How do you solve $e^{x} < 4.5$ $e^x<4.5$ ?

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Answer 1 · 2018-04-16T11:08:16

$(- \infty, ln (\frac{9}{2}))$ $color(blue)((-oo, ln(9/2))$

$e^{x} < \frac{9}{2}$ $e^x<9/2$

We can express $\frac{9}{2}$ $9/2$ as $e^{ln (\frac{9}{2})}$ $e^(ln(9/2))$

$:.$

$e^x< e^(ln(9/2))$

For:

$a^x< a^y$

$x < y$ if $a >1$

$:.$

$e > 1$

So:

$x< ln(9/2)$

$color(blue)((-oo, ln(9/2))$

How do you solve e^x<4.5ex<4.5e^x<4.5?