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Solving Exponential Equations with the Different Bases - Concept 75,628 views

Teacher/Instructor Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company


Sometimes we are given exponential equations with different bases on the terms. In order to solve these equations we must know logarithms and how to use them with exponentiation. We can access variables within an exponent in exponential equations with different bases by using logarithms and the power rule of logarithms to get rid of the base and have just the exponent.

We're now going to talk about solving exponential equations when our bases are different. Okay so right here I have a exponential equation and what we're trying to do is solve for x okay? For this particular problem we know that 8 and 16 both share the base 2 so we can rewrite these both as powers of 2 so this becomes 2 cubed to 2x and 16 is 2 to the fourth to the x+4. Using the powers of logarithms multiply powers 2 to the 6x equals 2 to the 4x+16, our bases are the same and so then we can just set our exponents equal 6x is equal to 4x+16, 2x is equal to 16, x is equal to 8. So when our bases have at least a power in common these are pretty easy to solve you get their base is the same so their exponents equal.
Life isn't always that easy okay? So we're going to talk about now is when we have bases that don't share a power okay? Here we have 7 and 12 there are 2 differnt ways of doing this okay. The first way is the one that I want us to get comfortable with and that is basically to find a way to somehow bring this exponent down okay? What we're going to use is the power rule from logarithms okay? We can take the log of both sides it doesn't matter what log we do as long is the same so for this one I'll use the natural log, if you want to use log base 10 it would work just fine so if I take the natural log of both sides okay? Once we have a, taking the natural log is just an operation. I can add 4 to both sides that's okay, I can divide by 2 on both sides that's okay as long as we take the natural log of both sides it's just like anything else okay? So once we have the natural log out in front, we can take this exponent down on to the front so what we really have here then is x natural log of 7 is equal to the natural log of 12. Natural log of 7 is just a number okay it's an ugly number it's one we don't know we can plug it into our calculator and find it out but it's just a number so we can divide by that okay? And what we end up with is x is equal to natural log of 12 over natural log of 7 okay? This is what simply called calculator ready form because natural log is a log we can put in our calculator so we can fairly easily just plug in natural log of 12 over natural log of 7 in order to figure out what x is okay?
Going to the other way which is a way some of you may want, starts doing these but eventually we're going to want to sort of we do off that because it's not going to always work okay? So I have the same exact problem over here okay? 7x is 7 to the x is equal to 12, if you remember this is called exponential form okay we have 7 to a power is equal to 12 we could fairly easily put this in to logarithmic form by bringing the 7 down around and what we would end up with is x is equal to log base 7 of 12 okay? So now we have log base 7 of 12 we don't know how to evaluate that because log base 7 isn't on our calculator. So what we can do is use the change of base formula in order to put this in our calculator, remember the change of base formula we drop down the base and make its own log so this would end up being x is equal to, could choose our base I'm going to do log base 10 in this case the common log, log base 12 over log base 7, you could do natural log if you wanted to but using our logarithmic form, we were able to get the same exact answer as we did over here just a slightly different form remember the change of base says that these two things are equal, so whenever we're solving exponential equations where our bases aren't the same or we can get them to be the same we have to use logarithms in order to solve them.

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