Inequality: log(base 2)x >= log(base 1/2) (x-1). Then x lies in the interval....

[ankitpaul]

log(base 2)x >= log(base 1/2) (x-1). Then x lies in the interval....

 

[Deleted member 4993]

log(base 2)x >= log(base 1/2) (x-1). Then x lies in the interval....

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[Deleted member 4993]

log(base 2)x >= log(base 1/2) (x-1). Then x lies in the interval....

Hint:
Log₂[1/2] = ?

 

[pka]

log(base 2)x >= log(base 1/2) (x-1). Then x lies in the interval....

\(\displaystyle {\large\log _{{2^{ - 1}}}}(x - 1) = \dfrac{{ - \log \left( {x - 1} \right)}}{{\log (2)}}\)