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for section 1H problem 3b: log(y+1)=x^2 + log(y1)
I can't seem to figure out why, when rewritten in the solutions, the x^2 becomes negative. What am I missing?
 one year ago
 one year ago
for section 1H problem 3b: log(y+1)=x^2 + log(y1) I can't seem to figure out why, when rewritten in the solutions, the x^2 becomes negative. What am I missing?
 one year ago
 one year ago

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matheusoliveiraBest ResponseYou've already chosen the best response.3
Could you please copy the original problem? I cant find it on the problem sets nor the exams.
 one year ago

lau22Best ResponseYou've already chosen the best response.0
1H3 Solve the following for y: b) log(y+1)=x2 +log(y−1) this is exactly as it's written here: http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/1.differentiation/partbimplicitdifferentiationandinversefunctions/problemset2/MIT18_01SC_pset1prb.pdf in the solutions, when it is rewritten, somehow the x^2 becomes negative and I can't see why: log(y + 1) − log(y − 1) = −x2 as shown in the solutions here: http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/1.differentiation/partbimplicitdifferentiationandinversefunctions/problemset2/MIT18_01SC_pset1sol.pdf
 one year ago

matheusoliveiraBest ResponseYou've already chosen the best response.3
I really believe the answer is wrong.Did it 2 times now and the solution for y is: \[y=\frac{ 10^{x ^{2}} +1 }{10^{x ^{2}} 1 }\]
 one year ago

lau22Best ResponseYou've already chosen the best response.0
thank you very much! thought I was crazy, but I got that same answer, too! I appreciate the help!
 one year ago
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