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Is there any law of exponents that applies to this equation?

How can show that the LHS gets converted into the RHS

$(e^x −e^{−x})^2 =(e^x +e^{−x})^2 −4$

Ali Naqvi
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  • The key relation you overlooked may be $e^{x}e^{-x} = 1$ for all $x \in \mathbb{R}$. – Yes May 30 '15 at 05:01
  • You can solve the resulting equation by noting that $X = e^{2x}$ and $e^{-2x} = \frac{1}{X}$. – Jared May 30 '15 at 05:27

6 Answers6

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Let $x \in \mathbb{R}.$ Then $$(e^{x} + e^{-x})^{2} = e^{2x} + 2 + e^{-2x} = e^{2x} - 2 + e^{-2x} + 4 = (e^{x} - e^{-x})^{2} + 4.$$

Yes
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$$(e^x-e^{-x})^2=(e^x)^2-2e^{x}e^{-x}+(e^{-x})^2=(e^{x})^2-2+(e^{-x})^2$$

Can you take it from there?

Teoc
  • 8,700
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Here is one way: since $c^2-d^2=(c+d)(c-d)$, \begin{align} (e^x+e^{-x})^2-(e^x-e^{-x})^2=4e^{x}e^{-x}=4. \end{align}

Martin Argerami
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Since we know the following two formulas:

$(a+b)^2=a^2+b^2+2ab. . . . . (1)$

$(a-b)^2=a^2+b^2-2ab. . . . . (2)$

subtracting equation (2) from (1) we have

$(a-b)^2-(a+b^2)=-4ab$

$\implies (a-b)^2=(a+b)^2-4ab. . . . . (3)$

now put $a=e^x$ and $b= e^{-x}$ in equation (3), you will required result

absolute friend
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$$\begin{align}(e^x-e^{-x})^2&=e^{2x}+e^{-2x}-2e^xe^{-x}\\&=e^{2x}+e^{-2x}+2e^xe^{-x}-4e^xe^{-x}\\&=(e^x+e^{-x})^2-4\end{align}$$

And keep in mind $e^xe^{-x}=1$

marwalix
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$(a \pm b)^2 = a^2 \pm 2ab + b^2$
$(e^x)(e^{-x})=e^{x-x}=e^0=1$
$a=a+0=a+b-b$ (not a rule of exponents but still important)

This gives us:
$(e^x-e^{-x})^2=(e^x)^2-2(e^x)(e^{-x})+(e^{-x})^2$
$(e^x-e^{-x})^2=e^{2x}-2+e^{-2x}$
$(e^x-e^{-x})^2=e^{2x}-2-2+2+e^{-2x}$
$(e^x-e^{-x})^2=e^{2x}+2+e^{-2x}-4$
$(e^x-e^{-x})^2=(e^x)^2+2(e^x)(e^{-x})+(e^{-x})^2-4$
$(e^x-e^{-x})^2=(e^x+e^{-x})-4 \quad QED$

Note that this proof works regardless of the base; there's nothing about $e$ that changes the proof. It is of note, however, that:
$$\sinh(x)=\frac{e^x-e^{-x}}2$$
$$\cosh(x)=\frac{e^x+e^{-x}}2$$

LordBidoof
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