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bo 3⁻³=1/3³=1/27
log <¹/₂₇> 3 = -⅓ ,
bo (¹/₂₇)^⁻¹/₃ = 1/∛(¹/₂₇)=1/⅓=3
log przy podstawie 1/3 z 27 = -3,
bo (⅓)⁻³=1/(⅓)³=1/(¹/₂₇)=27
log przy podst 27 z pierwiastek z 3=⅙,
bo 27^⅙=(27^⅓)^½=(∛27)^½=3^½=√3
log przy podst pierwiastek z 3 z lliczby 27=9
bo (√3)⁹=((√3)³)³=3³=27