# Change of Base Formula Calculator

Welcome to Omni's **change of base formula calculator**, where we'll learn **how to change the base of a log function**. In essence, a logarithm can sometimes be difficult to calculate if the numbers are not compatible. In such cases, we can turn to **the change of base rule** to make our lives easier and make the expression more approachable.

**So, what is the change of base formula?** Well, let's jump right into the article and find out!

## What is a logarithm?

Soon after we learn the basics of addition, they introduce us to **multiplication**. This new operation simplifies notation whenever we need to add the same number several times. But what if we go one step further and now want to **multiply by the same number a few times**? Again, there is a new way of writing that - the exponent. And instead of going further, let's stay with this one for a minute.

Useful as they are, it's sometimes useful to have addition, multiplication, and exponent **work both ways**. In other words, often, instead of adding, we'd like to subtract. Instead of multiplying, we'd like to divide. And instead of taking the power, we'd like to... Hm, **what would we like to do?**

**The logarithm is the inverse operation to taking an exponent**, i.e., the inverse function to the exponential one. However, we need to be careful here since taking the root can also be considered the inverse of an exponent. The difference is that with the root, it's not an exponential function that we're inverting but a polynomial one. Just to be on the safe side, **let's see an example**.

Taking the `k`

-th power of a number `n`

means:

`nᵏ = n·n·n·...·n`

,

with `n`

repeated `k`

times. Then, **the** `k`

**-th root of** `nᵏ`

is `n`

, i.e.,

`ᵏ√(nᵏ) = n`

.

On the other hand, **the logarithm with base** `n`

**of** `nᵏ`

is `k`

, i.e.,

`logₙ(nᵏ) = k`

.

In other words, **the two operations return two different values** (although both connected to the exponent). Just to be on the safe side, let's have **the proper, one-sentence logarithm definition** highlighted like so:

💡 `logₐ(b)` gives you the power to which you'd need to raise `a` in order to obtain `b` . Note, however, that in general, this can be a fractional exponent. |

Before we move further, let us have a pretty bullet list with **a few vital points of information about our new friend, the logarithm function**.

- There are
**two very special cases of the logarithm**which have unique notation: the natural logarithm and the logarithm with base`10`

. They are denoted`ln(x)`

and`log(x)`

(the second one simply without the small`10`

), and their bases are, respectively, the Euler number`e`

and (surprise, surprise!) the number`10`

. **The logarithm function is defined only for positive numbers.**In other words, whenever we write`logₐ(b)`

, we require`b`

to be positive.- Whatever the base,
**the logarithm of**`1`

**is equal to**`0`

. After all, whatever we raise to power`0`

, we get`1`

. **Logarithms are extremely important.**And we mean**EXTREMELY**important. Outside of mathematics, they're used in**statistics**(e.g., the lognormal distribution)**economy**(e.g., the the GDP index),**medicine**(e.g., the QUICKI index), and**chemistry**(e.g., the half-life decay). Also, quite**a few physical units**are based on logarithms, for instance, the Richter scale, the pH scale, and the dB scale.

Useful as they are, it may appear that **logarithms are difficult to calculate**. After all, we need to "*guess*" to what power we need to raise one number to obtain another. But in fact... Well, actually, no. **It indeed is quite tricky.**

For instance, we can easily see that `log₃(9)`

is `2`

because if we raise `3`

to the second power, we'll obtain `9`

. But what is `log₃(16)`

? Surely, **it won't be an integer anymore**, and it's really hard to guess a fractional exponent.

Fortunately, there are ways to make our lives a little easier, and one of them is **the log change of base**. It may not always give us the answer straight away, but surely, it'll get us one step closer.

## What is the change of base formula?

As we've mentioned in **it's not always easy to find the value of a log function**, and it's often because its base and argument are not "*compatible*". The logarithmic change of base is a way to make at least one of them simpler.

