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## Algebra 2 (Eureka Math/EngageNY)

### Course: Algebra 2 (Eureka Math/EngageNY)>Unit 3

Lesson 6: Topic B: Lesson 13: Changing the base

# 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 log, start base, 2, end base, left parenthesis, 50, right parenthesis. 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 log, start base, 2, end base, left parenthesis, 50, right parenthesis, 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 start color #0d923f, x, end color #0d923f.
• 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 $\log_2(50)$log, start base, 2, end base, left parenthesis, 50, right parenthesis

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 log, start base, 2, end base, left parenthesis, 50, right parenthesis to start color #1fab54, 10, end color #1fab54.
To do this, we apply the change of base rule with b, equals, 2, a, equals, 50, and x, equals, 10.
\begin{aligned}\log_\blueD{2}(\purpleC{50})&=\dfrac{\log_{\greenD{10}}(\purpleC{50})}{\log_{\greenD{10}}(\blueD2)} &&{\gray{\text{Change of base rule}}} \\\\ &=\dfrac{\log(50)}{\log(2)} &&{\gray{\text{Since} \log_{10}(x)=\log(x)}} \end{aligned}
We can now find the value using the calculator.
start fraction, log, left parenthesis, 50, right parenthesis, divided by, log, left parenthesis, 2, right parenthesis, end fraction, approximately equals, 5, point, 644

Problem 1
Evaluate log, start base, 3, end base, left parenthesis, 20, right parenthesis.

Problem 2
Evaluate log, start base, 7, end base, left parenthesis, 400, right parenthesis.

Problem 3
Evaluate log, start base, 4, end base, left parenthesis, 0, point, 3, right parenthesis.

## Justifying the change of base rule

At this point, you might be thinking, "Great, but why does this rule work?"
log, start base, b, end base, left parenthesis, a, right parenthesis, equals, start fraction, log, start base, x, end base, left parenthesis, a, right parenthesis, divided by, log, start base, x, end base, left parenthesis, b, right parenthesis, end fraction
Let's start with a concrete example. Using the above example, we want to show that log, start base, 2, end base, left parenthesis, 50, right parenthesis, equals, start fraction, log, left parenthesis, 50, right parenthesis, divided by, log, left parenthesis, 2, right parenthesis, end fraction.
Let's use n as a placeholder for log, start base, 2, end base, left parenthesis, 50, right parenthesis. In other words, we have log, start base, 2, end base, left parenthesis, 50, right parenthesis, equals, n. From the definition of logarithms it follows that 2, start superscript, n, end superscript, equals, 50. Now we can perform a sequence of operations on both sides of this equality so the equality is maintained:
\begin{aligned} 2^n &= 50 \\\\ \log(2^n) &= \log(50)&&{\gray{\text{If }A=B\text{, then }\log(A)=\log(B)}} \\\\ n\log(2)&=\log(50)&&{\gray{\text{Power Rule}}} \\\\ n &= \dfrac{\log(50)}{\log(2)} &&{\gray{\text{Divide both sides by} \log(2)}} \end{aligned}
Since n was defined to be log, start base, 2, end base, left parenthesis, 50, right parenthesis, we have that log, start base, 2, end base, left parenthesis, 50, right parenthesis, equals, start fraction, log, start base, x, end base, left parenthesis, 50, right parenthesis, divided by, log, start base, x, end base, left parenthesis, 2, right parenthesis, end fraction 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 start fraction, log, left parenthesis, 81, right parenthesis, divided by, log, left parenthesis, 3, right parenthesis, end fraction without using a calculator.

Challenge problem 2
Which expression is equivalent to log, left parenthesis, 6, right parenthesis, dot, log, start base, 6, end base, left parenthesis, a, right parenthesis?

## 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.
• how do you simplify log 1/e of x with base change rule
• 1/e is e^-1. Therefore, the expr. becomes -ln(x)
• What can I do during an exam? I CAN'T USE A CALCULATOR! HELP! The theory was very useful though. :-)
• See the [I need help!] right below Challenge Problem 1.
• 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)
• can't we find the value of logarithm without any calculator or log book and do it manually
• Some logs are easy to solve, such as log_2(8). but most log functions would take a lot of work. Sometimes even the ones that look simple are kinda challenging, such as log_4(8).
The way exponents work is you raise the number to the numerator-th power, and then denominator-root it. For example, with log_4(8), you would raise 4 to the 3rd power(64) and then square root it to get 8. So the answer is 3/2. The problem is that unless they are really friendly numbers, rooting and powering is a real pain most of the time, and there will probably be some really nasty numbers. Especially with rooting things by hand; it gets complicated SO quickly. conclusion: for logs with numbers sure, do it by hand, but if the answer seems like it'd be more complicated than 3/4 or something, probably best not.