Question

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

Answer 1

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

Explanation:

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)#