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## Algebra 2

### Course: Algebra 2>Unit 8

Lesson 4: The change of base formula for logarithms

# Evaluating logarithms: change of base rule

Sal approximates log₅(100) by rewriting it as log(100)/log(5) using the change of base rule, then evaluates with a calculator. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• I posted an answer to this question with an alternate proof to the change of base formula that I find more intuitive. I did this because neither the comment nor question sections have enough space for it. •   Let logₐ(b) = s. Then we wish to show that, for any value of x, logₐ(b) = logₓ(b) / logₓ(a). Let's say logₓ(a) = t and logₓ(b) = u.

Change each of these to the exponent notation:

logₐ(b) = s means aˢ = b.

logₓ(a) = t means xᵗ = a.

logₓ(b) = u means xᵘ = b.

So we know that aˢ = xᵘ, because they both equal b. We also know a = xᵗ now, so we can substitute xᵗ for a in the equation aˢ = xᵘ, and get (xᵗ)ˢ = xᵘ. This simplifies to xˢᵗ = xᵘ using the exponent properties, which implies st = u, or s = u/t. Recall that logₐ(b) = s, logₓ(a) = t, and logₓ(b) = u. So we've now shown with s = u/t that logₐ(b) = logₓ(b) / logₓ(a). This completes the proof.
• Where can I download a graphing calculator for my computer like what Sal is using? • So why do we always take log(base 10) ?
Why we dont take any other number ? • You can do that. In fact base 10, base 2 and base e (natural logarithm) are only the most common.
The limitation is your calculator. Simple calculators only have a function for base 10 and base e and that's it.
• I didn't get that last part of the video, where Clog(a) = log(b) turns into the very last part of what Khan was writing about, logb/loga. How does that work out? Please and thank you :) • this makes no sense :C • Certainly! Let's go through the explanation of the change of base rule for evaluating logarithms:

*Change of Base Rule for Logarithms*:

The change of base rule is a property of logarithms that allows us to rewrite the logarithm of any base as the logarithm of another base. Specifically, for any positive real numbers "a," "b," and "c," where "a" and "b" are the bases of the logarithms and "c" is the argument, we have:

logₐ(c) = log_b(c) / log_b(a)

*Explanation*:

Suppose we want to evaluate a logarithm with a certain base "a," but our calculator only has a logarithm function with a different base "b." The change of base rule provides a way to convert the logarithm with base "a" into an equivalent logarithm with base "b."

Let's use an example to illustrate the process:

*Example*:
Evaluate log₅(100) using a calculator that only has a logarithm function with base 10.

*Step 1*:
Apply the change of base rule:
log₅(100) = log(100) / log(5)

*Step 2*:
Evaluate the logarithms on the right-hand side using the calculator with base 10 logarithm (log) function:
log(100) ≈ 2 (since 10^2 = 100)
log(5) ≈ 0.69897 (approximately)

*Step 3*:
Substitute the values back into the expression:
log₅(100) ≈ log(100) / log(5) ≈ 2 / 0.69897 ≈ 2.86135

*Result*:
The approximation of log₅(100) using the change of base rule and a calculator with base 10 logarithm function is approximately 2.86135.

By using the change of base rule, we can evaluate logarithms with different bases using a calculator that supports a specific base for logarithms. This rule provides a helpful technique for logarithmic calculations when using calculators or software that may not have specific logarithm functions for every base.

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The change of base rule is a useful logarithmic property that allows us to evaluate logarithms of any base by using a calculator with a different base logarithm function. This property states that for any positive real numbers "a," "b," and "c," where "a" and "b" are the bases of the logarithms and "c" is the argument:

logₐ(c) = log_b(c) / log_b(a)

In the given example, Sal wants to approximate log₅(100) using a calculator that has a logarithm function with base 10 (log), but not with base 5 (log₅). So, Sal uses the change of base rule to convert log₅(100) into an expression that can be evaluated using the calculator:

log₅(100) = log(100) / log(5)

Now, Sal can use a regular calculator to find the logarithms of 100 and 5, which are:

log(100) ≈ 2 (since 10^2 = 100)
log(5) ≈ 0.69897 (approximately)

Now, substitute these values back into the expression:

log₅(100) ≈ log(100) / log(5) ≈ 2 / 0.69897 ≈ 2.86135

So, the approximation of log₅(100) using the change of base rule and a calculator with base 10 logarithm function is approximately 2.86135.
• How can you do this WITHOUT a calculator? • Can you put any number as a base in the logarithmic base conversion formula? • How would i solve 3^(x − 4) = 6 using the change of base formula. • How can I evaluate “log10 5” without a calculator? I know there has to be a way.

Also, why can’t (log10 100)/(log10 5) be written as
“log10 100 - log10 5”? The answer to expression is 1.3010. • We know that log_10 1 = 0 and log_10 10 = 1. From this, we can deduce this: 0 < log_10 5 < 1. We can specify more from this because log(√10) = 1/2. This means
1/2 < log_ 10 5) < 1 because √10 < 5. Using this method, you would be able to get close to log_10 5, but it would be very tedious work. Also,
(log_10 100)/(log_10 5) is something very different from log_10 100 - log_10 5 because if we say
A = log_10 100 and
B = log_10 5, we can come up with the following equations:
log_10 100 / log_10 5 = A / B, and
log_10 100 - log_10 5 = A - B.
These two equations can't ever be equal to each other unless A and B are both zeros. 