# Solve for x.

• Nov 21st 2006, 07:39 PM
Affinity
Solve for x.
ln(x^2 - 5)=0

I did it this way.. e^(ln[x^2 - 5])=e^0

x^2 - 5=1

x^2 = 6

which x = the square root of 6.

Would anyone kindly check this to see if I did it correctly? Thank you!
• Nov 21st 2006, 07:44 PM
putnam120
you work is correct.
but u missed a solution.
not that if $x^2=6$ then we have $x=\pm\sqrt{6}$.
• Nov 21st 2006, 07:49 PM
ThePerfectHacker
Quote:

Originally Posted by Affinity
ln(x^2 - 5)=0

I did it this way.. e^(ln[x^2 - 5])=e^0

x^2 - 5=1

x^2 = 6

which x = the square root of 6.

Would anyone kindly check this to see if I did it correctly? Thank you!

If $\ln (x^2-5)=0$
Thus,
$x^2-5=1$
Thus,
$x^2=6$
Thus,
$x=\pm \sqrt{6}$
Both of these are in domain of function