Question

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Answer 1

log 4 [ x ] = log 2^2 [ x] = (1/2) log 2 [ x]

log 8 [ x ] = log 2^3  [ x ] = (1/3) log 2 [ x]

więc

log 2 [ x] + log 4 [ x ] + log 8 [ x ] < 2 ;    x > 0

log 2 [ x ] + (1/2) log 2 [ x ] + (1/3) log 2 [ x ]  < 2   

log 2  [ x ] + log 2 [ x^(1/2) ] + log 2 [ x^(1/3) ]  <  loq 2  [ 4]

log 2 [ x* x^(1/2) * x^(1/3) ]  <  log 2 [ 4]

x * x^(1/2) * x^(1/3)  < 4

x^( 1 + 1/2 + 1/3)  < 4

x^(11/6) < 4

x^11 < 4^6

x^11 < 4 096

x < p 11 st ( 4 096 )

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p 11 st  - pierwiastek 11 stopnia

 

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II  sposób :

log 2 [ x ] + (1/2) log 2 [ x] + (1/3) log 2 [ x ] < 2

log 2 [ x ] = y

y + (1/2) y + (1/3) y  < 2

(11/6) y < 2   /  * ( 6/11)

y <  12/11

czyli

log2 [ x ]  <  12/11

log 2 [ x ]  <  log 2 [ 2 ^(12/11) ]

x < 2^(12/11)

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