How do you solve $7-2e^x=5$ ?

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Answer 1 · 2015-05-22T23:52:15

Using $ln$ is a useful way!

First, isolate $e^x$ in one side:

$e^x=(5-7)/(-2)$
$e^x=1$

Now, apply $ln$ on both sides, to remove $x$ from the exponent, following the logarithm property $lna^n=n*lna$

$lne^x=ln1$ (use a calculator for $ln=1$ )
$xlne=ln1$

Now we have two definitions of $ln$ :

$lne=1$ (because $e^1=e$ )
and $ln1=0$ (because $e^0=1$ )

Thus,

$x*1=0$
$x=0$

How do you solve 7-2e^x=57-2e^x=5?