The change of base formula, as its name suggests, is used to change the base of a logarithm. We might have noticed that a scientific calculator has only “log” and “ln” buttons. Also, we know that “log” stands for a logarithm of base 10 and “ln” stands for a logarithm of base e. But there is no option to calculate the logarithm of a number with any other bases other than 10 and e. The change of base formula solves this issue. Also, it is used in solving several logarithms problems. Let us learn the change of base formula along with its proof and a few solved examples.

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## What Is Change of Base Formula?

The change of base formula is used to write a logarithm of a number with a given base as the ratio of two logarithms each with the same base that is different from the base of the original logarithm. This is a property of logarithms. You can see the change of base formula here.

### Change of Base Formula

The change of base formula is:

logbb a = [logcc a] / [logcc b]

In this formula,

- The argument of the logarithm in the numerator is the same as the argument of the original logarithm.
- The argument of the logarithm in the denominator is the same as the base of the original logarithm.
- The bases of both logarithms of numerator and denominator should be the same and this base can be any positive number other than 1.

**Note: **Another form of this formula is, logbb a · logcc b = logcc a, which is also widely used in solving the problems.

## Change of Base Formula Derivation

Let us assume that

logbb a = p,logcc a = q, and logcc b = r.

By converting each of these into exponential form, we get,

a = b^{p}, a = c^{q}, and b = c^{r}.

From the first two equations,

b^{p} = c^{q}

Substituting b = c^{r} (which is from third equation) here,

(c^{r})^{p }= c^{q}

c^{rp }= c^{q} (Using a property of exponents, (a^{m})^{n} = a^{mn})

pr = q

p = q / r

Substituting the values of p, q, and r here,

logbb a = [logcc a] / [logcc b]

## Logarithm change of base rule intro

Learn how to rewrite any logarithm using logarithms with a different base. This is very useful for finding logarithms in the calculator!

## The change of base rule

We can change the base of any logarithm by using the following rule:

As always, the arguments of the logarithms must be positive and the bases of the logarithms must be positive and not equal to 1 in order for this property to hold!

## Examples Using Change of Base Formula

**Example 1: **Evaluate the value of log6464 8 using the change of base formula.

**Solution:**

We will apply the change of base formula (by changing the base to 10). Note that log1010 is same as log.

log_{64} 8 = [log 8] / [log 64]

= [log 8] / [log 8^{2}]

= [log 8] / [2 log 8] [∵ log a^{m} = m log a]

= 1 / 2

**Answer: **log_{64} 8 = 1 / 2.

**Example 2: **Caluclate log99 8 using the calculator. Round your answer to 4 decimals.

**Solution:**

We cannot calculate log_{9} 8 directly using the calculator because it doesn’t have a button named log_{9}. Thus, we apply the change of base formula first.

log_{9} 8 = [log 8] / [log 9]

= [0.903089…] / [0.95424…]

≈ 0.9464

**Answer: **log_{9} 8 ≈ 0.9464.

**Example 3: **Evaluate the value of log_{3} 2 · log_{4} 3 · log_{5} 4.

**Solution:**

By alternate form of the change of base formula, log_{b} a ⋅ log_{c} b = log_{c} a. We apply this twice to evaluate the given expression.

log_{3 }2 · log_{4 }3 · log_{5} 4

= log_{4} 2 · log_{5} 4

= log_{5} 2

**Answer: **log_{3} 2 · log_{4} 3 · log_{5} 4 = log_{5} 2.

## Justifying the change of base rule

At this point, you might be thinking, “Great, but *why* does this rule work?”

## FAQs on Change of Base Formula

### What Is Change of Base Formula?

The change of base formula is used to change the base of a logarithm. It has two forms.

- log
_{b}a = [log_{c}a] / [log_{c}b] - log
_{b}a ⋅ log_{c}b = log_{c}a

### How To Derive Change of Base Formula?

### What Are the Applications of Change of Base Formula?

The change of base formula is mainly used to change the base of a logarithm to any desired base. This is many used to calculate the logarithms with any other base than 10 and “e” because the calculator has options to calculate the logarithms with bases 10 (log button) and e (ln button) only.

