Solving Exponential Equations with the 'Same' Base - Problem 3

Solving a exponential equation. So for this particular example, what we're looking at is two exponential equations that are equal to each other one being a fraction, one not.

So what we need to do is make our bases the same somehow, so when our bases are the same, we can set our exponents equal. So the trick here is to get the same base. Both 9 which is our denominator here and 27 our power of 3. So we can rewrite them as such, the trick is to get this 9 out of the denominator and remember that negative exponents would do that for us.

So what we can do is we can say that this I know remember 9 to the -1 is equal to 1 over 9, so if you want to do in multiple steps what you can actually say is this is 9 to the -1 to the x and then remembering that 3² is 9, this then becomes 3² to the -1 to the x. You could just jump through to the negative second is, I forgot a parenthesis, you could just say 3 to the -2 is the same thing as 1 over 9, if you can see that in your head otherwise you can take a couple of steps to write it out.

So what we have right here is a bunch of powers, just remember that a power to power we multiply, so 2 to the negative first, this becomes -2 to the x, -2x, so this whole thing is 3 to the -2x. Rewriting 27 as a power of 3 becomes 3³, so those squared power to power multiply so that becomes the same thing then as 3 to the 6. Once our bases are the same, we set our exponents equal, so -2x is equal to 6 divide by -2, x is equal to -3.

So by getting our bases both to be the same, we could solve this exponential. One other note is that we didn't have to choose 3 to be our base, we could have if we wanted to chose 1/3 this one over. We would have instead made this exponent negative and kept this one positive either way would have been fine. In general I try to avoid fractions, but if you like fractions if you like 1/3 instead of 3, it would have worked just fine.