Properties of Logarithms - Online Tutor, Practice Problems & Exam Prep

Understanding logarithmic properties is essential for expanding and condensing log expressions. The product rule allows multiplication within logs to be expressed as addition of separate logs, while the quotient rule transforms division into subtraction. The power rule enables exponents to be moved in front of logs as multiplication. For complex bases, changing the base using the formula $log\left(m\right)/log\left(b\right)$ facilitates quick evaluations using calculators. Mastering these concepts enhances problem-solving skills in logarithmic functions.

Product, Quotient, and Power Rules of Logs

Hey, everyone. When working through problems dealing with logs, you're going to come across questions that ask you to expand log expressions, like, say, log base 2 of 3x, into multiple different logs. And we can do that using certain properties of logs. Now don't worry. We're not just going to have to memorize a bunch of brand new rules here because all of these properties of logs are actually properties of exponents that we already know and have used before. So, the same way we were able to graph log functions using what we knew about exponential functions, we can do the same thing here. So, I'm going to give you a quick refresher of these exponential rules and then show you their corresponding log rule and how to use that to expand log expressions. So let's go ahead and jump right in. Now remember when working with rules like this, we don't really care about the name. That name is not important. It's just important that we know how to use these rules, and the name is just a way to organize them. So, starting first with our product rule here, remember that whenever we had exponents of the same base being multiplied together, we could simply take those exponents and add them together. And we see something similar when working with logs. So, if I have some log of some base with 2 things being multiplied together, I can separate that out into 2 different terms being multiplied in a log, this simply means that we can add 2 terms being multiplied in a log, this simply means that we can add 2 different logs together. Now we see something similar happen when working with the quotient rule because with our exponents, we saw that whenever we had exponents of the same base being divided, we would simply take those exponents and subtract them. So when seeing a log with 2 things being divided, what do you think I'm going to do to 2 separate logs? Well, I'm going to end up subtracting them. So, the same way that we had division turn to subtraction with our exponent, the same thing is going to happen with our logs. So whenever we divide terms in a log, we see that we subtract 2 logs. So, multiplication becomes addition. Division becomes subtraction. So if I see something like log base 2 of 3x, like I have right here, I see this 3 and this x are being multiplied together. So since those things are being multiplied, I can turn this into the addition of 2 logs. So this would become a log base 2 of 3 plus log base 2 of x. So, see, you notice that that base stays the same, but I'm simply adding 2 logs together and I have separated that 3 and that x into that addition. Now if I'm given something like log base 5 of 5 divided by y, since that 5 and that y are being divided, this division turns to the subtraction of 2 separate logs. So this becomes log base 5 of 5 minus log base 5 of y using that quotient rule. Now we have one final log property to look at here, the power rule. And whenever we worked with exponents, we saw that if we took an exponent of some base, so b to the power of m, and raise that to another power like n, I would simply end up multiplying those 2 powers together. Now we see something slightly different when working with our log, but it's still going to see multiplication happening. So if I have log base b of m to the power of n, it's still going to turn out into multiplication. But I'm going to take this n and stick it on the front of that log. So log base m to the power of n becomes n times log base b of m. So I simply take that n and pull it out to the front and multiply it together. So anytime I'm raising a term to a power, I am simply going to end up multiplying the log by the power. So if I'm given something like the natural log of 7 to the power of 2, I'm going to take that 2 and stick it on the front of that, and this is going to become 2 times the natural log of 7. So now that we've seen these log properties, we have the tools we need to expand log expressions. Thanks for watching. And I'll see you in the next one.

