Compare the numbers X=(log2(√5+1))3 and Y=1+log2(√5+2).
First, note that 5>1, which gives 2√5>2 and further translates into 5+2√5+1=(√5+1)2>8, which implies √5+1>232, taking base 2 logarithm of both sides of the inequality we get:
log2(√5+1)>32
(log2(√5+1))3>(32)3=278
On the other hand, we have
52>√5
92=52+2>√5+2
∴
\displaystyle \log_29 -1\gt \log_2(\sqrt{5}+2)
\displaystyle 1+ \log_29 -1\gt 1+\log_2(\sqrt{5}+2)
\displaystyle \log_29 \gt 1+\log_2(\sqrt{5}+2)
If we can prove \displaystyle \dfrac{27}{8}\gt \log_29, 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 \displaystyle \dfrac{27}{8}\gt \log_29 and we're hence done as we can conclude X\gt Y.
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