
solving for X with logs
Hi could someone please let me know if this is correct?
Log4_X + log4_(x3)=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
(2X3)/4 = 1
2X3 = 4
2X = 4+3
X = 7/2
Thanks in advance

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?

I think so.
Produces the quadratic x^23X4=0?

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.


x^23X4=0
(x4)(x+1)
X=4 or 1
s.t x=4 into Log4_X + log4_(x3)=1
Log4_(4) + log4_(43)=Log4_4
Log4_(4)=1
log4_(1)= log1/log4=0
1+0=1

Quote:
log4_X + log4_(x3)=1
From here, I would just do this for x = 4 checking:
as required. It looks like that's essentially what you've done.
What happens when you substitute x = 1?

Log4_(1) + log4_(13)=1
Log4_(1) + log4_(4)=1=( log(1)/log4) + ( log(4)/log4)=0+(1)=1

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?

They are vertical asymptotes or poles where x=1, 4?

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 counterclockwise 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 realvalued 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.