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October 22nd, 2016, 02:32 AM   #1
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exponential inequality

values of parameter $a$ for which the inequality $a\cdot9^x+4(a-1)3^x+a>1$ is true
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October 22nd, 2016, 05:45 AM   #2
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This is sort of a trick question.

$9^x > 0 < 4 * 3^x\ for\ all\ real\ x.$

Can you take it from here?
Thanks from panky
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October 22nd, 2016, 06:10 AM   #3
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To jeffm would you like to explain me in detail, Thanks
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November 8th, 2016, 03:22 AM   #4
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Thanks Jeffm1 got it.

Denoting $3^x=t>0$, we need $a>\dfrac {4t+1}{t^2 + 4t+1}$ for all $t>0$. Since the function $t\mapsto \dfrac {4t+1}{t^2 + 4t+1}$ is decreasing on $[0,\infty)$, we need $a\geq \lim_{t\to 0} \dfrac {4t+1}{t^2 + 4t+1} = 1$, and so indeed the answer is $a\in [1,\infty)$
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