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Friday, April 29, 2016

Compare the numbers X=(log2(5+1))3 and Y=1+log2(5+2).

Compare the numbers X=(log2(5+1))3 and Y=1+log2(5+2).

First, note that 5>1, which gives 25>2 and further translates into 5+25+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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