Friday, April 29, 2016

Compare the numbers $X=(\log_2(\sqrt{5}+1))^3$ and $Y=1+\log_2(\sqrt{5}+2)$.

Compare the numbers $X=(\log_2(\sqrt{5}+1))^3$ and $Y=1+\log_2(\sqrt{5}+2)$.

First, note that $5\gt 1$, which gives $2\sqrt{5}\gt 2$ and further translates into $5+2\sqrt{5}+1=(\sqrt{5}+1)^2\gt 8$, which implies $\sqrt{5}+1\gt 2^{\frac{3}{2}}$, taking base 2 logarithm of both sides of the inequality we get:

$\log_2(\sqrt{5}+1)\gt \dfrac{3}{2}$

$(\log_2(\sqrt{5}+1))^3\gt \left(\dfrac{3}{2}\right)^3=\dfrac{27}{8}$

On the other hand, we have

[MATH]\frac{5}{2}\gt \sqrt{5}[/MATH]

[MATH]\frac{9}{2}=\frac{5}{2}+2\gt \sqrt{5}+2[/MATH]

[MATH]\therefore \log_2\left(\frac{9}{2}\right)\gt \log_2(\sqrt{5}+2)[/MATH]

[MATH] \log_29 -1\gt \log_2(\sqrt{5}+2)[/MATH]

[MATH]1+ \log_29 -1\gt 1+\log_2(\sqrt{5}+2)[/MATH]

[MATH] \log_29 \gt 1+\log_2(\sqrt{5}+2)[/MATH]

If we can prove [MATH]\dfrac{27}{8}\gt \log_29[/MATH], then we can say $X\gt Y$.

Note that $2^{27}=(2^5)^{5+2} \gt 4(3^3)^5 \gt 3^{16}=9^8$, this suggest [MATH]\dfrac{27}{8}\gt \log_29[/MATH] and we're hence done as we can conclude $X\gt Y$.

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