My Math Forum A problem with logarithms

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 December 12th, 2015, 05:56 AM #1 Member   Joined: Aug 2014 From: Lithuania Posts: 60 Thanks: 3 A problem with logarithms I need to solve this: $\displaystyle \log_2 (3x-x^2-2)>=3^{x+2}$ I know that $\displaystyle (3x-x^2-2)>0$, so $\displaystyle x\in(1;2)$. If $\displaystyle f(x)=\log_2 (3x-x^2-2)$ and $\displaystyle g(x)=3^{x+2}$, so $\displaystyle g(x)\in(27;81)$. Since $\displaystyle (3x-x^2-2)$ is a parabola which peaks at 1.5, so $\displaystyle f(x)=[-2;0)$. From this I can conclude that there are no solutions. Is this correct? Last edited by kaspis245; December 12th, 2015 at 06:29 AM.
 December 12th, 2015, 07:01 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 At $x = 1.5$, $3x - x^2 - 2 = 0.25$, so there are no solutions. Are you sure the problem you posted is what you are meant to be solving?

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