What is e^{-x}log_{e}(1+e^x) ?

The solution to e^{-x}log_{e}(1+e^x) is (ln(e^x+1))/(e^x)

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simplify e^{-x}log_{e}(1+e^x)

simplify

e^{- x} lo g_{e} (1 + e^{x})

\frac{ln ( e ^{x} + 1 )}{e ^{x}}

Solution steps

e^{- x} lo g_{e} (1 + e^{x})

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{x}} lo g_{e} (e^{x} + 1)

Simplify

lo g_{e} (1 + e^{x}) : ln (e^{x} + 1)

= \frac{1}{e ^{x}} ln (e^{x} + 1)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ln ( 1 + e ^{x} )}{e ^{x}}

Multiply:

1 \cdot ln (1 + e^{x}) = ln (1 + e^{x})

= \frac{ln ( 1 + e ^{x} )}{e ^{x}}

7.1 \cdot 1 0^{- 21}

(e^{\frac{2 π}{3}})^{2}

(2 x) (x)^{2}

- 3 t e^{- t}

e^{- 1200 (2)}

What is e^{-x}log_{e}(1+e^x) ?
The solution to e^{-x}log_{e}(1+e^x) is (ln(e^x+1))/(e^x)