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### Course: Algebra 2>Unit 8

Lesson 4: The change of base formula for logarithms

# 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!
Suppose we wanted to find the value of the expression ${\mathrm{log}}_{2}\left(50\right)$. Since $50$ is not a rational power of $2$, it is difficult to evaluate this without a calculator.
However, most calculators only directly calculate logarithms in base-$10$ and base-$e$. So in order to find the value of ${\mathrm{log}}_{2}\left(50\right)$, we must change the base of the logarithm first.

## The change of base rule

We can change the base of any logarithm by using the following rule:
Notes:
• When using this property, you can choose to change the logarithm to any base $x$.
• 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!

## Example: Evaluating ${\mathrm{log}}_{2}\left(50\right)$‍

If your goal is to find the value of a logarithm, change the base to $10$ or $e$ since these logarithms can be calculated on most calculators.
So let's change the base of ${\mathrm{log}}_{2}\left(50\right)$ to $10$.
To do this, we apply the change of base rule with $b=2$, $a=50$, and $x=10$.
$\begin{array}{rlrl}{\mathrm{log}}_{2}\left(50\right)& =\frac{{\mathrm{log}}_{10}\left(50\right)}{{\mathrm{log}}_{10}\left(2\right)}& & \text{Change of base rule}\\ \\ & =\frac{\mathrm{log}\left(50\right)}{\mathrm{log}\left(2\right)}& & \text{Since}{\mathrm{log}}_{10}\left(x\right)=\mathrm{log}\left(x\right)\end{array}$
We can now find the value using the calculator.
$\frac{\mathrm{log}\left(50\right)}{\mathrm{log}\left(2\right)}\approx 5.644$

Problem 1
Evaluate ${\mathrm{log}}_{3}\left(20\right)$.

Problem 2
Evaluate ${\mathrm{log}}_{7}\left(400\right)$.

Problem 3
Evaluate ${\mathrm{log}}_{4}\left(0.3\right)$.

## Justifying the change of base rule

At this point, you might be thinking, "Great, but why does this rule work?"
${\mathrm{log}}_{b}\left(a\right)=\frac{{\mathrm{log}}_{x}\left(a\right)}{{\mathrm{log}}_{x}\left(b\right)}$
Let's start with a concrete example. Using the above example, we want to show that ${\mathrm{log}}_{2}\left(50\right)=\frac{\mathrm{log}\left(50\right)}{\mathrm{log}\left(2\right)}$.
Let's use $n$ as a placeholder for ${\mathrm{log}}_{2}\left(50\right)$. In other words, we have ${\mathrm{log}}_{2}\left(50\right)=n$. From the definition of logarithms it follows that ${2}^{n}=50$. Now we can perform a sequence of operations on both sides of this equality so the equality is maintained:
Since $n$ was defined to be ${\mathrm{log}}_{2}\left(50\right)$, we have that ${\mathrm{log}}_{2}\left(50\right)=\frac{{\mathrm{log}}_{x}\left(50\right)}{{\mathrm{log}}_{x}\left(2\right)}$ as desired!
By the same logic, we can prove the change of base rule. Just change $2$ to $b$, $50$ to $a$ and pick any base $x$ as the new base, and you have your proof!

## Challenge problems

Challenge problem 1
Evaluate $\frac{\mathrm{log}\left(81\right)}{\mathrm{log}\left(3\right)}$ without using a calculator.

Challenge problem 2
Which expression is equivalent to $\mathrm{log}\left(6\right)\cdot {\mathrm{log}}_{6}\left(a\right)$?

## Want to join the conversation?

3) Evaluate log_4(0.3)

When I put the equation into my calculator, it came up with -0.868483 (ect). So I rounded the 4 up to a 5 since the number to the right of it was 8, and in turn, rounded up the 8 to a 9, to come up with -0.869. But when I put that in as my answer, Khan said I was incorrect. As soon as I changed the answer to -0.868, however, I was told it was correct. Isn't that inaccurate?
• "To the nearest thousandth" means to three digits right of the decimal. You cannot round something by first rounding to other decimal places.

For example, say we're told to round 45 to the nearest hundred. If we do what you did here, we first round to 50, then round 50 up to 100. But 45 is closer to 0 than to 100, so this is inaccurate. We need to just round down to 0 in the first place.
• How is log(50)/log(2) equal to ln(50)/ln(2)? Since the power to which base (10) is raised to give us 50 is not the same power to which base (e) is raised to give us 50. Same goes for log(2) and ln(2).
• You're correct that log(50)≠ln(50) and log(2)≠ln(2). That doesn't mean that their ratios cannot be the same, for the same reason that 2≠4 and 3≠6 doesn't mean that 2/3≠4/6.

Say log(50)/log(2)=x. Then log(50)=xlog(2)
log(50)=log(2^x) by logarithm properties
50=2^x, raising both sides to the 10th power. Do you see how we end up with this same equation regardless of the base of the logarithms?
• what if there is a number in front of LOG
• If there is a number in front of the log symbol, it is a coefficient. When you see the expression a*log_b(c), you would first find the log base b of c, and then multiply the result by a. Hope this helps.
• Why can't we just plug in the logs on a calculator, (on a TI-84 you can calculate logs with other bases) you can just do for example if you wanted log₂(8) you can just do log(8, 2) on the calculator and it outputs 3.
• Many calculators (less expensive than a TI-84) only have ln or log base 10. The change of base rule enables you to use these calculators to get the result.
• to solve the challenge problem it is possible by setting log(81)/log(3) to x.... then solve for x and by using the exponent rule in reverse you get x=4
• You can use the change of base rule in revers to convert log(81)/log(3) into log3(81). Then find the exponent that when applied to 3 = 81, which is 4.
• Hello, everyone.

Maybe a silly question, but I don't remember previously hearing about it (or I forgot): why can't the argument be negative?

Is it a rule for all logarithms or only for the rule to hold?
If so, why does the rule require this?

• log(a) is the power to which you must raise 10 in order to get a. But raising 10 to any power yields a positive number. Specifically, there is no real number x such that 10^x is negative. So logarithms of negative numbers don't exist in the real numbers.

(You can make logarithms of negatives work using imaginary/complex numbers, but it's a complicated matter, beyond K-12 use of complex numbers.)
• Is log equivalent to a numerical value (like pi) ?
• A logarithm is a function. This means it will operate on a set of numbers using a set of rules. π is a constant number. This means it has a fixed, unchanging (but ever growing) value (it is a real number, in spite of the infinite digits)