Expand & Condense Log Expressions

Hey, everyone. We just learned all of the rules needed in order to expand log expressions, but two things are going to happen. We're going to be asked to expand much more complicated log expressions like, say, log2 (3xy2), and we're also going to be asked to do the exact opposite of expanding log expressions and actually condense expressions with multiple logs down to a single log. Now you don't have to worry because we're going to take these rules that we already know, and I'm going to walk you through how to use multiple of these rules and how to use them in their reverse direction because each of these properties can be applied in both directions depending on what your goal is, whether taking a single log to multiple logs or condensing multiple logs back down into a single log. Now you'll be able to expand and condense any log expression that gets thrown your way, so let's go ahead and get started. Now we're going to just jump right into an example here. And in our first example, we see that we have log2 (3xy2), and we want to go ahead and expand this log expression as much as possible. So the first thing I noticed here is that I have 3 different things being multiplied together. This is actually 3 times x times y2. Now since this is multiplication, that clues me in that I need to go ahead and use the product rule because I know that the product rule takes multiplication and turns it into the addition of multiple logs. So let's go ahead and expand this log out into multiple logs using that product rule. So this is a log2 (3) plus log2 (x) plus log2 (y2). Now I have these multiple logs, but I want to go ahead and walk through each of these single logs to make sure that I can't expand them any further. So I have log2 (3). I can't expand that anymore, so that's going to remain as a log2 (3). And then log2 (x), I also cannot expand that anymore, so that will remain the same as well. And then finally, we come to our last term, log2 (y2). Now looking at this log, I know that I have this exponent here. And whenever I have an exponent, that tells me that I can go ahead and use the power rule because the power rule tells me that I can take that exponent and pull it to the front of my log in order to expand it. So this 2, I can go ahead and pull to the front of my log, and this last term becomes 2 times log2 (y). Now none of these terms can be expanded anymore, so this is my final answer. log2 (3) + log2 (x) + 2×log2 (y). We've taken this one log and expanded it out as much as we can. Now let's look at condensing a log expression. Now we do have to consider a couple of additional things whenever we're condensing logs. And the first is that we always want to make sure that the base has to be the same. So here we have 2 times the natural log of x minus the natural log of x plus 2. So both of these are natural logs. They have that same base of e. If this was, say, the natural log and log2, I couldn't condense those logs together because they have different bases. Now, one other thing we want to consider when condensing logs is that the power rule is always going to get applied first in order to get the correct answer. So with that in mind, let's go ahead and condense this log expression. So we have 2 times the natural log of x minus the natural log of x plus 2. Knowing that I need to apply the power rule first, I'm going to go ahead and look at these terms and see how I can apply that power rule. So I have 2 times the natural log of x. Since I have this 2 multiplying that, I know that using the power rule, I can take that thing that's multiplying it and make it into the exponent. So this becomes the natural log of x squared, pulling that 2 into the exponent of what I'm taking the natural log of. Now that second term, I can't apply the power rule, so it's going to remain the same for now, the natural log of x plus 2. Now how else can we condense this expression? Well, I have these terms being subtracted. So since they're being subtracted, that tells me that I should go ahead and use the quotient rule because I know that using the quotient rule, I can take subtraction and I can turn it into division of a single log expression. So this becomes the natural log of x squared divided by what's being subtracted, x plus 2. So now I have taken these multiple logs and condensed it down into a single log, so this is my final answer, the natural log of x squared divided by x plus 2. Now that we have seen how to expand and condense log expressions, let's go ahead and get some more practice. Thanks for watching, and let me know if you have questions.

Write the log expression as a single log.

$\log_2\frac{1}{9x}+2\log_23x$

• A. $\log_2x$
• B. $\log_2\frac{1}{3x}$
• C. $\log_21$
• D. $\log_23x$

Write the log expression as a single log.

$\ln\frac{3x}{y}+2\ln2y-\ln4x$

• A. $\ln\frac{3xy}{4}$
• B. $\ln\left(12x^2\right)$
• C. $\ln\left(\frac32\right)$
• D. $\ln\left(3y\right)$

Write the single logarithm as a sum or difference of logs.

$\log_3\left(\frac{\sqrt{x}}{9y^2}\right)$

• A. $2\log_3x-2-\log_39y$
• B. $\frac12\log_3x-2-2\log_3y$
• C. $\frac12\log_3x+2\log_33y$
• D. $\frac12\log_3x-2\log_39y$

Write the single logarithm as a sum or difference of logs.

