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### Course: Algebra 2>Unit 8

Lesson 4: The change of base formula for logarithms

# Logarithm properties review

Review the logarithm properties and how to apply them to solve problems.

## What are the logarithm properties?

Product rule${\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)$
Quotient rule${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)$
Power rule${\mathrm{log}}_{b}\left({M}^{p}\right)=p{\mathrm{log}}_{b}\left(M\right)$
Change of base rule${\mathrm{log}}_{b}\left(M\right)=\frac{{\mathrm{log}}_{a}\left(M\right)}{{\mathrm{log}}_{a}\left(b\right)}$

## Rewriting expressions with the properties

We can use the logarithm properties to rewrite logarithmic expressions in equivalent forms.
For example, we can use the product rule to rewrite $\mathrm{log}\left(2x\right)$ as $\mathrm{log}\left(2\right)+\mathrm{log}\left(x\right)$. Because the resulting expression is longer, we call this an expansion.
In another example, we can use the change of base rule to rewrite $\frac{\mathrm{ln}\left(x\right)}{\mathrm{ln}\left(2\right)}$ as ${\mathrm{log}}_{2}\left(x\right)$. Because the resulting expression is shorter, we call this a compression.
Problem 1
Expand ${\mathrm{log}}_{2}\left(3a\right)$.

Want to try more problems like this? Check out this exercise.

## Evaluating logarithms with calculator

Calculators usually only calculate $\mathrm{log}$ (which is log base $10$) and $\mathrm{ln}$ (which is log base $e$).
Suppose, for example, we want to evaluate ${\mathrm{log}}_{2}\left(7\right)$. We can use the change of base rule to rewrite that logarithm as $\frac{\mathrm{ln}\left(7\right)}{\mathrm{ln}\left(2\right)}$ and then evaluate in the calculator:
$\begin{array}{rl}{\mathrm{log}}_{2}\left(7\right)& =\frac{\mathrm{ln}\left(7\right)}{\mathrm{ln}\left(2\right)}\\ \\ & \approx 2.807\end{array}$
Problem 1
Evaluate ${\mathrm{log}}_{3}\left(20\right)$.

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• How did a mathematician find e? What's its origin?
• in log_1(1)=x, doesn't x = infinity?
• Why would you need to use ln?
• The natural log function, ln, is the log with a base of Euler's number, e.

Here is an example of when it can be used:
e^x = 2
--> To solve for x, we would take the ln of both sides. This is because x is the exponent of e, and the e and natural log will cancel out when put together.
ln(e^x) = ln(2)
x = ln(2)

This is the most common way I've seen the natural log used, but there are no doubt other ways to use it.
• why e^𝝿i+1=0?
How did Euler proof this equation?
• It is a specific case of his formula, e^(i*x) = cos(x) + i*sin(x). The proof of this formula requires Calculus level math though (e.g. power series or differentiation).
• Is ln the same thing as log base 10?
• What is log_15 (9), if log_15 (5) = a

Could someone explain the steps to solve this?
• I can explain how to check it, at least, though I'm not sure that this is how you would originally come to this answer.
log_15(5)=a, and we want to see whether log_15(9)=2-2a. We can begin by finding a.
log_15(5)=a
log(5)/log(15)=a Use the change of base rule.
a=approx. 0.5943
Next we can find 2-2a:
2-2*approx.0.5943 Try to use the entire answer you got for a, instead of a rounded one, if you can.
2-approx. 1.1886=approx. 0.8114
Now we can find log_15(9), and see if it equals 0.8114.
log_15(9)=log(9)/log(15) Change of base rule.
log(9)/log(15)=approx. 0.8114
log_15(9)=2-2a True.
This shows that it is indeed the case that if log_15(5)=a, then log_15(9)=2-2a, but it does not seem like it is the way that you would come to figure it in the first place.
Perhaps if you figured that a=approx. 0.5943, and that log_15(9)=approx. 0.8114, you might just happen to notice that 0.8114+2(0.5943)=approx. 2. I have not figured anything better than this for this question. Maybe someone else will. For now, I hope you have a good day. Keep going!