Logarithms

Roots and indices are the same thing, and here we will see that logarithms are also indices.

Definition

Let $a$ and $x$ be real numbers, with $a>1$ and $x>0$. Then the logarithm of $x$ to the base $a$, denoted ${\log }_{a}x$, is the number $y$ such that $x=a^y$.

The logarithm is the power $a$ has to be risen to in order to obtain $x$.

Remember, $x$ has to be a positive number. The log of zero or a negative number isn’t defined.

Examples

• ${\log }_{2}8$ Since $2^3=8$, that means ${\log }_{2}8=3$.
• ${\log }_{3}3$ Since $3^1$ = 3, ${\log }_{3}3$ must be $1$.

Breakdown

It’s pretty simple when you break it down, so let’s try a very simple one below.

• ${\log }_{2}16$

What do we need to raise 2 to the power of, to get 16?

• $2^4$ = 16

We raise it to the power of 4, which means ${\log }_{2}16$ = $4$.

Simple, right? Let’s try a scarier looking one.

The Scarier One

• ${\log }_5(\frac{1}{25})$

It’s not that bad I promise! We just need to do the same thing we did last time. What do we need to raise 5 to the power of to get $\frac{1}{25}$?

• $5^{-2}$ = 0.04 or $\frac{1}{25}$

So now we know ${\log }_5\frac{1}{25}=-2$.

Easy, right?

Why not try it yourself with a few of these examples? Click the little arrow to see the answer once you’re finished.

Q1)

${\log }_{10}1$

${\log }_{10}1=0$, since $10^0=1$

This one might be a bit confusing if you haven’t read this article.

Q2)

${\log }_8\frac{1}{2}$

${\log }_8\frac{1}{2}=\frac{1}{-3}$, since $\frac{1}{2}=8^{-\frac{1}{3}}$