I have that:
$$f(x) = \ln\left(1-\frac{1}{x^2}\right)$$
I need to prove that $f(2)+f(3)+f(4)=\ln\left(\frac58\right)$
Indeed, I proved that $f(2)+f(3)+f(4)=\ln(3/4)+\ln(8/9)+\ln(15/16)$ But couldn't derive $\ln(5/8)$ using it.
Any hint is welcome!
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2I think it should be $\ln(5/8)$. – David Mitra Feb 02 '14 at 12:09
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ah yes you're right! 5/8 – realandcomplex Feb 02 '14 at 12:11
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Use $\ln a+\ln b=\ln(ab)$. – David Mitra Feb 02 '14 at 12:12
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Ah thank you David I solved it now! :) – realandcomplex Feb 02 '14 at 12:15
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$$f(x)=\ln\left(1-\dfrac1{x^2}\right)=\ln\left(\dfrac{x^2-1}{x^2}\right)$$ $$\Rightarrow f(2)=\ln\left(\dfrac34\right)\quad f(3)=\ln\left(\dfrac{8}{9}\right)\quad f(4)=\ln\left(\dfrac{15}{16}\right)$$
Property $\bf1$: $$\log{(ab)}=\log(a)+\log(b)$$
$$\begin{align}f(2)+f(3)+f(4)&=\ln\left(\dfrac34\right)+\ln\left(\dfrac{8}{9}\right)+\ln\left(\dfrac{15}{16}\right)\\ & \overset{\mathbf{Pr. 1}}{=} \ln\left(\dfrac34\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\right)\\ &=\ln\left(\dfrac58\right). \end{align}$$