How do you solve $7*e^(x-3)=57$ ?

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Answer 1 · 2016-10-27T02:13:26

$x=5.097141119$

$x=5.1$

We start by rearranging to have $e^(x-3)$ by its self.

$7*e^(x-3)=57$

$e^(x-3)=57/7$

the inverse function of e^x is ln(x) thus,

$ln(e^(x-3))=x-3$

so,

$x-3=ln(57/7)$

$x=ln(57/7)+3$

$x=ln(57)-ln(7)+3$

$x=5.097141119$

$x=5.1$

so to check,

$7*e^(5.097141119-3)$

$=57$

How do you solve 7*e^(x-3)=577*e^(x-3)=57?