General Discussion

Hey so I have a test tomorrow and I noticed some questions on my homework that look like this:
log(x-3) = log(7x-23) - log(x+1)
I don’t know how to do that? Could someone explain how I solve something like that?

Thanks!

i definitely know that when you have a log minus a log that means division

so it would become log(7x-23)/(x+1) with the log like, encompassing the whole thing

i think for what to do after, you just get rid of the logs...? and then solve for x, i think the main thing they want you to know here is that division rule

@SashaFiredrake forgot to ping

@REOSpeedwagon Thank you! Just one question - when you say get rid of the logs, what do you mean? Just ignore them entirely? Or turn them into exponential things somehow?

@SashaFiredrake i know theres a way to remove them by like, raising the entire equation to be a power of e but i dont quite remember how it works, ive always just ignored them ;;

@REOSpeedwagon Ok, thank you so much! I don’t know the e thing, but I can definitely just ignore them!

@SashaFiredrake ofc!! yeah i dont really get it either lol

Ooo I can help some!
So REOSpeedwagon is right about the left side reducing to log(7x-23/x-1).

I'm assuming the base isn't given, so they're both base 10 (at least if I remember my common practice). To get rid of the logs, raise both sides to be a power of 10 and cancel those right out.

Then you can just simplify away until you solve for x! First step for that would be multiplying both sides by (x-1), of course.

Hope this helps!

ETA: the reason why exponentiation cancels out the logarithms is because they're inverses! So 10^x is 10 multiplied by itself x times, while logx is how many times 10 must be multiplied by itself to produce x. Example: log100 is 2, since 10x10 (aka 10^2) = 100.

... sorry if I just made things overly confusing. I tried.

@rhapsodicDream Thank you! That wasn’t confusing, it was super helpful! Made much more sense than whatever’s happening in my math class.

Slipping a thing in here - I don't think you need to raise by a power of ten to cancel logs? for instance if you have log(x) = log(y) you can just cut the logs straight out to have x = y

okay as i was writing this i realised yes, that's right; technically right before cancelling you'd have 10^x = 10^y, but you can just jump steps instead to make it faster.