### How To Use Change of Base Formula?

## Basic Log Rules & Expanding Log Expressions

You have learned various rules for manipulating and simplifying expressions with exponents, such as the rule that says that *x*^{3} × *x*^{5} equals *x*^{8} because you can add the exponents. There are similar rules for logarithms.

In less formal terms, the log rules might be expressed as:

1) Multiplication inside the log can be turned into addition outside the log, and vice versa.

2) Division inside the log can be turned into subtraction outside the log, and vice versa.

3) An exponent on everything inside a log can be moved out front as a multiplier, and vice versa.

Warning: Just as when you’re dealing with exponents, the above rules work *only* if the bases are the same. For instance, the expression “log_{d}(*m*) + log_{b}(*n*)” cannot be simplified, because the bases (the “d” and the “b”) are not the same, just as *x*^{2} × *y*^{3} cannot be simplified because the bases (the *x* and *y*) are not the same.

## Expanding logarithms

Log rules can be used to simplify (or, more correctly, to “condense”) expressions, to “expand” expressions, or to solve for values. We’ll start with expansion.

- Expand log
_{3}(2*x*).

When the instructions say to “expand”, they mean that they’ve given me one log expression with lots of stuff inside it, and they want me to use the log rules to take the log apart into many separate log terms, each with only one thing inside its particular log. That is, they’ve given me *one* log with a *complicated* argument, and they want me to convert this to *many* logs, each with a *simple* argument.

In this case, I have a “2*x*” inside the log. Since “2*x*” is multiplication, I can take this expression apart, according to the first of the log rules above, and turn it into an addition outside the log:

log_{3}(2*x*) = log_{3}(2) + log_{3}(*x*)

Then the answer they are looking for is:

log_{3}(2) + log_{3}(*x*)

Note: Do not try to evaluate “log_{3}(2)” in your calculator. While you would be correct in saying that “log_{3}(2)” is just a number (and we’ll be seeing later how to rearrange this expression into something that you *can* evaluate in your calculator), what they’re actually looking for here is the “exact” form of the log, as shown above, and not a decimal approximation from your calculator.

If you give “the answer” as being the decimal approximation, you should expect to lose points.

- Expand log
_{4}(^{ 16}/)._{x}

I have division inside the log. According to the second of the log rules above, this can be split apart as subtraction outside the log, so:

log_{4}(^{ 16}/_{x}) = log_{4}(16) – log_{4}(*x*)

The first term on the right-hand side of the above equation can be simplified to an exact value, by applying the basic definition of what a logarithm is. In this case, I’m using the fact that the power required on 4 to create 16 is 2; in other words, since 4^{2} = 16, then:

log_{4}(16) = 2

Then the original expression expands fully as:

log_{4}(^{ 16}/_{x} ) = **2 – log _{4}(x)**

Always remember to take the time to check to see if any of the terms in your expansion (such as the log_{4}(16) above) can be simplified.

- Expand log
_{5}(*x*^{3}).

The exponent inside the log can be taken out front as a multiplier:

log_{5}(*x*^{3}) = 3 · log_{5}(*x*) = **3log _{5}(x)**

The examples above are very simple uses of the log rules, as applied to the expansion of log expressions. On the next page, we’ll take a look at the sort of exercises you’ll be seeing in your homework and on the next test.

The second term above, with just a “5” inside, is as “expanded” as it can get, because there’s only just the one thing inside the log. And, because 5 is not a power of 2, there’s no simplification I can do. So that part of the expansion is done; I’ll just be carrying the “log(5)” along for the ride to the final answer.

In the first term, though, there’s still more than just one thing inside the log. In particular, I see that there’s an exponent inside the log. However, I can’t take the exponent out front yet, because that power is only on the *x*, not the 8. I have to remember that the rule says that I can only take the exponent out front if it is “on” everything inside the log. So I first need to isolate that part of the argument that has the power on it.

