MHB Prove A < B with Log Inequality $\pi\approx3.1416$

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Inequality Log
AI Thread Summary
The discussion centers on proving that A is less than B, where A is defined as A = 1/log5(19) + 2/log3(19) + 3/log2(19) and B as B = 1/log2(π) + 1/log5(π). Calculations show A is approximately 1.999 and B is approximately 1.565 with the original definition of B, but redefining B leads to B being approximately 2.011, which complicates the proof. The relationship log_ab = ln(b)/ln(a) is used to analyze A and B further, indicating that A < 2 and B > 2 under certain conditions. The conversation highlights the nuances in defining B and the implications for the inequality A < B.
Albert1
Messages
1,221
Reaction score
0
$\pi\approx3.1416$

$A=\dfrac{1}{log_5 19}+\dfrac{2}{log_3 19}+\dfrac{3}{log_2 19}$

$B=\dfrac{1}{log_2\pi}+\dfrac{1}{log_3\pi}$

edit :$B=\dfrac{1}{log_2\pi}+\dfrac{1}{log_{\color{red}5}\pi}$
$Prove: \,\, A < B$
 
Last edited:
Mathematics news on Phys.org
Albert said:
$\pi\approx3.1416$

$A=\dfrac{1}{log_5 19}+\dfrac{2}{log_3 19}+\dfrac{3}{log_2 19}$

$B=\dfrac{1}{log_2\pi}+\dfrac{1}{log_3\pi}$

$Prove: \,\, A < B$
Is this correct? My calculator gives $A\approx 1.999$ and $B\approx 1.565$. If you define $B=\dfrac{1}{log_2\pi}+\dfrac{1}{log_{\color{red}5}\pi}$ then you get $B\approx 2.011$, which makes for a more interesting problem.

[sp]Using the relation $\log_ab = \dfrac{\ln b}{\ln a}$, you find that $A = \dfrac{\ln 360}{\ln 19} <\dfrac{\ln 361}{\ln 19} =2$ (because $361 = 19^2$). But, using my definiton of $B$, $B = \dfrac{\ln 10}{\ln\pi} > 2$ because $\pi^2<10.$[/sp]
 
Opalg said:
Is this correct? My calculator gives $A\approx 1.999$ and $B\approx 1.565$. If you define $B=\dfrac{1}{log_2\pi}+\dfrac{1}{log_{\color{red}5}\pi}$ then you get $B\approx 2.011$, which makes for a more interesting problem.

[sp]Using the relation $\log_ab = \dfrac{\ln b}{\ln a}$, you find that $A = \dfrac{\ln 360}{\ln 19} <\dfrac{\ln 361}{\ln 19} =2$ (because $361 = 19^2$). But, using my definiton of $B$, $B = \dfrac{\ln 10}{\ln\pi} > 2$ because $\pi^2<10.$[/sp]
sorry ! a typo :o
the original post has been edited
$B=\dfrac{1}{log_2\pi}+\dfrac{1}{log_{\color{red}5}\pi}$
and your solution is :cool:
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top