How do you solve $e^{x} > 30$ $e^x>30$ ?

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How do you solve $e^{x} > 30$ $e^x>30$ ?

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Answer 1 · 2018-07-08T10:57:57

$x > ln (30)$ $x > ln(30)$

Explanation:

It may be helpful to look at a graph of $e^{x}$ $e^x$ . Notice that because it is an exponential function with a base greater than 1, it is monotonically increasing, so:

$e^{a} > e^{b} \Rightarrow a > b$ $e^a > e^b => a > b$

To clarify, I'll use a concrete example.

$e^{2} > e^{0} \Rightarrow 2 > 0$ $e^2 > e^0 => 2 > 0$

Getting back to our original problem, we want to find when $e^{x}$ $e^x$ is greater than $30$ $30$ . Rewrite $30$ $30$ as $e^{ln (30)}$ $e^ln(30)$

$e^{x} > e^{ln (30)}$ $e^x > e^ln(30)$

But if $e^{x} > e^{ln (30)}$ $e^x > e^ln(30)$ :

$x > ln (30)$ $x > ln(30)$

Technically, you could also take the natural logarithm on both sides since that is also monotonically increasing.

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How do you solve $e^{x} > 30$ $e^x>30$ ?

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