How do you solve $e^(2x+1)=27$ ?

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Answer 1 · 2015-09-21T03:53:20

It is $x=1/2(3*ln3-1)$

Taking logarithms to both sides of the equation we have that

$e^(2x+1)=27=>lne^(2x+1)=ln3^3=>(2x+1)=3*ln3=> x=1/2(3*ln3-1)$

How do you solve e^(2x+1)=27e^(2x+1)=27?