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$$-\ln(x-\sqrt{x^2-1})=\ln(x+\sqrt{x^2-1})$$

I'm having issues showing that the left hand side equals the right. I think I'm missing some sort of easy issue but I'm overlooking it. Going backwards on the left-hand side isn't working for me.

How would I go about showing this?

JCollins
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5 Answers5

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Remember that for logarithms, we have the property that $$x\ln(A) = \ln(A^x).$$ That means that $$-\ln(x-\sqrt{x^{2}-1}) = \ln\left(\frac{1}{x-\sqrt{x^{2} - 1}}\right).$$ Multiplying the numerator and denominator by the conjugate, we get \begin{align*} -\ln(x-\sqrt{x^{2}-1}) &= \ln\left(\frac{1}{x-\sqrt{x^{2} - 1}}\right)\\ &=\ln\left(\frac{1}{x-\sqrt{x^{2} - 1}}\cdot\frac{x+\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}}\right)\\ &=\ln\left(\frac{x+\sqrt{x^{2}-1}}{x^{2} +x\sqrt{x^{2}-1}-x\sqrt{x^{2}-1}- x^{2}+1}\right)\\ &=\ln\left(x+\sqrt{x^{2}-1}\right). \end{align*}

DMcMor
  • 9,407
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Note that $\sqrt{x^2-1}<|x|$ so for the logarithm and the square root to be defined we need $x\ge 1$.

Let's make the change of variable $x=\cosh(u)$ with $u\ge 0$.

$\ln(x+\sqrt{x^2-1})=\ln(\cosh(u)+\sqrt{\sinh(u)^2})=\ln(\cosh(u)+\sinh(u))=\ln(e^u)=u$

$-\ln(x-\sqrt{x^2-1})=-\ln(\cosh(u)-\sinh(u))=-\ln(e^{-u})=-(-u)=u$


In fact this is the logarithmic form of the hyperbolic function : $\operatorname{arccosh}(x)=\ln(x+\sqrt{x^2-1})$, this is why it works so well.

zwim
  • 28,563
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Hint: Use $$x-\sqrt{x^2-1} = \frac{1}{x+\sqrt{x^2-1}}$$ which is true because $$(x+\sqrt{x^2-1})\cdot (x-\sqrt{x^2-1}) = x^2-(x^2-1)=1$$

Lazy Lee
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$ln(x-\sqrt{x^2-1})+ln(x+\sqrt{x^2-1})=ln((x-\sqrt{x^2-1})(x+\sqrt{x^2-1}))=ln(1)=0$.

Tsemo Aristide
  • 87,475
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You can think of it from the following steps:

  • $-\ln{(x)} =\ln{(x^{-1})}= \ln{(\dfrac{1}{x})} $
  • multiply $\dfrac{1}{x-\sqrt{x^2-1}}$ above and below by $x+\sqrt{x^2-1}$, which doesn't alter the result.
  • use the identity $(a+b)\cdot(a-b) =a^2-b^2$ to expand the denominator; it leads to some nice cancellations
A. Fenzry
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