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$$\begin{align} u(x_1x_2) & = -a\ln{a\over (a+b)x_1} -b\ln {b\over (a+b)x_2} \\ & = a\ln x_1 + b\ln x_2 + \text{constant}. \end{align}$$

Could somebody please explain how this is simplified? I'm confused with the steps needed to give us these signs.

(Image of original formula as supplied by OP)

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up vote 4 down vote accepted

$$u(x_1,x_2) = -a(\ln a - \ln(a+b)x_1) - b(\ln b - \ln(a+b)x_2) =$$

$$ =-a\ln a + a\ln(a+b)x_1 -b\ln b + b\ln(a+b)x_2 =$$

$$= -a\ln a -b\ln b + a\ln(a+b) +a\ln x_1 +b\ln(a+b) + b\ln x_2 = b\ln x_1 + b\ln x_2 + C$$


$$C = -a\ln a -b\ln b + (a+b)\ln(a+b)$$

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Try using \log or \ln between signs of dollars to properly write the above – DonAntonio Oct 15 '12 at 3:43
Im sorry. forgot about this! Thanks – ILoveMath Oct 15 '12 at 4:03

Recall the following identities $$\ln(pq) = \ln(p) + \ln(q)$$ $$\ln \left( \dfrac1q \right) = - \ln(q)$$ $$\ln \left( \dfrac{p}{q} \right) = \ln(p) - \ln(q)$$ Make use of the above identities to get your answer.

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