Question

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?

Answer 1

Could you please copy the original problem? I cant find it on the problem sets nor the exams.

Answer 2

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

Answer 3

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

Answer 4

thank you very much! thought I was crazy, but I got that same answer, too! I appreciate the help!

Answer 5

yw :D