SOLUTION: I'm having trouble solving for 'x' in this equation {{{log4(x-2)+ log4(x+6)=2}}}. I have tried re-arranging it to make an exponential equation, but I end up confusing myself even
Algebra.Com
Question 98545This question is from textbook Calculus with Applications Brief Version
: I'm having trouble solving for 'x' in this equation . I have tried re-arranging it to make an exponential equation, but I end up confusing myself even more.
Thank you for looking at this.
This question is from textbook Calculus with Applications Brief Version
Found 2 solutions by stanbon, bimanewton:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
log4(x-2)+ log4(x+6)=2
-----------
Note: Keep the "base 4" in mind
============
log(x-2)(x+6) = 2
(x-2)(x+6) = 4^2
x^2+4x-12 = 16
x^2+4x-28=0
x = [-4 +- sqrt(16-4*-28]/2
x = [-4 +- sqrt(128)]/2
----------------
x = [-4 + 8sqrt(2)]/2 = -2+4sqrt2
OR
x = [-4 - 8sqrt(2)]/2 = -2-4sqrt2
====================
Cheers,
Stan H.
Answer by bimanewton(8) (Show Source): You can put this solution on YOUR website!
x1=
=
x2=
=
RELATED QUESTIONS
I hope someone can clarify this. I'M CONFUSED!!!
2(log4 x + logx 4)=5
My working:... (answered by Alan3354)
I need help solving this I have tried it for about 20 mins now!
Log... (answered by Alan3354)
Solve for x:
log4^8 + log4^6 =... (answered by Nate)
I need help in solving this equation: 2log x - log4 = 1/2... (answered by scott8148)
log4(x^2)-log4(4x-4)=0
I could really use some help solving this... (answered by lwsshak3)
Good evening, I have some difficulties with this logarithmic equation:... (answered by Boreal,Theo,MathTherapy)
Solve for x:
2^(3x-1) = 4^(2x)
I used logarithms to start this problem but I'm... (answered by josmiceli,MathTherapy)
Solve for x
Log4^x+... (answered by solver91311)
log4(x)+log4(x-6)=2 (answered by nerdybill)