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

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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.
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$\pi\approx3.1416$

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

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

edit :$B=\dfrac{1}{log_2\pi}+\dfrac{1}{log_{\color{red}5}\pi}$
$Prove: \,\, A < B$
 
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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:
 
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