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### Course: Algebra 2>Unit 8

Lesson 6: Solving exponential models

# Logarithms: FAQ

## What are logarithms?

Logarithms are the inverse operation of exponentiation. We can use logarithms to find the exponent to which a given base must be raised in order to produce a particular result. For example, ${\mathrm{log}}_{2}8=3$, because ${2}^{3}=8$.

## What is the constant $e$‍ ?

The constant $e$ is a very important number in mathematics. It is an irrational number, which means it cannot be written exactly as a fraction or a decimal, but we often approximate it as $e\approx 2.71828$. It is the base of the natural logarithm, $\mathrm{ln}$.
As $n$ gets larger and larger, the sequence ${\left(1+\frac{1}{n}\right)}^{n}$ gets closer and closer to $e$.
The constant $e$ appears in various other contexts in mathematics and science, such as in statistics and in the study of exponential growth and decay.

## What are some properties of logarithms?

There are a few key properties of logarithms that we use frequently:
• ${\mathrm{log}}_{b}1=0$ for any base $b$
• ${\mathrm{log}}_{b}b=1$ for any base $b$
• ${\mathrm{log}}_{b}xy={\mathrm{log}}_{b}x+{\mathrm{log}}_{b}y$
• ${\mathrm{log}}_{b}\frac{x}{y}={\mathrm{log}}_{b}x-{\mathrm{log}}_{b}y$
• ${\mathrm{log}}_{b}{x}^{n}=n{\mathrm{log}}_{b}x$

## What is the change of base formula for logarithms?

The change of base formula allows us to convert a logarithm from one base to another. It states that ${\mathrm{log}}_{b}x=\frac{{\mathrm{log}}_{c}x}{{\mathrm{log}}_{c}b}$ where $b$ and $c$ are any two bases.
Let's say we want to evaluate the logarithm ${\mathrm{log}}_{5}100$. We can use the change of base formula to convert this logarithm to a different base, like base $10$:
${\mathrm{log}}_{5}100=\frac{{\mathrm{log}}_{10}100}{{\mathrm{log}}_{10}5}$
Now, we can use a calculator (or our knowledge of logarithms) to evaluate the two base-$10$ logarithms on the right side of the equation:
${\mathrm{log}}_{5}100\approx \frac{2}{0.6989\text{…}}\approx 2.861$.

## How can we use logarithms to solve exponential equations?

Logarithms can be a really useful tool for solving exponential equations. For example, say we want to solve ${2}^{x}=9$. We can take the logarithm of both sides, and use the properties of logarithms to isolate the variable:
$\begin{array}{rl}{2}^{x}& =9\\ \\ {\mathrm{log}}_{10}{2}^{x}& ={\mathrm{log}}_{10}9\\ \\ x{\mathrm{log}}_{10}2& ={\mathrm{log}}_{10}9\\ \\ x& =\frac{{\mathrm{log}}_{10}9}{{\mathrm{log}}_{10}2}\\ \\ x& \approx 3.167\end{array}$

## How are logarithms and radicals used in the real world?

Logarithms and radicals are used in lots of different ways in the real world. They are important in fields like finance, engineering, and science. For example, the Richter scale, which measures the magnitude of earthquakes, uses a logarithmic scale. This means that a magnitude 6 earthquake is actually ten times stronger than a magnitude 5 earthquake!

## Want to join the conversation?

• I think there's a typo in "How can we use logarithms to solve exponential equations?". it says "solve 2^x=10" and then it solves 2^x=9. so unless I'm extremely bad at reading numbers, you might want to change that if possible. I don't mean to be rude, but it did confuse me and might proceed to confuse people later on.
• on the ection that says "What is the change of base formula for logarithms?" on the next paragraph it says that log_c(x)/log_c(b) for any value of b and c.
i was wonering if even 0 is included or negative values, because i think with 0 it gets undifined.
thanks
• You are correct in that it does not work for 0 or negative numbers. The shape of a log function has a vertical asymptote at x=0, so it is always positive.
• for the definition of e, (1+1/n)^n, one thing I don't get it is that since the exponent is n, can we expand the formula into 1^n + 1/n^n, and let n be a huge number like infinity and then this will be 1 + a very small number and the result will be 1 instead of 2.7......? also if we don't expand the formula but still let n be infinity, this will be 1 plus a very small number to a huge power, and I think the result will be close to 1 or 2, why is 2.7...?
• I see where you are coming from. The problem is that (1+1/n)^n does not equal 1^n+(1/n)^n. Remember that you can't just raise everything you're adding in the parentheses to the power. For example, if you want to find (a+b)^2 it is not a^2+b^2, instead it is a^2+2ab+b^2. Hope this clears things up!
• This page says that log_9/log_2 = 3.167. My TR-83 calculator says it equals 3.1699... What gives?
• Whoever did the rounding for that made a mistake. You were correct to say that it was 3.1699. I have a TI-84 Plus Color Edition graphing calculator. I entered the equation log(9)/log(2) and it answered 3.169925001, confirming that your answer was indeed correct. Rounded to 5 digits, this value is indeed 3.1699. However, rounded to 4 digits, it should have been 3.170. Whoever did the rounding for this put the 7 in the wrong spot.
(1 vote)
• In the last part, how did you get 2^x = 9