Using the $\ln(\cdot)$ for $(1-e^{-x})$

Question

The given function:

$$B= A(1-(e^{-x}))$$

Now, I want to 'destroy' the e-function by taking the logarithm of it.

First, since $\ln(ab) = \ln(a) + \ln(b)$ we get that $\ln(b) = \ln(a) + \ln(1-e^{-x})$.

Using $\ln(1) =0$, $\ln(b) = \ln(A) +x$.

Is this the right way to handle the function?

Thank you.

Answer 1

No it is not true that $\ln(a+b)=\ln a+\ln b$ which you used to split up $\ln(1-e^{-x})$

A better approach is to first solve for the $e^{-x}$

$$e^{-x} = 1-\frac{B}{A}$$

Now take the logarithm.

Answer 2

No, you can't handle the function like this as $\ln(a+b)\neq\ln(a)+\ln(b)$ and thus $\ln(1-e^{-x})\neq \ln(1)-\ln(e^{-x})$.

Try instead the following (assuming that $A\neq 0$): $$B=A\cdot \left(1-e^{-x}\right)\iff \frac{B}{A}=1-e^{-x}\iff \frac{B}{A}-1=-e^{-x} \iff -\frac{B}{A}+1=e^{-x}$$

Now you can use $\ln$ (assuming that $-\frac{B}{A}+1>0$).

Answer 3

In order to eliminate e-function, it can be done as follows $$B=A\cdot (1-e^{-x})$$ $$(1-e^{-x})=\frac{B}{A} $$ $$(1-e^{-x})=\frac{B}{A} $$ $$e^{-x}=1-\frac{B}{A}$$ $$e^{-x}=\frac{A-B}{A}$$ Now, taking logarithm as follows $$\ln e^{-x}=\ln\left|\frac{A-B}{A}\right|$$ $$-x\ln e=\ln\left|\frac{A-B}{A}\right|$$ $$x=-\ln\left|\frac{A-B}{A}\right|$$ $$x=\ln\left|\frac{A}{A-B}\right|$$

Answer 4

You seem to have accidentally used the incorrect formula $\ln(a+b)=\ln(a)+\ln(b)$. This is confusing, because the distributive property $c(a+b)=ca+cb$ looks so similar.

Someone once asked a student why she had accidentally assumed that $(a+b)^2=a^2+b^2$. The answer: "I thought we could do that any time we had parentheses." So it's a good idea to emphasize: the distributive property is a property of the multiplication, not parentheses! The confusion could come from the fact that multiplication isn't written out (as we write $ab$ rather than $a\times b$). The distributive property says that: $$c\times(a+b)=(c\times a)+(c\times b)$$ (Some of those parentheses are redundant.) Note that it only talks about how multiplication behaves with addition; it does not mention anything about exponents or logs. The parentheses are just there to mark what operations are done first, nothing else.

(It's a good idea for you to try to prove the distributive property and see why it's true.)