What are possible values of x if $\frac{{ln e}^{2 x}}{2} < x ln (3) ?$ $lne^(2x)/2<xln(3)?$ ?

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What are possible values of x if $\frac{{ln e}^{2 x}}{2} < x ln (3) ?$ $lne^(2x)/2<xln(3)?$ ?

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Answer 1 · 2015-11-03T23:27:41

Question asked for 'values' so you have to state all of them.
$x \in] 0, \infty]$ $x in ]0,oo]$ Notation for the range of zero to infinity but excluding 0

Explanation:

$ln (e^{2 x})$ $ln(e^(2x))$ has the same value as $2 x ln (e)$ $2xln(e)$
$ln (e) = 1$ $ln(e) = 1$ so we now have:

$\frac{2 x}{2} < x ln (3)$ $(2x)/2 < x ln(3)$

$x < x ln (3)$ $x < x ln(3)$

$0 < x ln (3) - x$ $0 < xln(3) - x$

$0 < x (ln (3) - 1)$ $0< x(ln(3) -1)$

$\frac{0}{ln (3) - 1} < x$ $0/(ln(3) -1) < x$

$x > 0$ $x > 0$

Answer 2 · 2015-11-03T23:32:36

This inequality is true for any positive real number $x \in (0; + \infty)$ $x in (0;+oo)$

Explanation:

$\frac{ln (e^{2 x})}{2} < x ln (3)$ $ln(e^(2x))/2 < xln(3)$

$\frac{2 x ln e}{2} < x ln 3$ $(2xlne)/2 < xln3$

$x < x ln 3$ $x < xln3$

$x (1 - ln 3) < 0$ $x(1-ln3) < 0$

$x > 0$ $x>0$

I changed the sign of inequality in last expression because $ln 3 \approx 1.098$ $ln3~~1.098$ , so last step was the division by a negative number ( $1 - ln 3$ $1-ln3$ )

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What are possible values of x if $\frac{{ln e}^{2 x}}{2} < x ln (3) ?$ $lne^(2x)/2<xln(3)?$ ?

2 Answers

Explanation:

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