If {1/log pi to base 3 +1/log pi base 4} >x then x can be 2 3 3.5 π

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If $\frac{1}{l o g_{3} π} + \frac{1}{l o g_{4} π} > x$ , then the biggest possible integer $x$ can take -

\frac{1}{l o g_{3} π} + \frac{1}{l o g_{4} π} > x

p = \frac{1}{{log}_{3} π} + \frac{1}{{log}_{4} π} + 1

log 3 \int - log 3 {cot}^{- 1} (\frac{e^{x} - 1}{e^{x} + 1}) d x =

4^{{log}_{2} log x} = log x - {(log x)}^{2} + 1

x

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