Algebra: need steps to understand "In 0.38. = - 0.9676"

Erm... sorry, but I can't understand your question. It looks to me like you wrote "In 0.38..." Is that perhaps a typo and meant to be "ln 0.38" (with a lower-case L instead of an upper-case I)? Assuming that's the case:

ln(x) is a standard mathematical notation to take the natural logarithm of x. This is defined as the logarithm, base e, of x, where e is the mathematical constant:

\(\displaystyle \displaystyle e=\sum _{n=0}^{\infty }\:\frac{1}{n!}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2\cdot 3}+...\approx 2.71828\)

For more information, see this page. If my guess was incorrect, and you really meant to post "In 0.38," please share with us what your class/teacher/textbooks means by that notation, along with any and all work you've done on this problem, even if you know it's wrong. Thank you.
 
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In 0.38. = - 0.9676

It is true that ln(0.38) equals -0.9676, rounded to four places.

What is the question?

If you're trying to ask how to approximate ln(0.38), we use a scientific calculator for that.
 
In 0.38. = - 0.9676
need help with steps so I can understand it.
ln(a) = b is the same statement as e^b = a. There is NOTHING to understand in that statement as it is just a definition. For the record ln (a) is short hand for loge(a). So your equation ln 0.38. = - 0.9676 is just saying that e-.9676=0.38.
 
\(\displaystyle \displaystyle e=\sum _{n=0}^{\infty }\:\frac{1}{n!}=\frac{1}{1}+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2\cdot 3}+...\approx 2.71828\)

That should be


\(\displaystyle \displaystyle e \ = \ \sum _{n=0}^{\infty }\:\frac{1}{n!} \ = \ \frac{1}{1} + \frac{1}{1}+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2\cdot 3} \ + \ ... \ \approx \ 2.71828\)
 
That should be

\(\displaystyle \displaystyle e \ = \ \sum _{n=0}^{\infty }\:\frac{1}{n!} \ = \ \frac{1}{1} + \frac{1}{1}+\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2\cdot 3} \ + \ ... \ \approx \ 2.71828\)

Ah yes, you're correct. I forgot about the 0 factorial term. Thanks for pointing it out to me, I'll edit my post accordingly. :)
 
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