How do you solve $ln (1 - e^{- x})$ $Ln(1-e^-x)$ ?

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Answer 1 · 2018-01-19T06:05:07

$ln (e^{x} - 1) - x$ $ln(e^x-1)-x$

We can write the argument as a fraction. We first need to get rid of the negative exponent.

$ln (1 - e^{-} x) = ln (1 - \frac{1}{e^{x}}) = ln (\frac{e^{x} - 1}{e^{x}})$ $ln(1−e^−x) = ln(1-1/e^x) = ln((e^x-1)/e^x)$

From here, you use the identity that $ln (\frac{x}{y}) i s ln (x) - ln (y)$ $ln(x/y) is ln(x) - ln(y)$

Which simplifies to $ln (e^{x} - 1) - ln (e^{x}) = ln (e^{x} - 1) - x$ $ln(e^x-1) - ln(e^x) = ln(e^x-1)-x$ .

How do you solve ln(1−e−x)Ln(1-e^-x)?