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I've been given the Cobb-Douglas utility function:

$\ u(q_1, q_2)=a\ln q_1+b\ln q_2=q_1^aq_2^b \ $

If I want to prove homothetic preferences, I use the following condition:

$\ u(\lambda q_1, \lambda q_2)=\lambda u(q_1,q_2) \ $

According to my calculation, this yields:

$\ u(\lambda q_1, \lambda q_2)=(\lambda q_1)^a(\lambda q_2)^b \ $

$\ =\ \lambda^{a+b}q_1^aq_2^b \ $

Is it correct to say that as a result, preferences are homothetic if and only if $\ a+b=1 \ $?

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This $$ a\ln q_1+b\ln q_2=q_1^aq_2^b $$ is not true, only $$ a\ln q_1+b\ln q_2=\ln(q_1^aq_2^b) $$ is true.

$$ u(\lambda q_1, \lambda q_2)=\lambda u(q_1,q_2) $$ is a sufficient, but not necessary condition for homotheticity of $u$. You can find more on this by searching for questions with the word homothetic on this site.

It is indeed true that $$ \lambda^{a+b}q_1^aq_2^b = \lambda q_1^aq_2^b $$ if $a+b = 1$.

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