$\log_5\left(\frac{5\left(2x+3\right)^2}{x^3}\right)$

• A. $5+2\log_5\left(2x+3\right)-\log_53x$
• B. $2\log_5\left(2x+3\right)-3\log_5x$
• C. $1+2\log_5\left(2x+3\right)-3\log_5x$
• D. $\log_5\left(2x+3\right)-\log_5x$

Change of Base Property

At this point, we've evaluated logs by hand. And we've used different rules and properties in order to expand and condense log expressions. But sometimes you're just going to need to quickly evaluate a log by plugging it into your calculator. But if you're given something like log base pi of 9, well, that doesn't look quite so simple to just type in our calculator and go. So here I'm going to show you that whenever you're given a log with some base that doesn't look so easy to deal with, like, say, pi, you can simply change the base to be whatever you need it to be. So here I'm going to walk you through exactly how to do that and let's not waste any time here and just jump right in.

So if I have some log, logb(m), but my base b is, say, pi or 27 or 500, some number that is not so easy to deal with, so I want to change it, I can go ahead and change it by taking my log and turning it into a fraction. So if my original base was b but I wanted a base of a, I would simply take loga(m) and divide it by loga(b). So whatever I was originally taking the log of m is going to go on the top and then my original base b goes on the bottom. That's one way to remember this, base goes on the bottom.

So let's say that I'm given some log, like log5(2), and I want to change that to have a base of 10. I could go ahead and change that into a fraction. And that fraction would be log10(2), whatever I was originally taking the log of, divided by log10(5). Now from here, since there actually is a button for log base 10, my common log, on my calculator, I can quickly and easily type this in my calculator and come up with an answer.

Now because log base 10, your common log, has a button on your calculator and so does the natural log, log base e, you're most often going to want to change your base to be either 10 or e so that you can quickly type it in your calculator and get an answer. So we just saw that whenever we changed our base to 10, this would become log10(m) divided by log10(b). Now log base 10 can, of course, just be written as log because it's the common log. So, this is really just log of M divided by the log of b. Now we see the same thing whenever we're faced with changing our base to e. We know that log base e is just the natural log. So this could simply be written as the natural

Now you can change any log of any base to either of these two options. Most often, it will be specified to you which one you want to change it into. But if it's not, you can use either one and it doesn't really matter. So with that in mind, let's go ahead and work through some examples here. So with our first example, we see log7(31). So log7(31) and for a and b here, we want to go ahead and use common logs, so changing our base to 10. So looking at my first base here, log7(31), I want to change to log base 10. So doing that, I can take log base 10 in my fraction and just plug in whatever I need to plug in on the top and bottom. So I'm going to take whatever I was originally taking the log of on top. So this is log base 10 of 31 and then divided by log base 10 of my original base 7. Now, of course, log base 10 can just be written as log. So this is just log(31)log(7) and I can go ahead and type this into my calculator. And when I do, I'm going to go ahead and get an answer of 1.76, and I have fully evaluated that log.

Now let's look at another example. Here we have logπ(9), and we want to again change this into a log base 10 because we're still using common logs here. So logπ(9), I know that I can turn this into log(9)log(π). So whatever I was originally taking the log of goes on the top and then, of course, my base goes on the bottom. So I can go ahead and plug this into my calculator, log of 9 over the log of pi, and I'll end up getting an answer of 1.92. And I've evaluated that log.

Now looking at our 3rd example, we again have logπ(9). But now we want to go ahead and use a natural log, so changing our base instead to e. So knowing that base e loge(e) is just the natural log, I know that this will turn into ln(9)ln(π). So, again, whatever we were originally taking the log of goes on the top and then my original base goes on the bottom. So I can go ahead and type this in my calculator, natural log of 9 divided by the natural log of pi, and I'm going to end up getting 1.92. Now earlier, I said it doesn't matter what you change your base into, and that's because we're going to get the same answer regardless. Whenever we changed our base to 10, we got 1.92 and then whenever we changed our base to e, we also got 1.92. So it doesn't really matter. You're going to get the same answer regardless.