**So what is the change of base formula?** Well, here it is:

As you can see, it takes a single logarithmic expression and transforms it into a fraction of two new ones, but **with a different base**. Arguably, we still have to **take caution when using the log change of base**. After all, if we translate one expression that we can't calculate into two expressions that we can't calculate, then what's the point?

Still, it may happen that **the logarithm change of base is the simplest way to get the answer**. For instance, if we'd like to find `log₂₇(9)`

, then if you're not a big-headed mathematician, you might not see the solution straight away. However, we can observe that `27`

and `9`

are respectively, the cubic and the square root of `3`

(i.e., that `3³ = 27`

and `3² = 9`

). Therefore, we could **use the change of base rule** to have both these numbers inside logarithms with base `3`

. From there, it's a piece of cake.

Well, **we began to see numbers**, that must be a good sign, right? And indeed, it is because that means that we're ready to leave theory behind and **move on to examples**!

Ready? **We know we are!**

## Example: how to change the base of log

Say that you have the end of the semester getting closer and closer, and that can mean only one thing - **a mathematics test** that summarizes everything you've learned the last few months.

Being the good student that you are, **you decide to start studying early and systematically**. The first topic at hand concerns **logarithms**. Needless to say, they're not the most straightforward thing there is in mathematics. But they don't seem too bad, either. Still, if it's going to be on the test, it seems like a good idea to spare a few moments on the subject, **especially the change of base rule**. You choose two examples to practice how much you remember: change `log₂₇(9)`

to base `3`

and change `log₅(1000)`

to base `10`

.

Firstly, let's see **how easy the task is when we have the change of base formula calculator at hand**. There, we have three variable fields: `x`

, `a`

, and `b`

. According to the formula given at the top, we determine that `x`

**denotes the number inside our logarithm**, `a`

**is its base**, and `b`

**is the new base** that we'd like to obtain.

Therefore, for the first case, i.e., when we want to change `log₂₇(9)`

to base `3`

, we have to put into the calculator:

`x = 9`

, `a = 27`

, `b = 3`

.

Similarly, when we change `log₅(1000)`

to base `10`

, then we have:

`x = 1000`

, `a = 5`

, `b = 10`

.

Observe how, when we input the first two values, the change of base formula calculator **already gives us the value** of our logarithm. Once we input the third, it additionally gives **the step-by-step application of the change of base rule**. As simple as that, we got the answer!

Let's now see **how to change the base of the log ourselves using all that we've learned in this article**. We begin with making `log₂₇(9)`

into an expression with base `3`

.

According to the logarithmic change of base from a fraction of two others, both with base `3`

instead of the `27`

that we have. The nominator is the **log of the argument** (the big number, the one in the brackets). And the denominator is **that of the original base**. So that's:

`log₂₇(9) = log₃(9) / log₃(27)`

.

Now, we observe that `3² = 9`

and `3³ = 27`

, so:

`log₂₇(9) = log₃(9) / log₃(27) = 2/3`

.

Well, **that wasn't so bad, was it?** Encouraged, let's face the second example: changing `log₅(1000)`

to base `10`

. Again, we use the formula and obtain:

`log₅(1000) = log₁₀(1000) / log₁₀(5)`

.

After some head-scratching, we can see that `10³ = 1000`

, so

`log₅(1000) = log₁₀(1000) / log₁₀(5) = 3 / log₁₀(5)`

.

**But what do we do with** `log₁₀(5)`

**?**

Unfortunately, **nothing too clever**. As we've already said in , **finding logarithms is, in general, no easy matter**. We could play around with some approximations, but this could take ages. Instead, **we usually use external tools for such computations**, just like Omni's change of base formula calculator.

Nevertheless, we still got an expression that's not half bad and, most importantly, **practiced our knowledge of the change of base rule**. We can happily check logarithms off the list of things to study for the test. Surely, this must contribute at least thirty percent of the material on the test, right?