The 8 is multiplied onto the *x*^{4}, so I can split the factors inside the log by converting to added logs:

log_{2}(8*x*^{4}) – log_{2}(5) = log_{2}(8) + log_{2}(*x*^{4}) – log_{2}(5)

Since 8 is a power of 2 (namely, 2^{3}), I can simplify the first log to an exact value. Because 2^{3} = 8, then log_{2}(8) = 3, so I get:

log_{2}(8) + log_{2}(*x*^{4}) – log_{2}(5)

= 3 + log_{2}(*x*^{4}) – log_{2}(5)

Okay; now I’m finished with the first term, too; I’m only left with the middle term to expand, with the exponent inside its log.

The variable *x* has the exponent (which is now “on” everything inside its log), so I can use a log rule and move the exponent out in front of the log as a multiplier:

log_{2}(8) + log_{2}(*x*^{4}) – log_{2}(5)

= 3 + 4log_{2}(*x*) – log_{2}(5)

Each log now finally contains only one thing, and the first log term has been simplified to a numerical value, so this expression is fully expanded. Then my final answer is:

3 + 4log_{2}(*x*) – log_{2}(5)

In following my work in the steps of the above computations, you may have felt that I was being a bit confusing, carrying the “log_{2}(5)” and “3” along as I did other steps. But it is important to not drop bits of an exercise as one goes along. In whatever manner you decide to do your work — maybe including doing some of the steps, or at least portions of some of them, off to the side on scratch paper — make sure that all the steps in your final result make sense.

For instance, if I remove the talking between the steps, the previous example is worked out as follows:

= log_{2}(8) + log_{2}(*x*^{4}) – log_{2}(5)

= 3 + log_{2}(*x*^{4}) – log_{2}(5)

= 3 + 4log_{2}(*x*) – log_{2}(5)

Just be careful not to try to do too many things in any one step, at least when you’re just getting started.

This is a gawd-awful mess! To do the expansion, I’ll be using the log rules, and I’ll be taking care not to try to do anything “in my head” or too much all at once.

The first thing I see, inside the log, is that I’ve got one complicated expression that’s divided by another complicated expression. To start my expansion, then, I’ll split the division inside the log into subtraction of logs outside.

Inside each of the two log terms I’ve got now, I find multiplication. So my next step will be to take apart the multiplications inside as addition of logs outside. To make sure I don’t mess up my signs, I’ll be sure to put grouping symbols around the results of each split.

Don’t think that this example is “too complicated” to show up on the next test. In fact, you should *expect* to see at least one question *at least* this “complicated”!

### Condensing Log Expressions

The logs rules work “backwards”, so you can condense (“compress”?) strings of log expressions into one log with a complicated argument. When they tell you to “simplify” a log expression, this usually means they will have given you lots of log terms, each containing a simple argument, and they want you to combine everything into one log with a complicated argument. “Simplifying” in this context usually means the opposite of “expanding”.

There is no standard definition, in this context, for “simplifying”. You have to use your own good sense. If they give you a big complicated thing and ask you to “simplify”, then they almost certainly mean “expand”. If they give you a string of log terms and ask you to “simplify”, then they almost certainly mean “condense”.

Pay particular attention to how I grouped the log terms according to sign. This can be very important, and is where many students get lost and then lose points. Don’t try to convert addition outside to multiplication inside, or subtraction outside to division inside, until you’ve made sure that all the “plus” terms are together up front, followed by all the “minus” terms. Then you can combine by multiplication within each set, and then finish by converting the big “minus” that’s subtracted from the big “plus” into one big division inside one log.

## Trick Questions Based on Log Rules

The basic log rules are as follows:

1) log_{b}(*mn*) = log_{b}(*m*) + log_{b}(*n*)

2) log_{b}(^{m}/_{n}) = log_{b}(*m*) – log_{b}(*n*)

3) log_{b}(*m ^{n}*) =

*n*· log

_{b}(

*m*)

You should expect to need to know the basic log rules, because there is a certain type of question that the teacher can put on the test to make sure you *know* the rules, and know how to *use* the rules.

