What is the logaritmic form of $v=Ae^{Bi}$

Question

I am reading a scientific paper, which uses a model of the form $v=Ae^{Bi}$ and then it says that this model has the following logarithmic form

$\ln (v) = Bi + ln(A)$ where A is a constant. But the right logarithmic form of the equation above isn't $\ln (v) = Bi + Bi*ln(A)$ instead?

I'm I doing somehting wrong?

Answer 1

Work out the steps:

$$v=Ae^{Bi}$$ $$\ln{(v)}=\ln{(Ae^{Bi})}$$ $$\ln{(v)}=\ln{(A)}+\ln{(e^{Bi})}$$ $$\ln{(v)}=\ln{(A)}+Bi$$

You break over the multiplication inside the logarithm before you pull down the exponent. Because the $Bi$ is only an exponent on the $e$.