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- Oct 4th 2008, 05:17 AM #1

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- Oct 4th 2008, 05:27 AM #2

- Oct 4th 2008, 06:09 AM #3

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- Oct 4th 2008, 08:37 AM #4

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- Oct 4th 2008, 11:06 PM #5
$\displaystyle 2^{\frac6x} \geq 5^3$

Take the inequality to the power of $\displaystyle \frac x6$:

$\displaystyle \left(2^{\frac6x}\right)^{\frac x6} \geq \left(5^3\right)^{\frac x6}~\implies~2 \geq 5^{\frac x2}$ Now square the both sides of the inequality:

$\displaystyle 4 \geq 5^x~\implies~\log_5(4) \geq x$

To get an approximate result use the base-change-formula:

$\displaystyle x\leq \log_5(4)$ and $\displaystyle \log_5(4)=\dfrac{\ln(4)}{\ln(5)}\approx 0.861...$