Warning: If you discover one of these trick questions on your next test, you will *not* be able to “cheat” your way to the answer with your calculator. If you don’t know the log rules, you’re toast.

Okay, let’s dive into these trick problems.

- Let log
_{b}(2) = 0.3869, log_{b}(3) = 0.6131, and log_{b}(5) = 0.8982. Using these values, evaluate log_{b}(10).

The trick to doing this exercise is to notice that they’ve asked me to find something (namely, the log of ten) which can be created out of what they’ve given me (namely, the logs of two and five).

I just need to figure out how to apply the log rules to allow for the substitutions I’m going to need to make. So:

Since 10 = 2 × 5, and since they’ve given me logs of 2 and 5, then it will be helpful to me to turn the 10 into the product of 2 and 5, and then use a log rule to split the log with multiplication inside into the addition of two log expressions. In other words:

log_{b}(10) = log_{b}(2 × 5)

= log_{b}(2) + log_{b}(5)

Since they gave me values for log_{b}(2) and log_{b}(5), I can now substitute and evaluate:

log_{b}(2) + log_{b}(5)

= 0.3869 + 0.8982

= 1.2851

Then:

log_{b}(10) = 1.2851

- Let log
_{b}(2) = 0.3869, log_{b}(3) = 0.6131, and log_{b}(5) = 0.8982. Using these values, evaluate log_{b}(9).

They’ve asked me for the log of 9; they’ve given me (among other things) the log of 3. Since 9 = 3^{2}, then I can convert “9” to “3^{2}“, and then apply the log rule that moves a power inside a log to being a multiplier in front of the log.

log_{b}(9) = log_{b}(3^{2})

= 2 · log_{b}(3)

Since I have the value for log_{b}(3), then I can substitutie and evaluate:

2 · log_{b}(3)

= 2 · 0.6131

= 1.2262

Then:

log_{b}(9) = 1.2262

- Let log
_{b}(2) = 0.3869, log_{b}(3) = 0.6131, and log_{b}(5) = 0.8982. Using these values, evaluate log_{b}(7.5).

This one is a bit more complicated, but, after fiddling with the numbers for a bit, I notice that 7.5 = 15 ÷ 2. “So what?”, you ask. Well, I can create 15 from 3 and 5, and they’ve given me values for the logs of 3 and 5. And the division by 2 is easy to deal with, because they’ve given me a value for the log of 2. So:

log_{b}(7.5)

= log_{b}(15 ÷ 2)

= log_{b}(15) – log_{b}(2)

And 15 = 5 × 3, so:

log_{b}(15) – log_{b}(2)

= [log_{b}(5) + log_{b}(3)] – log_{b}(2)

= log_{b}(5) + log_{b}(3) – log_{b}(2)

And now I can substitute and evaluate:

log_{b}(5) + log_{b}(3) – log_{b}(2)

= 0.8982 + 0.6131 – 0.3869

= 1.1244

Then:

log_{b}(7.5) = 1.1244

- Let log
_{b}(2) = 0.3869, log_{b}(3) = 0.6131, and log_{b}(5) = 0.8982. Using these values, evaluate log_{b}(6).

Since 6 = 2 × 3, then:

log_{b}(6) = log_{b}(2 × 3)

= log_{b}(2) + log_{b}(3)

Since I have these values, I can evaluate:

log_{b}(2) + log_{b}(3)

= 0.3869 + 0.6131

= 1.0000

Then:

log_{b}(6) = 1.0000

Hmm… that was interesting. I got that log_{b}(6) = 1. Using The Relationship for logs, I get:

log_{b}(6) = 1

b^{1} = 6

b = 6

So now I know that their mysterious unnamed base “b” was actually 6. But they will not usually give you problems that let you figure out the base in this manner. In fact, they may give you questions for which the math doesn’t even work out. For instance:

- Let log
_{b}(2) = 0.4, log_{b}(3) = 0.6, log_{b}(4) = 0.75, and log_{b}(7) = 1.3. Using these values, evaluate log_{b}(21).

