As we know that $Q*P=const.$ for Cobb-Douglas preferences, we can thus conclude that $\frac{dQ/Q}{dP/P}$ is always $-1$:
$$ QP=const. \implies 0=d(PQ)=Q\ dP+P\ dQ \implies \frac{dQ}{Q}=-\frac{dP}{P} $$
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Sign up to join this communityAs we know that $Q*P=const.$ for Cobb-Douglas preferences, we can thus conclude that $\frac{dQ/Q}{dP/P}$ is always $-1$:
$$ QP=const. \implies 0=d(PQ)=Q\ dP+P\ dQ \implies \frac{dQ}{Q}=-\frac{dP}{P} $$
This will be a weird answer:
As we know that $Q*P=const.$ for Cobb-Douglas preferences.
$$ QP=const. \implies 0=d(PQ)=Q\ dP+P\ dQ \implies \frac{dQ}{Q}=-\frac{dP}{P} $$ thus we can conclude that $\frac{dQ/Q}{dP/P}$ is always $-1$.
Yes.
I have to include at least 30 characters in an answer, so let me repeat: Yes.