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

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

#### Activity overview:

Subscribe! http://www.freemathvideos.com Want more math video lessons? Visit my website to view all of my math videos organized by course, chapter and section. The purpose of posting my free video tutorials is to not only help students but allow teachers the resources to flip their classrooms and allow more time for teaching within the classroom. Please feel free to share my resources with those in need or let me know if you need any additional help with your math studies. I am a true educator and here to help you out. Welcome ladies and gentlemen. So what I'd like to do is show you how to solve logarithmic equations when dealing with subtraction, or using the quotient property of logarithms. And I wanted to make a separate video for this because I think these problems are the ones that usually students get the most kind of stuck on when they're working through the problems. And basically because now we're dealing with fractions or a quotient part.

But again, these problems are just like the ones we just did. The only thing is just instead of using the product property for the logarithms, we're now going to use the quotient property. And again our objective is the same. We're either going to want to get one logarithm all by itself, so therefore we can convert it to exponential form, or we're going to get logarithms isolated on both sides so we can use the one to one property. And we're going to use both of those solving techniques in this video.

So the first one, you can see have two logarithms on the same side. Again, I don't want to add this logarithm to the other side and use the one to one property because I can't, because that 3 is there. The one to one property only works when a logarithm is isolated to another logarithm.

So what I need to do is use the quotient property. Remember when you have two logarithms with the same base, subtract them from one another. Therefore, you can rewrite them as the quotient of those values. So therefore I have log base 2 of x plus 2 divided by x minus 5 equals 3.

Now us look might look pretty confusing here. But again, what we can do is now that we have a logarithm isolated, now we can rewrite it in exponent form. So that's going to look like-- here goes my sneeze. Bless you. Thank you. So I have 2 cubed equals x plus 2-- I think I'm going to sneeze again-- x minus 5.

All right, now 2 cubed is going to be 8, right? Now what I need to do is I can't just solve for x. I can't have x in the numerator and denominator. I have to get this x off of the denominator. So to do that, I'm going to multiply by x minus 5 on both sides.

By doing that, that now divides the 1. And I can apply distributive property here. And I get 8x minus 40 equals x plus 2. Now I can go ahead and solve. So I get the x by themselves. And let's see, I get 7x equals 42 divided by 7 divided by 7x equals 6.

So therefore I have one solution. I just want to make sure though I go back and check it to make sure it's not extraneous. So when I plug 6 in for x, I get 8, which works. When I plug 6 in for x over there, I get 1. Remember, as long as you plug them in, as long as it doesn't make the value of the logarithm negative, you're good to go. So that works. So we'll do this nice little check mark.

On the next one, we're going to do the exact same thing. I'll just rewrite this as log base 4 of x over x minus 1 is equal to 1/2. Rewrite this in exponential or power, x over x minus 1.

Now this one gets a little bit fun because remember, 4 to the 1/2 power is-- I didn't write it in there. 4 to the 1/2 power is equal to the square root of 4 to the first power, which is just 2. So therefore I'll multiply by x minus 1 on both sides. And I get 4 to the 1/2 power is 2. So it's going to be x minus 1 times 2 equals x. Apply distributive property. 2x minus 2 equals x.

And now I can go ahead and solve. So I subtract an x, subtract an x, plus a 2, plus a 2. So I get x equals positive 2. 2x, everything looks pretty good. OK, all right, fair enough. Good job.

So now on the x one, now what you can see in these last two problems, I don't just have logarithms on one side. I have logarithms on both sides. So again, I am going to be using the one to one property. The biggest mistake that I see students is they just cancel out all the logs. No, you're not cancelling the logs. The one to one property states, when you have a logarithm equal to another logarithm, the values of those two logarithms are equivalent.

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