Logarithms question help me please?

2009-12-26 3:37 pm

Log(2t)-2log(2t).

i think the final answer is Log(2/(9t)) could you show me how you got there and explain if possible
i have an exam and i need to revise this question as well as a few others

Thanks

回答 (4)

[匿名]

2009-12-26 3:42 pm

✔ 最佳答案

log(2t) - 2log(2t)
= log(2t) - log((2t)^2) (since a log(b) = log(b^a) )
=log(2t / (2t^2)) = log(1/(2t)) = log(1) - log(2t) (since log(a/b) = log(a) - log(b))
=log(1) -log(2) - log(t) (since log(a*b) = log(a) + log(b))
= -log(2) - log(t) (since log(1) = 0.)

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An ESL Learner

2009-12-26 4:16 pm

log(2t) - 2log(2t)
= log(2t) - log[(2t)^2]
= log(2t) - log(4t^2)
= log[2t/(4t^2)]
= log[1/(2t)]

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