Log of e raised to an exponent

Question

I have a textbook which states the following:

$y=e^{(-\lambda x)}$

it then takes the log of both sides and comes up with:

$log \ y = - \lambda x$

Why is the right side what it is? Shouldn't it be:

$log \ y = (- \lambda x)(log \ e)$

?

Note: "$\log$" means the natural logarithm (in mathematics). In some other fields of study it may mean log base 10 or log base 2. — GEdgar, Oct 19 '19 at 20:31
Is the Log in the textbook actually referring to $Log_e$ and not $log_{10}$? Because if that is the case, then I understand. — user523841, Oct 19 '19 at 20:33
Yes. the book is refering to $\log$ and $\ln$ or $\log_e$. This is pretty commonplace. There's utterly no reason to take the $\log_{10}$ of both sides and every reason (to undo) to take the $\ln$. — fleablood, Oct 20 '19 at 01:05
If you did do $\log_{10}$ of both sides you could continue. $\log_{10}y = -\lambda x\log_{10}e$ so $\frac {\log_{10}y}{\log_{10}e} = -\lambda x$ and $\frac {\log_{10}y}{\log_{10}e} = \ln y$. — fleablood, Oct 20 '19 at 01:08

score 1 · Answer 1 · answered Oct 19 '19 at 20:30

1

Some texts and softwares use $\log$ for natural logarithm instead of $\ln$

answered Oct 19 '19 at 20:30

3

yes, for those who do mathematics, the base-$10$-logarithm is so unimportant that “the” logarithm is the base-$e$-logarithm. – Lubin Oct 19 '19 at 20:34
That makes sense, thanks! I always assumed that Log means $Log_{10}$ – user523841 Oct 19 '19 at 20:36

score 1 · Answer 2 · answered Oct 20 '19 at 00:21

1

After reading the comments, I realized that the textbook is referring to $log_e$ and not $log_{10}$ which is what I have thought my entire life.

answered Oct 20 '19 at 00:21

user523841

I thought that too! In highschool, we were taught that $log$ is the “common logarithm” which is $log_{10}$ and $ln$ is the “natural logarithm” which is $log_{e}$. One day a college professor used $log$ to refer to the natural logarithm and I wondered why he didn’t write “$ln$”. – Radial Arm Saw Sep 19 '21 at 00:28

score 0 · Answer 3 · answered Oct 20 '19 at 00:58

For logarithms, $$x = a^b \rightarrow \log x = b \log a$$

In this case, you have the natural logarithm ($\log_{e}$, more commonly noted as $\ln$), and the property $$\ln e = 1$$

Hence, $y = e^{-\lambda x} \rightarrow \ln y = \ln e (-\lambda x) \rightarrow \ln y = (1) (-\lambda x) \rightarrow \ln y = -\lambda x$.

3 Answers3