok if the one of the solutions results in a negative log nor zero. However, if the original problem is a logarithm then it cannot be a negative.
Example, $\log -3$ is incorrect. However $10^{-3}$ is totally fine!

Log both sides first

Sometimes (especially if you have an $e$) simply taking the log of both sides is the easiest way to simplify it, because it removes it!

$$\begin{array}{rl}{e}^{x+1}& =2\\ & \downarrow \\ \cancel{\ln }({\cancel{e}}^{x+1})& =\ln (2)\\ & \downarrow \\ x+1& =0.69314718056\end{array}$$

Ignore!

if both sides of the equation are logs, just ignore them!

$$\begin{array}{rl}\log (2x+1)& =\log (3x+6)\\ & \downarrow \\ 2x+1& =3x+6\end{array}$$