C Cari_R New member Joined Oct 10, 2015 Messages 3 Oct 11, 2015 #1 I have this question: y= ln(x) - ln(x+2) + ln(4-x^2), express x in terms of y. I have come as far as : 2x-x^2= e^y but how do I get to x= 1 ± √1-e^y ? (the answer that is given in the back of the book) Thank you for your help!

I have this question: y= ln(x) - ln(x+2) + ln(4-x^2), express x in terms of y. I have come as far as : 2x-x^2= e^y but how do I get to x= 1 ± √1-e^y ? (the answer that is given in the back of the book) Thank you for your help!

S Subhotosh Khan Super Moderator Staff member Joined Jun 18, 2007 Messages 18,149 Oct 11, 2015 #2 Cari_R said: I have this question: y= ln(x) - ln(x+2) + ln(4-x^2), express x in terms of y. I have come as far as : 2x-x^2= e^y but how do I get to x= 1 √1-e^y ? (the answer that is given in the back of the book) Thank you for your help! Click to expand... 2x-x^2= e^y x^2 - 2x + e^y = 0 Now solve the quadratic equation!

Cari_R said: I have this question: y= ln(x) - ln(x+2) + ln(4-x^2), express x in terms of y. I have come as far as : 2x-x^2= e^y but how do I get to x= 1 √1-e^y ? (the answer that is given in the back of the book) Thank you for your help! Click to expand... 2x-x^2= e^y x^2 - 2x + e^y = 0 Now solve the quadratic equation!