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!

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- Nov 21st 2006, 07:39 PMAffinitySolve 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 PMputnam120
you work is correct.

but u missed a solution.

not that if then we have . - Nov 21st 2006, 07:49 PMThePerfectHacker