Could some one tell me how to obtain numerical value of $a$ and $b$ from equations below:
$$\frac{1+exp(b)}{1+exp(0.9a+b)}=0.95$$
$$\int_{0}^{\infty} \frac{1+exp(b)}{1+exp(a x+b)}=1$$
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$\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$– user9660Commented Jan 27, 2015 at 18:58
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2$\begingroup$ Is this Question about the Software Mathematica? If so please complement your Question with Code. Else Mathematics satisfies your needs better. $\endgroup$– user9660Commented Jan 27, 2015 at 18:58
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$\begingroup$ This is a mathematica question. $\endgroup$– barejCommented Jan 27, 2015 at 19:45
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1 Answer
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eq1 = (1 + E^b)/(1 + E^(9/10 a + b)) == 95/100;
a0 = a /. Solve[eq1, a] /. C[1] -> 0
f[b1_, x_] := (1 + E^b1)/(1 + E^(x a0 + b1)) /. b -> b1
Quiet@FindRoot[NIntegrate[f[b, x], {x, 0, Infinity}] == 1, {b, -20}]
(* {b -> -29.4444} *)
answered Jan 27, 2015 at 19:33
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$\begingroup$ Something is a little bit unclear to me. What is the reason of using
C[1] -> 0? $\endgroup$– barejCommented Jan 28, 2015 at 6:18 -
1$\begingroup$ @barej It should be evident if you type
Solve[eq1, a]and take a look at the result $\endgroup$ Commented Jan 28, 2015 at 13:34