e^-1/2ln3
log5 (4)+log5 (50) -log5 (8)
please explain. Thanks.
How to do these question without a calculator?
2011-12-28 5:53 pm
回答 (6)
2011-12-28 5:55 pm
✔ 最佳答案
e^-1/2ln3 = e^ln3^(-1/2) = 3^(-1/2) = 1/sqrt(3)
2011-12-29 2:09 am
1.
e^-1/2 ln3 = is the antilog of -1/2ln3 = 3^-1/2 = 1/sqrt3 = (sqrt3)/3
2
=log5(4*50/8) = log5(25) = 2 (,,, i.e. 25 = 5^2)
e^-1/2 ln3 = is the antilog of -1/2ln3 = 3^-1/2 = 1/sqrt3 = (sqrt3)/3
2
=log5(4*50/8) = log5(25) = 2 (,,, i.e. 25 = 5^2)
2011-12-29 2:09 am
a^log[a](b) = b
e^((-1/2) * ln(3)) =>
(e^ln(3))^(-1/2) =>
3^(-1/2) =>
1 / 3^(1/2) =>
1 / sqrt(3) =>
sqrt(3) / 3
log[5](4) + log[5](50) - log[5](8) =>
log[5](4 * 50 / 8) =>
log[5](25) =>
log[5](5^2) =>
2 * log[5](5) =>
2 * 1 =>
2
e^((-1/2) * ln(3)) =>
(e^ln(3))^(-1/2) =>
3^(-1/2) =>
1 / 3^(1/2) =>
1 / sqrt(3) =>
sqrt(3) / 3
log[5](4) + log[5](50) - log[5](8) =>
log[5](4 * 50 / 8) =>
log[5](25) =>
log[5](5^2) =>
2 * log[5](5) =>
2 * 1 =>
2
2011-12-29 2:07 am
e^(-1/2 ln3)
(e^ ln 3)^(-1/2)
3^(-1/2)
1/ 3^(1/2)
1 / â3
â3 / 3
1.732/3
.577
log5 (4)+log5 (50) -log5 (8)
log5(4 * 50 / 8)
log5(25)
log5(5^2)
2 log5(5)
2
(e^ ln 3)^(-1/2)
3^(-1/2)
1/ 3^(1/2)
1 / â3
â3 / 3
1.732/3
.577
log5 (4)+log5 (50) -log5 (8)
log5(4 * 50 / 8)
log5(25)
log5(5^2)
2 log5(5)
2
2011-12-29 2:01 am
log_5 (4) + log_5 (50) - log_5 (8) =
log_5 (4 * 50) - log_5 (8) =
log_5 (200 / 8) =
log_5 25 =
log_5 5^2 =
2 log_5 5 =
2
this uses properties of logs:
log_a a = 1 (since a^1 = a)
log M + log N = log (MN)
log M - log N = log (M/N)
log a^r = r log a
log_5 (4 * 50) - log_5 (8) =
log_5 (200 / 8) =
log_5 25 =
log_5 5^2 =
2 log_5 5 =
2
this uses properties of logs:
log_a a = 1 (since a^1 = a)
log M + log N = log (MN)
log M - log N = log (M/N)
log a^r = r log a
2011-12-29 1:59 am
log5 (4)+log5 (50) -log5 (8)
= log5 [ (4)(50)/8] = log5 [200/8]
Let x = log5 (200/8)
5^x = 200/8
5^x = 25
x=2
log5 (4)+log5 (50) -log5 (8) = 2
= log5 [ (4)(50)/8] = log5 [200/8]
Let x = log5 (200/8)
5^x = 200/8
5^x = 25
x=2
log5 (4)+log5 (50) -log5 (8) = 2
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https://hk.answers.yahoo.com/question/index?qid=20111228095303AAba3vp