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How do you simplify #Ln(1-e^-x)#?

Answer 1

Theodore Amidon

We can write the argument as a fraction after getting rid of the negative exponent:

#ln(1-e^-x)=ln(1-1/e^x)=ln((e^x-1)/e^x)#

From here, use #ln(a/b)=ln(a)-ln(b)# and #ln(e^x)=x#:

#=ln(e^x-1)-ln(e^x)=color(blue)(ln(e^x-1)-x#

I don't know if this is a simplification per se, but it's definitely a valid way to rewrite the function.

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Answer 2

Elias Ackerman

To simplify (\ln(1-e^{-x})), first, recognize that (1 - e^{-x}) is a common expression. Then, use the properties of logarithms to simplify the expression.

Apply the property of logarithms: (\ln(ab) = \ln(a) + \ln(b))
Recognize (1 - e^{-x}) as a product: (1 - e^{-x} = 1 \cdot (1 - e^{-x}))
Use the property to split the expression: (\ln(1) + \ln(1 - e^{-x}))
Since (\ln(1) = 0), the expression simplifies to (\ln(1 - e^{-x})).

Therefore, (\ln(1-e^{-x})) is already in its simplest form.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.