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lau22

  • 3 years ago

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

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  1. matheusoliveira
    • 3 years ago
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    Could you please copy the original problem? I cant find it on the problem sets nor the exams.

  2. lau22
    • 3 years ago
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    1H-3 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/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-b-implicit-differentiation-and-inverse-functions/problem-set-2/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/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-b-implicit-differentiation-and-inverse-functions/problem-set-2/MIT18_01SC_pset1sol.pdf

  3. matheusoliveira
    • 3 years ago
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    I really believe the answer is wrong.Did it 2 times now and the solution for y is: \[y=\frac{ 10^{x ^{2}} +1 }{10^{x ^{2}} -1 }\]

  4. lau22
    • 3 years ago
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    thank you very much! thought I was crazy, but I got that same answer, too! I appreciate the help!

  5. matheusoliveira
    • 3 years ago
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    yw :D

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