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 December 12th, 2015, 05:56 AM #1 Member   Joined: Aug 2014 From: Lithuania Posts: 62 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: 20,926 Thanks: 2205 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? Tags logarithms, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zendetax Algebra 1 September 23rd, 2015 06:37 AM azad Algebra 2 September 19th, 2014 02:54 PM ron246 Algebra 5 May 22nd, 2014 06:59 AM Mr Davis 97 Algebra 1 May 3rd, 2014 06:34 PM qcom Algebra 2 March 20th, 2010 09:28 PM

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