How do you solve $2 = e^{5 x}$ $2 = e^(5x)$ ?

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How do you solve $2 = e^{5 x}$ $2 = e^(5x)$ ?

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Answer 1 · 2018-06-05T10:48:36

$x = 0.1386 (4 d p)$ $x=0.1386(4dp)$

Explanation:

$e^{5 x} = 2$ $e^(5x)=2$

taking natural logs of both sides

$5 x = ln 2$ $5x=ln2$

$\Rightarrow x = \frac{1}{5} ln 2$ $=>x=1/5ln2$

$x = 0.1386294361$ $x=0.1386294361$

Answer 2 · 2018-06-05T10:52:37

$x \approx 0.1386 to 4 dec. places$ $x~~0.1386" to 4 dec. places"$

Explanation:

$using the laws of logarithms$ $"using the "color(blue)"laws of logarithms"$

$∙ x {log x}^{n} \Leftrightarrow n log x$ $•color(white)(x)logx^nhArrnlogx$

$∙ x {log}_{b} b = 1$ $•color(white)(x)log_b b=1$

$take the ln (natural log ) of both sides$ $"take the ln (natural log ) of both sides"$

$ln 2 = {ln e}^{5 x}$ $ln2=lne^(5x)$

$5xcancel(lne)^1=ln2$

$x=ln2/5~~0.1386" to 4 dec. places"$

Answer 3 · 2018-06-06T02:14:22

$x~~0.139$

Explanation:

$ln$ (Natural Log) cancels with base- $e$ , so we can take the natural log of both sides. We have

$ln2=cancel(lne)^(5x)$

$ln2=5x$

$x=ln2/5$

$x~~0.139$

Hope this helps!

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