Asked by Bola
log8-log4/log4-log2
Answers
Answered by
oobleck
assuming the usual carelessness with parentheses,
(log8-log4)/(log4-log2)
= (3log2 - 2log2)/(2log2-log2)
= 1
or,
log(8/4) / log(4/2) = log2/log2 = 1
(log8-log4)/(log4-log2)
= (3log2 - 2log2)/(2log2-log2)
= 1
or,
log(8/4) / log(4/2) = log2/log2 = 1
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