How do you solve $-3e^(7a+9)+6=-6$ ?

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Answer 1 · 2016-10-08T08:08:15

$3e^(7a+9)+6==-6$ does not have any solution.

$3e^(7a+9)+6==-6$

$hArr3e^(7a+9)==-6-6=-12$ and

$e^(7a+9)==-12/3=-4$

But no power of $e$ can be negative, as for all $x's$ , $e^x$ is positive.
graph{e^x [-10, 10, -5, 5]}

Hence, $3e^(7a+9)+6==-6$ does not have any solution.

How do you solve -3e^(7a+9)+6=-6-3e^(7a+9)+6=-6?