Need solution for this?
2016-05-30 8:48 am
Given that Log (base10) (2) =x and Log (base10) (3)= y, the value of log (base10) (60) is expressed as
回答 (3)
2016-05-30 8:52 am
log₁₀(2) = x
log₁₀(3) = y
log₁₀(60)
= log₁₀(2×3×10)
= log₁₀(2) + log10(3) + log₁₀(10)
= x + y + 1
log₁₀(3) = y
log₁₀(60)
= log₁₀(2×3×10)
= log₁₀(2) + log10(3) + log₁₀(10)
= x + y + 1
2016-05-30 8:57 am
let log(x) be the usual base 10
60 = 2•3•10
x = log(2)
y = log(3)
1 = log(10) ; 10^1 = 10
log(abc) = log(a)+log(b)+log(c)
log(60) = log(2)+log(3)+log(10)
= x + y + 1
60 = 2•3•10
x = log(2)
y = log(3)
1 = log(10) ; 10^1 = 10
log(abc) = log(a)+log(b)+log(c)
log(60) = log(2)+log(3)+log(10)
= x + y + 1
2016-05-30 8:52 am
log₁₀(2) = x.
log₁₀(3) = y.
log₁₀(60) = log₁₀(2·3·10) = log₁₀(2) + log₁₀(3) + log₁₀(10).
log₁₀(10) = 1, so log₁₀(60) = x + y + 1.
log₁₀(3) = y.
log₁₀(60) = log₁₀(2·3·10) = log₁₀(2) + log₁₀(3) + log₁₀(10).
log₁₀(10) = 1, so log₁₀(60) = x + y + 1.
收錄日期: 2021-04-18 14:54:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160530004822AAa8ZgA