How do you solve $e^{x} - 9 = 19$ $e^x-9=19$ ?

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Answer 1 · 2017-08-30T08:22:18

Real solution:

$x = ln 28$ $x = ln 28$

Complex solutions:

$x = ln 28 + 2npii" "n in ZZ$

Given:

$e^x-9 = 19$

Add $9$ to both sides to get:

$e^x = 28$

Take the natural log of both sides to get:

$x = ln(28)$

More generally note that:

$e^(x+2npii) = e^x*e^(2npii) = e^x" "$ for any integer $n$

So if we include complex solutions, we have:

$x = ln 28+2npii" "n in ZZ$

How do you solve e^x-9=19ex−9=19e^x-9=19?