---
title: "Advanced Algebra: Logs and Sequences"
---

## Advanced Algebra: The Magic if Logs and Sequences

(Why you only need to memorize the **geometric** sequence/sum formulas!)

### Log Rule Recap

$$
\begin{align}
\log(m \cdot n) &= \log(m) + \log(n) \tag{1} \\
\log(m^n) &= n\log(m) \tag{2}
\end{align}
$$

### Sequence Rule Recap

$$
\begin{align}
a_n &= a_1 + (n-1)d \tag{Arithmetic} \\
a_n &= a_1r^{n-1} \tag{Geometric}
\end{align}
$$

From our cheatsheet equation labeled $\text{(Geometric)}$, we know that an **explicit rule** for the terms of a **geometric** sequence will look like

$$
a_n = a_1r^{n-1}
$$

Now, to bring **logs** back in (for finals review!), let's look at what happens if we just take the log of both sides of this rule:

$$
\log(a_n) = \log(a_1r^{n-1})
$$

There's not much we can do with the left side, but on the right side, we remember our first log rule $(1)$ above, which tells us that we can split the $a_1$ part from the $r^{n-1}$ part!

$$
\log(a_n) = \log(a_1) + \log(r^{n-1}).
$$

Then, we remember our other log rule, equation $(2)$ above, which that tells us that we can bring the $n-1$ down **out of the exponent**:

$$
\log(a_n) = \log(a_1) + (n-1)\log(r)
$$

But look what we have now, does this look familiar? Check the equation labeled $\text{(Arithmetic)}$... We have exactly the equation for the explicit rule of an **arithmetic sequence**! If we define $\widehat{a_n} = \log(a_n)$, $\widehat{a_1} = \log(a_1)$, and $\widehat{r} = \log(r)$ (meaning, we put a little hat on the variable to indicate that it's the **log** of the variable), then the equation looks even more like the arithmetic sequence rule:

$$
\widehat{a_n} = \widehat{a_1} + (n-1)\widehat{r}
$$

So yeah, just wanted to make that linkage so that (a) you can derive the arithmetic sequence formula from the geometric sequence formula, in case you forget one of them, and (b) to bring **logs** back into mind for finals studying!