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December 12th, 2015, 05:56 AM   #1
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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.
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December 12th, 2015, 07:01 AM   #2
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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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