Solving a Logarithmic Equation with Multiple Logs - Problem 2

Solving a logarithmic equation where we have more than one log on a single side. Whenever we have more than one log, what we’re always going to have to do is use our properties of logs to condense everything down, once we have a single log we can either put everything into exponential form or know that if we have two logs equal to each other whatever is on the inside has to be the same.

This one, this problem, we’re adding in between our logs, so for condensing that tells us we have to multiply, so we end up with log base 2 of x times x plus 4 is equal to 5. Now we have a single log and a constant on the side, we just need to put everything into exponential form.

I’m going to skip a step, I’m going skip multiplying these that are inside the log and just do it as we put everything into exponential form, so this becomes x² plus 4x is equal to 2 to the 5th, 2 to the 5th, 2, 4, 8, 16, 32, so this ends up being x² plus 4x is equal to 32.

Now we’re solving a quadratic so we brig that 32 around, x² plus 4x minus 32 is equal to zero and factor to solve. So we end up with x plus 8 and x minus 4, leaving us with -8 and +4.

Whenever we’re solving a logarithm equation what we have to do is make sure that our answers actually work in the equation that we started with. What I mean by work is do they actually go in and are they actually in the domain? Remember we can’t take the log of a negative number. So when we take -8, plug this in, we can’t take the log of a negative number and we’re left with the log base 2 of -8. That’s not going to work. So -8 is actually an extraneous solution, it’s not going to work.

Now we need to try the same thing with 4. Log base 2 of 4, that’s fine, log base 2 of 8, that’s fine, so that actually works. So go back and check, one of our solutions ends up working, one of them doesn’t, but by condensing our two logarithms down to 1, we were able to solve this logarithmic equation.