The properties of logarithms tell us that the logarithm of a product is the sum of the logarithms,
the logarithm of a quotient is the difference of the logarithms,
and the logarithm of a base raised to a power is the product of that power with the logarithm of the base.
Here's some examples. We want to expand an expression that looks like this: (log₃4)(x). I see that this is multiplication right here:
(4)(x). So, I apply property one to logarithms to write this as log₃4 + log₃x. So, I say,
the logarithm of this product is the sum of the logarithms. And it's true for any value of x that I want to substitute in here, that this is defined for;
if I simplify this expression and this expression, I'll get exactly the same number. Let's look at our next problem: log₆5 over x.
This is a quotient, so I'm going to use my second property of logarithms, to say, log₆5 - log₆x.
And property two for logarithms says that the logarithm of a quotient is the difference of the logarithms.
So, here's property one: log of a product is the sum of the logs. Property two: the log of a quotient is the difference of the logs.
Here's another problem: log₉∛z; I'll rewrite this as log₉z to the 1/3. Then, I use my third property of logarithms
to write this as 1/3 log₉z; so, exponents become coefficients with property three for logarithms.
The logarithm of this base raised to this exponent is the product of that exponent with the logarithm of the base.
So, there is property three used to simplify that expression. Next, I have this; I'm going to rewrite this as
log₁₀x²y ÷ z^½. Now, as I expand this, I'm going to get log₁₀x²
+ log₁₀y - log₁₀z^½. So here, this is multiplication, so the logarithm of that product is the sum
of the logarithms. It's division with this, so I have the logarithm—that becomes the difference of the logarithms;
the only thing I have left to do is to take these exponents and write them out in front. I'll do that—
and end up with 2 log₁₀x + log₁₀y - 1/2 log₁₀z. So there, I've extended this expression as much as possible,
using my properties of logarithms. You see each of my exponents becomes a coefficient; where I had multiplication, it turns into an addition.
and where I had a quotient, it turned into subtraction. Now, let's go the other way, and take one that's in expanded form,
and contract it back to a single logarithm. Write as a single logarithm: logₐx + logₐz.
I see I have the sum of two logs here with the same base. So, that's going to be the logarithm to that base of just that product.
So, the sum of these logs ends up to be the log of that product. Now, as I move down to problem 6, you see I have a more complicated expression.
What I'm going to do is rewrite this, by first taking the exponent, or the coefficients back up as exponents.
So, I'll write this as log base₂x³ when this coefficient goes back to that exponent; + log₂y^½
when that coefficient goes back to an exponent; and then, here I'm going to even take this one with me.
This is just my preference to be doing this problem; not everybody likes to do it this way, but I do.
So, I can think of this subtraction sign as the coefficient of -1—I'll take that back to that exponent on z.
Now, it's all addition right here, with logarithms of the same base; so I'm going to use property one for logarithms, to write this as
(log₂x³)(y^½)(z⁻¹). And then, if I want to simplify this further
and write my exponent of 1/2 as a square root, and my exponent of -1 as a reciprocal,
this will become log₂x₃√y ÷ z.
So, notice my exponent of 1/2; I write it as √y; my exponent of -1, I make the reciprocal of the base, so it is 1 over z;
and here I have a single logarithm that is equal to this original, more complicated expression; so, I contracted that expression
down to a single expression here by using my properties of logarithm. Let's look at one last problem. Here, we have a logarithmic equation
that we want to solve: log₃x + log₃x - 2 = 1. Well, here I have the sum of two logarithms,
so I'm going to write that as the logarithm of a product. So, (log₃x)(x - 2) = 1.
So I see, using my properties of logarithms, I can write this as a single logarithm. Now, this is just a single logarithm statement,
so I can write it in exponential form as 3¹ is equal to this. So, change this from logarithmic form to the exponential form
using the definition for logarithms. 3¹ = (x)(x) - 2. So, here I used my properties of logarithms, right here,
to contract this down to one single logarithm. Then, I use my definition for logarithms to change this from a logarithmic form
into exponential form. And look, all I have left is a nice quadratic equation. It should be pretty simple to solve; let's just go through the steps:
3 = x² - 2x when I multiply this out here. I'll put this in standard form by adding -3 to both sides;
x² - 2x - 3 = 0. I want to factor that expression on the right. We're factoring the product of two binomials:
so, x - 3 and x + 1; is that right? Yes, I got that right. Set those two factors equal to 0: x - 3 = 0, or x + 1 = 0.
Solve this, I have x = 3; solve the other one, and I have x = -1. Now, I need to check both of these solutions
because my original equation involved logarithms. Even though I ended up solving an exponential equation right here,
it came from an equation that involved logarithms; and remember, the domain for any logarithmic function has to be a positive number.
So, when I look at this solution, x = -1—if I substitute it back in here, I have log₃-1; that's impossible.
There is no expression like that. So, this solution, x = -1 is extraneous. If I go to this solution, x = 3, and substitute it
in either place, I get a defined expression; and x = 3 will make this a true statement.
So, logarithmic equations there, an example of one that we saw using the properties of logarithms, the definition of logarithms;
and we have to go back and check both of our solutions to make sure one of them isn't extraneous.