Thread: 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!

you work is correct.
but u missed a solution.
not that if then we have .

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
Thus,

Thus,

Thus,

Both of these are in domain of function