Hi could someone please let me know if this is correct?
Log4_X + log4_(x-3)=1
Log4_X + log4_X - Log4_3=1
Change base ln
lnX/ln4 + lnX/ln4 - ln3/ln4 = lne
As all ln the cancel all and solve with algebra
X/4 +X/4 - 3/4 = 1
(2X-3)/4 = 1
2X-3 = 4
2X = 4+3
X = 7/2
Thanks in advance
May 20th 2011, 06:15 AM
You should ALWAYS check your answer by plugging it back into the original equation. In this case, it doesn't fit. I would rather solve your problem this way:
Can you continue from here?
May 20th 2011, 06:21 AM
I think so.
Produces the quadratic x^2-3X-4=0?
May 20th 2011, 06:24 AM
Yep. Looks good to me. You should definitely check both answers against the original equation to make sure you're not doing something weird like taking the logarithm of a negative number or some such verboten action.
Neither -1 nor -4 are in the domain of the log4 function. That is, log4(-1) and log4(-4) are undefined. What does that tell you?
May 20th 2011, 09:53 AM
They are vertical asymptotes or poles where x=-1, -4?
May 20th 2011, 09:57 AM
Neither. The technical explanation is that there is a branch cut discontinuity in the complex plane. That is, you have two surfaces representing the function, and they don't meet on the negative real axis or at zero. It's a lot like a parking garage: you're heading clockwise on one floor, and your friend is heading counter-clockwise on the next floor up. You're not going to meet, are you? That's what this is like.
But that's more than you probably need to know right now. What you need to know right now is that, for real-valued logarithm functions with a positive base, the domain is the positive real numbers. You can't plug zero or a negative number into the logarithm functions you're dealing with.