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

[Albert1]

$\pi\approx3.1416$

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

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

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

 

[Opalg]

$\pi\approx3.1416$

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

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

$Prove: \,\, A < B$

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 definition of $B$, $B = \dfrac{\ln 10}{\ln\pi} > 2$ because $\pi^2<10.$[/sp]

 

[Albert1]

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 definition of $B$, $B = \dfrac{\ln 10}{\ln\pi} > 2$ because $\pi^2<10.$[/sp]

sorry ! a typo
the original post has been edited
$B=\dfrac{1}{log_2\pi}+\dfrac{1}{log_{\color{red}5}\pi}$
and your solution is