How do you solve $- 3 e^{4 x} + 11 = 2$ $-3e^(4x)+11=2$ ?

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Answer 1 · 2016-11-05T10:40:38

$x = \frac{1}{4} ln 3$ $x=1/4ln3$

$- 3 e^{4 x} + 11 = 2$ $-3e^(4x)+11=2$

$- 3 e^{4 x} = 2 - 11 = - 9$ $-3e^(4x)=2-11=-9$

$3 e^{4 x} = 9$ $3e^(4x)=9$

$e^{4 x} = 3$ $e^(4x)=3$

$4 x = ln 3$ $4x=ln3$

$x = \frac{1}{4} ln 3$ $x=1/4ln3$

How do you solve -3e^(4x)+11=2−3e4x+11=2-3e^(4x)+11=2?