Doing this evaluation is easy. They’ve given me the logs of 3 and of 7, and 21 = 3 × 7, so:

log_{b}(21) = log_{b}(3 × 7)

= log_{b}(3) + log_{b}(7)

= 0.6 + 1.3 = 1.9

Then:

log_{b}(21) = 1.9

However, look at the other information that they included in this exercise. The value of the log of 4 *should* follow from the value of the log of 2, because 4 = 2^{2}. But look:

log_{b}(4) = log_{b}(2^{2})

= 2 · log_{b}(2)

= 2 · 0.4 = 0.8

This is *not* the value they gave me for log_{b}(4). They’d said that log_{b}(4) = 0.75. What’s up?

Well, what’s up is that they wanted to see if I really knew the log rules well enough to go straight to evaluation. Clearly, these log “values” aren’t related to any actual base; the author of the exercise just picked somewhat random values. The “right” answer to the question is not the mathematically-valid one, since they weren’t working with any particular log anyway; the “right” answer is the result of a correct application of the log rules, which is what this sort of question is meant to test.

## The Change-of-Base Formula

There is one other log “rule”, but it’s more of a formula than a rule.

You may have noticed that your calculator only has keys for figuring the values for the common (that is, the base-10) log and the natural (that is, the base-*e*) log. There are no keys for any other bases. Some students try to get around this by “evaluating” something like “log_{3}(6)” with the following keystrokes:

**[**LOG**]** **[** 3 **] [** ( **] [** 6** ] [** ) **]**

Of course, they then get the wrong answer, because the above actually (usually) calculates the value of “log_{10}(3) × 6″. This is not what had been intended.

In order to evaluate a non-standard-base log, you have to use the Change-of-Base formula:

What this rule says, in practical terms, is that you can evaluate a non-standard-base log by converting it to the fraction of the form “(standard-base log of the argument) divided by (same-standard-base log of the non-standard-base)”. I keep this straight by looking at the position of things. In the original log, the argument is “above” the base (since the base is subscripted), so I leave things that way when I split them up:

Here’s a simple example of this formula’s application:

- Evaluate log
_{3}(6). Round your answer to three decimal places.

The argument is 6 and the base is 3. I’ll plug them into the change-of-base formula, using the natural log as my new-base log:

I would have gotten the same final answer if I had used the common log instead of the natural log, though the numerator and denominator of the intermediate fraction would have been different from what I displayed above:

As you can see, it doesn’t matter which standard-base log you use, as long as you use the same base for both the numerator and the denominator.

While I showed the numerator and denominator values in the above calculations, it is actually best to do the calculations entirely within your calculator. You don’t need to bother with writing out that intermediate step.

In fact, to minimize on round-off errors, it is best to try to do all the steps for the division and evaluation in your calculator, all in one go. In the above computation, rather than writing down the first eight or so decimal places in the values of ln(6) and ln(3) and then dividing, you would just do “ln(6) ÷ ln(3)” in your calculator.

You may get some simple (but fairly useless) exercises on this topic. Don’t begrudge them; they’re easy points, as long as you keep the change-of-base formula straight in your head. For instance:

- Convert log
_{3}(6) to an expression with logs having a base of 5.

I can’t think of any particular reason why a base-5 log might be useful, so I think the only point of these problems is to give you practice using change-of-base. Fine; I’ll plug-n-chug:

Since getting an actual decimal value is not the point in exercises of this sort (the converting using change-of-base is the point), just leave the answer as a logarithmic fraction.

While the above exercises were fairly pointless, using the change-of-base formula can be very handy for finding plot-points when graphing non-standard logs, especially when you are supposed to be using a graphing calculator.

- Use your graphing utility to graph
*y*= log_{2}(*x*).

