How do you solve 4.5>=e^(0.031t)?

Nov 1, 2016

$\implies t \le \ln \frac{4.5}{0.031}$

$\implies t \le 48.52$ to 2 decimal places

Explanation:

As soon as you see the word solve it indicates you need to find at least 1 value.

Note that $e$ is a predefined constant. So the only unknown is $t$

Also for all $a \in \mathbb{R} : {\log}_{a} \left(a\right) = 1$ and we can use this to 'get rid' of $e$
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Given:$\text{ } 4.5 \ge {e}^{0.031 t}$

Take logs of both sides

$\ln \left(4.5\right) \ge 0.031 t \ln \left(e\right)$

But $\ln \left(e\right) \to {\log}_{e} \left(e\right) = 1$ giving

$\ln \left(4.5\right) \ge 0.031 t$

$\implies t \le \ln \frac{4.5}{0.031}$