# Cobb-Douglas function homotheticity

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 \$$?

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 $$a+b = 1$$.