How do you solve $e^(k+7)=26$ ?

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Answer 1 · 2016-09-06T16:49:24

Take the natural log of both sides.

Take the natural log of both sides.
$ln e^(k+7) = ln 26$

Using the property of logs that allows exponents to become coefficients...
$(k+7)lne = ln26$

The natural log of e is 1, so..
$k+7 =ln26$

Subtract 7 from both sides.
$k = ln26 -7$

Using a calculator,
$k=-3.742$

How do you solve e^(k+7)=26e^(k+7)=26?