Simplifying a log function

Question

How come:

y=c1*e^(c2*t)

is simplified to:

ln(y)=ln(c1)+c2*t

?

What I got is:

ln(y) = ln(c1*e^(c2*t))
ln(y) = (c2*t) * ln(c1*e)
ln(y) = (c2*t) * (ln(c1) + ln(e))
ln(y) = (c2*t) * ln(c1)

i.e. I have a multiplication sign when it should be addition sign. I don't get how there should be an addition sign. Any explanation is appreciated!

Thanks in advance!

Answer 1

You have incorrectly used (or perhaps failed to use) the product rule of logarithms. Given positive numbers $a$ and $b,$ we have $$\ln(a\cdot b)=\ln(a)+\ln(b).$$ In particular, $$\ln(c_1e^{c_2t})=\ln(c_1)+\ln(e^{c_2t})=\ln(c_1)+c_2t.$$

Answer 2

Actually the problem is that you need to know not just the arithmetic operations but also the precedence rules. $ae^b = a \times ( e^b )$, which is why $\ln(ae^b) = \ln(a) + \ln(e^b)$ when $a > 0$. "$ae^b$" does not mean "$( a \times e )^b$", which is why it is incorrect that $\ln(ae^b) = b \ln(ae)$.