# solving for X with logs

• May 20th 2011, 06:09 AM
Neverquit
solving for X with logs
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
Log4_X= lnX/ln4
Log4_3=ln3/ln4
1=lne^1=lne

Now
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

• May 20th 2011, 06:15 AM
Ackbeet
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:

$\log_{4}(x)+\log_{4}(x-3)=1$

$\log_{4}(x(x-3))=\log_{4}(4^{1}).$

Can you continue from here?
• May 20th 2011, 06:21 AM
Neverquit
I think so.

• May 20th 2011, 06:24 AM
Ackbeet
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.
• May 20th 2011, 07:15 AM
Ackbeet
• May 20th 2011, 09:20 AM
Neverquit
x^2-3X-4=0
(x-4)(x+1)

X=4 or -1

s.t x=4 into Log4_X + log4_(x-3)=1
Log4_(4) + log4_(4-3)=Log4_4
Log4_(4)=1
log4_(1)= log1/log4=0
1+0=1
• May 20th 2011, 09:26 AM
Ackbeet
Quote:

log4_X + log4_(x-3)=1
From here, I would just do this for x = 4 checking:

$\log_{4}(4)+\log_{4}(4-3)=1+\log_{4}(1)=1+0=1,$

as required. It looks like that's essentially what you've done.

What happens when you substitute x = -1?
• May 20th 2011, 09:41 AM
Neverquit
Log4_(-1) + log4_(-1-3)=1

Log4_(-1) + log4_(-4)=1=(- log(1)/log4) + (- log(4)/log4)=0+(-1)=-1
• May 20th 2011, 09:42 AM
Ackbeet
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
Neverquit
They are vertical asymptotes or poles where x=-1, -4?
• May 20th 2011, 09:57 AM
Ackbeet
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.