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I have a utility function $u(x) = a - b e^{-x/20\,000}$ and it is given that $u(0)=0$ and $u(100\,000)=1$.

I am trying to show that $a = 1.0067837$ and $b = 1.0067837$. Here is what I tried:

\begin{align}u(0) &= a - b e^{-0/20\,000} = 0 &\implies a-be^1=0 &\implies a=be\\u(100\,000) &= a - b e^{100\,000/20\,000} = 1 &\implies a-be^5=1 &\implies a=1+be^5\end{align}

When $a=be=1+be^5$, taking $e = 2.72$, it doesn't get $b = 1.0067837$.

Where did I do wrong, and how can I correct it? Thank you.

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The utility function $u(x)=a-be^{-x/20\,000}$ with the conditions $u(0)=0$ and $u(100\,000)=1$ gives $$u(0)=a-be^{-0/20\,000}\implies 0=a-be^0=a-b\implies a=b\tag1$$ and $$u(100\,000)=a-be^{-100\,000/20\,000}\implies 1=a-ae^{-5}=a(1-e^{-5})\tag2$$ so that $a=b=(1-e^{-5})^{-1}=1.0067837$ as desired.

Where you went wrong:

  • $u(0) = a - b e^{-0/20\,000} = 0 \implies a-be^1=0$ should be $a-be^0=0$.

  • $u(100\,000) = a - b e^{100\,000/20\,000} = 1$ should be $a-be^{-100\,000/20\,000}$; you just forgot the minus sign.

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    $\begingroup$ Thank you so much for the teaching in details and so well thoughtful. wish you a brilliant weekend! $\endgroup$
    – Mark K
    Oct 26 '19 at 10:32
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    $\begingroup$ No problem, I appreciate your trying the problem before asking. Have a nice weekend too. $\endgroup$
    – TheSimpliFire
    Oct 26 '19 at 10:33
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set k=1 in a=be=1+be^5. therefore a=k+be^5. solve for k=a-be^5.

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    $\begingroup$ Welcome to OR.SE! Please use MathJax to typeset the math in your posts. Your post is a bit hard to follow as written. $\endgroup$
    – LarrySnyder610
    Nov 18 '20 at 17:51

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