master-solving-logarithmic-equations-using-the-product-property-of-logarithms

# Interactive video lesson plan for: Master Solving Logarithmic Equations using the product property of logarithms

#### Activity overview:

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Welcome ladies and gentlemen. So what I'd like to do is show you how to solve logarithmic equations. Now when solving logarithmic equation, previously what we did was solve a logarithmic equation just by rewriting them in exponential form. But you can see in each and every one of these, we have more than one logarithm. So we just can't quickly and easily reach switch it over to exponential form because exponential form-- to go from logarithmic to exponential form, we had to have a logarithm isolated.

Now if we have logarithms on the same side, we can still use the rules of logarithms to condense it down to one logarithm. And if we have a logarithm on both sides, we can also use the one to one property that we use in exponential form by taking the log of base x equals log base b to y. I'm about to sneeze. Since the bases are the same, you could just set the values of each logarithm equal to one another.

The next thing that we also want to do is, since we have two logarithms, we also want to make sure we can check for extraneous solutions or solutions that are not work. Because if you remember, when we're looking at the solution of a logarithm, here's a logarithm with no transformation. Now all of these are going to have transformations. So it is possible to have negative solutions.

However, in a logarithm with no solution, we can't take an x value and plug it into log base b of x. There is no negative x that we can plug. So whenever we take our solutions, when we plug them back into the original format, we can't take the log of a negative answer. And we'll get to this as we work along. But I wanted to kind of show you just the graph with no transformations. You can't take the log of a negative value.

So in the first case here, I have two logarithms on the same side. Now I don't want to add them to the other side to do my one to one property, because if I did that, I still have this two here. And the one to one property only works when you have a logarithm equal to a logarithm. Since there is that two here, we can't complete the one to one property.

So what I'm going to have to do is keep the logarithms here and condense them down to one logarithm. And I can do that using my rules of exponents, which is the product rule. And remember the product rule states if you have two logarithms, as long as you have the same base and you're adding them, you can rewrite them as the product of those two values.

Now I have a logarithm that's been isolated equal to a value. I can rewrite this in exponential form. So 4 squared equals negative x squared-- I'll apply distributive property here-- minus 10x.

Now I can rewrite this as 16 equals-- I'm going to factor. Let's keep that in there first. Actually, let's leave this out. I'll do one more. 16 equals negative x squared minus 10x.

Now you can see that I have a quadratic. My x that I need to solve for is being raised to the second power. So therefore I'm going to treat this as a quadratic equation where I can either solve by factoring, I can solve by the quadratic formula, completing the square, and so forth.

So the first thing I want to do is try to see if I can solve by factoring. So to do that, I need to get all the terms on the one side and set it equal to 0. So I have 0 equals negative x squared minus 10x minus 16.

I don't like factoring when my a is negative, so I'm going to factor out a negative 1. That's not going to affect my solution because I can divide out that negative 1 at any time. So I have 0 equals x squared plus 10x plus 16.

Now to go ahead and factor that, I'm just going to go ahead and rewrite this in factored form. So 0 equals-- let's see, what two numbers multiply to give me 16, add to give me 10. That's going to be x plus 8 as well as x plus 2.

Running out of space. So therefore now what I can do is use the zero product property, say x plus 8 equals 0 and x plus 2 equals 0. To solve, I get x equals negative 8 and x equals negative 2.

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