#
Solving Exponential Equations with the Different Bases - Problem 4
*
*8,409 views

Solving exponential equations where we can’t get out bases to be the same. So whenever we have exponential equations where we have bases that can’t be the same and we have variables in the exponents, what we have to do is use logarithms in order to get those exponents down to a level we can manipulate them. Doesn’t matter what logarithm we take just as long as they’re going to be the same.

For this particular example I’m going to take the natural log of both sides. We could just as easily take the common log, it doesn’t matter. We take the natural log of both sides and the reason I do that is in order to bring these exponents down and around using the law, the power rule for logarithms. What we end up with is x plus 1, times the natural log of 7 is equal to x minus 4 natural log of 9.

Now we need to solve for x, because that’s what we’re trying to find and in order to do that people tend to kind of forget how we solve for this type of problem. So I have another problem that we’re going to look at first.

What I have right here is ax minus 1 is equal to cx plus 1, and I want to solve this for x. I have two different xs, I need to somehow me them one. The first thing we need to do is distribute our a and our c. We end up with ax minus a is equal to cx plus c. We now need to get all of our xs to one side. I’m going to choose to bring my xs over here and bring everything else over to this side, leaving me with ax minus cx is equal to c plus a.

Next I need to factor out my x, so I’m making 2xs into one. Factor out the x, and then in order to solve for that last x, we need to divide by the term a minus c leaving us with x is equal to c plus a over a minus c.

This is the case when we’re dealing with just variables a, c, x and 1. Over here with the logarithms it’s the same exact idea. Log base 7, natural log of 7 and natural log of 9 are just numbers. They’re ugly numbers but they’re just numbers. What we have to do is first distribute those in. We end up with x ln7 plus ln7 is equal to x ln9 minus 4 ln9. We now just like we did in the other problem we need to get all of the xs to one side and everything else to the other side. So I’m going to choose to bring all my xs over here and bring everything else over to this side.

We have x ln7 minus x ln9 is equal to -4ln9, and now we’re minusing this over as well, minus ln7. We have 2xs, we need to make one, we have to factor it out.

Factor out the x and the other side stays the same. I’m going to not write it just save the step and then our last thing is we’re going to have to divide by the coefficient of x which is just the ln7 minus ln9 leaving us with this side over our coefficient, so we end up with x is equal to 4ln9 minus ln7 and I think I left my negative sign off, I did, over the coefficient here, ln7 minus ln9. And that 9 is not very pretty we’ll rewrite it, 9.

It’s a little bit of a tedious process but the main thing that we had to do is take the natural log of both sides, and then isolate our xs, factor out the x and divide by our coefficient.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete