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mariyatimoshenko
@mariyatimoshenko
August 2022
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Если lg5=a, lg7=b, то выразите log 2 по основанию 14 через а и в
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mmb1
Verified answer
Log(a) b = 1 / log(b) a
log(a) (b*c) = log(a) b + log(a) c
log(a) (b/c) = log(a) b - log(a) c
log(a) b = log(c) b / log(c) a
--------------------------------
log(14) 2 = 1/ log(2) 14
log(2) 14 = log(2) 2 + log(2) 7 = 1 + log(2) 7
log(2) 7 = lg 7 / lg 2 = lg 7 / (lg 10/5) = lg 7 / ( lg 10 - lg 5 ) = lg 7 / ( 1 - lg 5) = b/(1-a)
1 + log(2) 7 = 1 + b/(1-a) = (1+b-a)/(1-a)
log(14) 2 = (1-a)/(1+b-a)
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Answers & Comments
Verified answer
Log(a) b = 1 / log(b) alog(a) (b*c) = log(a) b + log(a) c
log(a) (b/c) = log(a) b - log(a) c
log(a) b = log(c) b / log(c) a
--------------------------------
log(14) 2 = 1/ log(2) 14
log(2) 14 = log(2) 2 + log(2) 7 = 1 + log(2) 7
log(2) 7 = lg 7 / lg 2 = lg 7 / (lg 10/5) = lg 7 / ( lg 10 - lg 5 ) = lg 7 / ( 1 - lg 5) = b/(1-a)
1 + log(2) 7 = 1 + b/(1-a) = (1+b-a)/(1-a)
log(14) 2 = (1-a)/(1+b-a)