If I were working by hand, I would use the definition of logs to note that:

- since 2
^{-2}= ¼, then log_{2}(¼) = -2 - since 2
^{–1}= ½, then log_{2}(½) = –1 - since 2
^{0}= 1, then log_{2}(1) = 0 - since 2
^{1}= 2, then log_{2}(2) = 1 - since 2
^{2}= 4, then log_{2}(4) = 2 - since 2
^{3}= 8, then log_{2}(8) = 3 - since 2
^{4}= 16, then log_{2}(16) = 4

And then I would draw my graph by hand.

(Why did I pick these particular *x*-values? Because anything smaller would have been too tiny to graph by hand, and anything larger would have led to a ridiculously wide graph. I picked the values that fit my needs.)

But, in this case, I’m supposed to be doing the graph with my graphing calculator. How can I do this? (Or what if I’d just like to use my graphing calculator’s “TABLE” feature to find some nice neat plot points?) I don’t have a “log-base-two” button. However, I can enter the given function into my calculator by using the change-of-base formula to convert the original function to something that’s stated in terms of a base that my calculator can understand. Flipping a coin, I choose the natural log:

(I could have used the common log, too. In that case, the function would have been “*y*1 = log(*x*)/log(2)”.)

In my graphing calculator, after adjusting the viewing window to show useful parts of the plane, the graph will look something like this:

By the way, you can check that the graph contains the expected “neat” points (that is, the points I would have calculated by hand, as shown above) to verify that the picture displays the correct graph:

## Use the change-of-base formula for logarithms

Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or \displaystyle e*e*, we use the **change-of-base formula** to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.

To derive the change-of-base formula, we use the **one-to-one** property and **power rule for logarithms**.

## Change of Base Formula or Rule

I have discussed most of the log rules in a separate lesson. However, I have intentionally left one out to discuss it here in detail. The log rule is called the **Change-of-Base Formula**.

If you are interested in why the Change-of-Formula works, click the following link to see the proof: Proofs of Logarithm Properties.

The logarithm that is using base-10 is known as the * common logarithm*, while the one using base-e

*e*is known as the

*.*

**natural logarithm**## The Number *e*

**Note:** The number e*e* is a mathematical constant which has a numerical value of e ≈ 2.71828. It is an irrational number because it cannot be expressed as a ratio of two integers or as a fraction. More so, the number e*e* is the base of the natural logarithm.

Therefore, the common logarithms and natural logarithms are utilizing the **standard bases** 10 and *e*, respectively.

Before we continue, I would like to point out some nuances or subtleties with regard to the mathematical expressions of common and natural logarithms.

## The LOG and LN Buttons of a Graphing Calculator

Most graphing calculators have functions or keys that directly calculate the logarithms of numbers in base-1010 and base-e*e*. Thus, you will only see two buttons: **LOG** for common logarithm and **LN** for natural logarithm.

It is obvious that a problem arises when we want to calculate the logarithm of a number using non-standard bases such 2, 3, 7, 0.5, and 0.25.

The logarithms above use NON-STANDARD bases because they are neither 10 nor the number *e*.

How do we proceed to punch in the numbers on a graphing calculator? As I mentioned before, most calculators are limited to only compute logarithms with base 10 and base e.

This is where the **Change-of-Base Formula** comes to rescue. It can convert a non-standard-base logarithm as a ratio of two logarithmic operations that use the standard base either 10 or the constant *e*.

**What is the Change-of-Base Formula**?

The **Change-of-Base Formula** is an instruction on how to rewrite or transform a given logarithmic expression as a ratio or fraction of two logarithm operations using any valid base.

That means, if we have a logarithm using a specific base, then we can turn this into an equivalent ratio or fraction of two logarithmic operations such that we can pick any base that we want. We can literally select any base as long as it is positive but not equal to 1.

But if we want to calculate or know the value of a logarithm, we should pick base-10 or base-*e* since most calculators have these function keys. The **log key** [log] computes the common (using base-10) logarithm while the **ln key** [ln] calculates the natural (using base-*e*) logarithm.

