How do you solve for x in $7.316 = e^{ln (2 x)}$ $7.316 = e^(ln(2x))$ ?

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Answer 1 · 2016-07-06T18:10:06

$x = 3.658$ $x = 3.658$

$ln (x) = {log}_{e} (x)$ $ln(x) = log_e(x)$

The function $ln (x) = a$ $ln(x) = a$ finds the value of a such that

$x = e^{a}$ $x = e^a$

Therefore, $e^{ln (x)} = e^{a} = x$ $e^(ln(x)) = e^a = x$

What I'm trying to say is that $e^{x}$ $e^x$ and $ln (x)$ $ln(x)$ are inverse operations, so they cancel out.

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

So $7.316 = 2 x \Rightarrow x = 3.658$ $7.316 = 2x implies x = 3.658$

How do you solve for x in 7.316=eln(2x)7.316 = e^(ln(2x))?