How do you solve $e^{2 x + 1} = π$ $e ^(2x + 1) = pi$ ?

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Answer 1 · 2015-12-14T00:30:06

$x = \frac{ln π - 1}{2}$ $x=(lnpi-1)/2$

Take the natural log of both sides.

$ln (e^{2 x + 1}) = ln π$ $ln(e^(2x+1))=lnpi$

$2 x + 1 = ln π$ $2x+1=lnpi$

$x = \frac{ln π - 1}{2}$ $x=(lnpi-1)/2$

How do you solve e2x+1=πe ^(2x + 1) = pi?