Let’s parse how the formula transformed the original logarithm into an equivalent expression as a ratio of two log operations.

The argument of the original logarithm becomes the argument of the logarithm of the numerator.

The base of the original logarithm becomes the argument of the logarithm of the denominator.

The logarithms in the numerator and denominator have the same base. The value of the base \large\color{green}c*c* is any base that we chose.

## Examples of the Change-of-Base Formula

The first two examples (Example #1 and #2) are perfect textbook problems because the argument and the base of a logarithm can be expressed as powers of a common number (a positive number that is not equal to 1) which serves as the new base when applying the change-of-base rule.

**Example 1:** Evaluate log_{4}8.

The first thing I recognized is that both the argument and its base can be expressed as a power of 2.

Here is the complete solution.

The Negative Rule of Exponent allows us to invert the fraction (reciprocal) but we must **switch** the sign of the exponent.

**Example 3:** Calculate the value of log_{5}(12). Round your answer to the nearest thousandth.

This is no longer a “nice” problem because the argument and the base of the logarithm cannot be expressed as powers of a common number. In other words, there is no scenario where we can express 5 and 12 as exponential numbers such that they have the same base.

To solve this, we can use the change-of-base rule to rewrite the original logarithm as a ratio of two logarithms of the base of our choosing. We have two options: use base-10 or base-*e*. It doesn’t matter which one we choose because the answer will be the same. For this problem, let’s use base-10.

Don’t forget to round your answer to three decimal places because we are asked to round it to the nearest thousandth.

Our calculator should confirm that our answer is correct.

**Example 4:** Calculate the value of log_{7}(9). Round your answer to the nearest hundredth.

In the previous example, we used base-10 to evaluate the logarithm. This time we will use the natural number *e* as the base of choice when applying the Change-of-Base formula.

Notice that we don’t need to write the natural logarithm as log* _{e}*(

*x*). We can omit that step and write it immediately as ln(

*x*). I added it as one of the steps below for clarity and emphasis purposes.

Let’s go ahead and apply the change-of-base rule to convert log_{7}(9) as a ratio or fraction of two natural log operations.

Also, make sure to round your answer to two decimal places since the problem requires as to express our final answer in the nearest hundredth.

Your calculator should output similar result like the one below.

**Example 5:** Change log_{6}(0.1) as a quotient of two natural logarithms. Calculate its value then round to the nearest tenth.

This problem requires us to change the given logarithm as a quotient of natural logarithms. That means we have no choice but to use the natural number *e* as the base when we apply the change-of-base formula. Don’t forget as well that we are told to round our answer to the nearest tenth (one decimal place).

Here’s our solution:

Our calculator agrees with our answer.

**Example 6:** Change log(7) as a quotient in terms of natural logarithms. Then compute its value. Round your answer to the nearest ten-thousandth.

I admit that although we can directly solve for the value of log(7) with a calculator since it has a **LOG **key, this problem wants us to take the long route. Not because it is a futile exercise of our time but more importantly, it is an opportunity for us to apply our solid grasp or understanding of the change-of-base formula.

Remember that when you see a log operation without a base, it is assumed to have a base of 10. Therefore, our very first step is to rewrite log(7) as log10(7) so that it is much easier to see what are the numbers we are dealing with during the change-of-base step.

It’s a great feeling when a calculator outputs the value that verifies our answer.

**Example 7:** Change ln(13) as a ratio in terms of common logarithms. Then calculate its value. Round your answer to the nearest ten-thousandth.

Just like example #6, there’s no need to apply the change-of-base formula because we can directly compute it with a calculator. However, the intention of this problem is to showcase our in-depth understanding of common and natural logarithms, and how to properly handle the formula.

So, let’s transform ln(13) in **LOG form** where the base is e, thus, log* _{e}*(13). Now, we use the change-of-base formula to express it as a ratio of two common logarithms. Remember, the common logarithm uses base-10

Yes! The calculator agrees with our answer.

✅ Math Formulas ⭐️⭐️⭐️⭐️⭐

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