How do you solve $7 e^{2 x} = 77$ $7e^(2x) = 77$ ?

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Answer 1 · 2016-05-28T14:57:51

$x = \frac{ln 11}{2} = 1.199$ $x=ln 11/2=1.199$ , nearly.

$e^{2 x} = \frac{77}{7} = 11$ $e^(2x)=77/7=11$

Inversely, $2 x = ln 11$ $2x=ln 11$ . So, $x = \frac{ln 11}{2} = 1.199$ $x =ln 11/2=1.199$ , nearly.

How do you solve 7e2x=777e^(2x) = 77?