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