Question

Can someone explain the role "e" has in this equation.

Answer 1

Solving for r

\[\ln (1+\frac{ r }{ 12 }) = \frac{ \ln1.8 }{ 60 } = 0.009796\]

\[(1+\frac{ r }{ 12 }) = e ^{0.009796}\]

Answer 2

I think I get it kind of.

The rule is \[e ^{lnx} = x\]

Answer 3

So I think they raised both sides by e to cancel the natural log on the left side, and got \[e ^{0.009796} \] on the right side

Answer 4

yup