MHB Is $(\log_2(\sqrt{5}+1))^3$ Greater Than $1+\log_2(\sqrt{5}+2)$?

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Comparison
AI Thread Summary
The discussion centers on comparing two expressions: X, defined as $(\log_2(\sqrt{5}+1))^3$, and Y, defined as $1+\log_2(\sqrt{5}+2)$. Participants are encouraged to analyze these values to determine which is greater. The thread highlights the importance of mathematical reasoning and invites contributions from others before revealing the author's solution. The conversation emphasizes collaborative problem-solving in mathematics.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Compare the numbers $X=(\log_2(\sqrt{5}+1))^3$ and $Y=1+\log_2(\sqrt{5}+2)$ and determine which is larger.
 
Mathematics news on Phys.org
anemone said:
Compare the numbers $X=(\log_2(\sqrt{5}+1))^3$ and $Y=1+\log_2(\sqrt{5}+2)$ and determine which is larger.

The golden ratio (which I'll denote here as $\varphi$) satisfies $1.61<\varphi<1.62$ and is equivalent to$\dfrac{\sqrt5+1}{2}$, so we have

$$(\log_2(2\varphi))^3=(1+\log_2\varphi)^3=1+3\log_2\varphi+3\log_2^2\varphi+\log_2^3\varphi$$

Subtracting $1$ and approximating with $\sqrt2$ we have

$$X-1>\dfrac32+\dfrac34+\dfrac18=2.375$$

Also,

$$Y-1=\log_2(\sqrt5+2)=\log_2(2\varphi+1)<\log_24.24<\log_217-2<2.375$$

as required.
 
Thanks greg1313 for participating and thanks for your creative solution!(Cool)

I will wait for a bit before showing my solution and also the other insightful solution, just in case there are others who wanted to take part in this challenge.:)
 
My solution:

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

$$\frac{5}{2}\gt \sqrt{5}$$

$$\frac{9}{2}=\frac{5}{2}+2\gt \sqrt{5}+2$$

$$\therefore \log_2\left(\frac{9}{2}\right)\gt \log_2(\sqrt{5}+2)$$

$$ \log_29 -1\gt \log_2(\sqrt{5}+2)$$

$$1+ \log_29 -1\gt 1+\log_2(\sqrt{5}+2)$$

$$ \log_29 \gt 1+\log_2(\sqrt{5}+2)$$

If we can prove $$\dfrac{27}{8}\gt \log_29$$, 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 $$\dfrac{27}{8}\gt \log_29$$ 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

$$\frac{5}{2}\gt \sqrt{5}$$

$$\frac{9}{2}=\frac{5}{2}+2\gt \sqrt{5}+2$$

$$\therefore \log_2\left(\frac{9}{2}\right)\gt \log_2(\sqrt{5}+2)$$

$$ \log_29 -1\gt \log_2(\sqrt{5}+2)$$

$$1+ \log_29 -1\gt 1+\log_2(\sqrt{5}+2)$$

$$ \log_29 \gt 1+\log_2(\sqrt{5}+2)$$

If we can prove $$\dfrac{27}{8}\gt \log_29$$, 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 $$\dfrac{27}{8}\gt \log_29$$ and we're hence done as we can conclude by now that $X\gt Y$.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Back
Top