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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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
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