Let's take a look at one final example here. We have log3(e). And I want to use natural logs for this one as well. So changing this to a base of e, I'm going to get ln(e)ln(3). Now we can actually do some further simplification here because the natural log we know is just log base e. So this is really log base e of e. Now whenever we have the same base as what we're taking the log of, I know that this just simplifies to 1. So this really just becomes 1 over the natural log of the square root of 3. Now I can go ahead and just plug that into my calculator. And when I do, I'm going to get an answer of 1.82, and I have fully evaluated that log.

Now that we know how to change the base of a log to be whatever we need it to be, let's get some more practice. Let me know if you have any questions.

Evaluate the given logarithm using the change of base formula and a calculator. Use the common log.

$\log_317$

• A. 0.39
• B. 2.58
• C. 1.23
• D. 0.48

Evaluate the given logarithm using the change of base formula and a calculator. Use the common log.

$\log_967$

• A. 1.91
• B. 0.52
• C. 0.95
• D. 1.83

Evaluate the given logarithm using the change of base formula and a calculator. Use the natural log.

$\log_841$

• A. 1.61
• B. 0.9
• C. 0.56
• D. 1.79

Evaluate the given logarithm using the change of base formula and a calculator. Use the natural log. 

$\log_23789$

• A. 0.08
• B. 11.89
• C. 3.58
• D. 0.30

Here’s what students ask on this topic:

How do you expand log base 2 of 3xy² using logarithmic properties?

To expand log base 2 of 3xy², we use the product and power rules of logarithms. First, recognize that 3xy² is a product of three terms: 3, x, and y². Using the product rule, we can write:

${\log }_2\left(3x{y}^2\right)={\log }_2\left(3\right)+{\log }_2\left(x\right)+{\log }_2\left(y^2\right)$

Next, apply the power rule to the term log base 2 of y²:

${\log }_2\left(y^2\right)=2{\log }_2\left(y\right)$

So, the fully expanded form is:

${\log }_2\left(3x{y}^2\right)={\log }_2\left(3\right)+{\log }_2\left(x\right)+2{\log }_2\left(y\right)$

How do you condense 2ln(x) - ln(x + 2) into a single logarithm?

To condense 2ln(x) - ln(x + 2) into a single logarithm, we use the power and quotient rules of logarithms. First, apply the power rule to 2ln(x):

$2\ln \left(x\right)=\ln \left(x^2\right)$

Next, use the quotient rule to combine the terms:

$\ln \left(x^2\right)-\ln (x+2)=\ln \left(\frac{x^2}{x+2}\right)$

So, the condensed form is:

$\ln \left(\frac{x^2}{x+2}\right)$

How do you change the base of log base 5 of 2 to base 10?

To change the base of log base 5 of 2 to base 10, we use the change of base formula:

${\log }_5\left(2\right)=\frac{{\log }_{10}\left(2\right)}{{\log }_{10}\left(5\right)}$

Since log base 10 is the common log, we can simplify this to:

$\log \left(2\right)/\log \left(5\right)$

Now, you can easily evaluate this using a calculator:

$\log \left(2\right)/\log \left(5\right)\approx 0.4307$

What is the power rule for logarithms and how is it used?

The power rule for logarithms states that if you have a logarithm of a number raised to a power, you can move the exponent to the front of the logarithm as a multiplier. Mathematically, it is expressed as:

${\log }_b\left(m^n\right)=n{\log }_b\left(m\right)$

For example, if you have log base 2 of 8³, you can use the power rule to simplify it:

${\log }_2\left(8^3\right)=3{\log }_2\left(8\right)$

This makes it easier to evaluate or further manipulate the logarithmic expression.

How do you evaluate log base 7 of 31 using a calculator?

To evaluate log base 7 of 31 using a calculator, you can use the change of base formula. Convert log base 7 of 31 to a base that your calculator can handle, such as base 10 or base e (natural log). The change of base formula is:

${\log }_7\left(31\right)=\frac{{\log }_{10}\left(31\right)}{{\log }_{10}\left(7\right)}$

Using common logs, this becomes:

$\log \left(31\right)/\log \left(7\right)$

Enter this into your calculator:

$\log \left(31\right)/\log \left(7\right)\approx 1.5132$