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Show that e−xe−x+1 = (1)/(e^x+1).

Mathematics

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Answer

Shown:

e−xe−x+1=(1)/(ex+1)e-xe-x+1 = (1)/(e^x+1)

Step-by-step solution

4 steps

The problem asks to show that e−xe−x+1=1ex+1\frac{e^{-x}}{e^{-x}+1} = \frac{1}{e^x+1}.

  1. Start with the left-hand side (LHS) of the equation.

    LHS=e−xe−x+1\text{LHS} = \frac{e^{-x}}{e^{-x}+1}

  2. To eliminate the negative exponent e−xe^{-x} from the expression, multiply both the numerator and the denominator by exe^x.

    LHS=e−x⋅ex(e−x+1)⋅ex\text{LHS} = \frac{e^{-x} \cdot e^x}{(e^{-x}+1) \cdot e^x}

  3. Simplify the numerator and distribute exe^x in the denominator.

    Recall that e−x⋅ex=e−x+x=e0=1e^{-x} \cdot e^x = e^{-x+x} = e^0 = 1.

    LHS=1e−x⋅ex+1⋅ex\text{LHS} = \frac{1}{e^{-x} \cdot e^x + 1 \cdot e^x} LHS=1e0+ex\text{LHS} = \frac{1}{e^0 + e^x} LHS=11+ex\text{LHS} = \frac{1}{1 + e^x}

  4. Rearrange the terms in the denominator to match the right-hand side (RHS) of the original equation.

    LHS=1ex+1\text{LHS} = \frac{1}{e^x + 1}

    This is equal to the RHS.
    Therefore, the identity is shown.

    Shown: e−xe−x+1=1ex+1\boxed{\text{Shown: } \frac{e^{-x}}{e^{-x}+1} = \frac{1}{e^x+1}}

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