a) og10 200 - log10 2
b) 2 log2 2 + log2 8
c) log5 5 = log5 25 - log2 8
d) 1/2 log2 64 + 1/3 log5 125
evaluate these logarithms please...?
2010-01-16 3:06 pm
回答 (9)
2010-01-16 3:47 pm
✔ 最佳答案
a ) log base 10-----------------------
log 200 - log 100
log (200/100)
log 2
b ) log base 2
---------------------
log 4 + log 8
log 32
5
c ) Assume
log5 5 + log5 5 - log2 8
1 + 1 - 3
- 1
d)
(1/2) (6) + (1/3) (3)
3 + 1
4
2016-05-26 3:07 pm
=-2
2010-01-16 3:30 pm
c. log5 5 = log5 25 - log2 8
That will come out to 1 = -1
Did you type the problem correctly?
That will come out to 1 = -1
Did you type the problem correctly?
2010-01-16 3:29 pm
a)
log_10(200) - log_10(2)
= log_10(200/2)
= log_10(100)
= log_10(10^2)
= 2log_10(10)
= 2
b)
2log_2(2) + log_2(8)
= 2 + log_2(2^3)
= 2 + 3log_2(2)
= 2 + 3
= 5
c)
log_5(5) ≟ log_5(25) - log_2(8)
1 ≟ log_5(5^2) - log_2(2^3)
1 ≟ 2log_5(5) - 3log_2(2)
1 ≟ 2 - 3
1 ≠ -1
Therefore, log_5(5) = log_5(25) - log_2(8) is not true.
d)
1/2[log_2(64)] + 1/3[log_5(125)]
= 1/2[log_2(2^6)] + 1/3[log_5(5^3)]
= 1/2[6log_2(2)] + 1/3[3log_5(5)]
= 6/2 + 3/3
= 3 + 1
= 4
log_10(200) - log_10(2)
= log_10(200/2)
= log_10(100)
= log_10(10^2)
= 2log_10(10)
= 2
b)
2log_2(2) + log_2(8)
= 2 + log_2(2^3)
= 2 + 3log_2(2)
= 2 + 3
= 5
c)
log_5(5) ≟ log_5(25) - log_2(8)
1 ≟ log_5(5^2) - log_2(2^3)
1 ≟ 2log_5(5) - 3log_2(2)
1 ≟ 2 - 3
1 ≠ -1
Therefore, log_5(5) = log_5(25) - log_2(8) is not true.
d)
1/2[log_2(64)] + 1/3[log_5(125)]
= 1/2[log_2(2^6)] + 1/3[log_5(5^3)]
= 1/2[6log_2(2)] + 1/3[3log_5(5)]
= 6/2 + 3/3
= 3 + 1
= 4
2010-01-16 3:22 pm
Remember that the log of a number to any base is the power to which the base must be raised to generate the number, e.g. log(10) 1000 = 3
a) log(100 200 - log(10) 2 = log(10) 200/2 = log(10) 100 = 2
b) 2 log(2) 2 + log(2) 8 = 2+3 = 5
c) log(5) 5 = log(5) 25 - log(2) 8 => 1=2-3. This is false, so I am guessing you missed the shift key when you typed it and you really meant:
c) log(5) 5 + log(5) 25 - log(2) 8 = 1 + 2 - 3 = 0
d) 1/2 log(2) 64 + 1/3 log(5) 125 = 3 + 1 = 4
a) log(100 200 - log(10) 2 = log(10) 200/2 = log(10) 100 = 2
b) 2 log(2) 2 + log(2) 8 = 2+3 = 5
c) log(5) 5 = log(5) 25 - log(2) 8 => 1=2-3. This is false, so I am guessing you missed the shift key when you typed it and you really meant:
c) log(5) 5 + log(5) 25 - log(2) 8 = 1 + 2 - 3 = 0
d) 1/2 log(2) 64 + 1/3 log(5) 125 = 3 + 1 = 4
2010-01-16 3:21 pm
a) log base 10 problem: log 200 = log 2 + log 100 So log 200 - log 2 = log 100 = 2 since 100 = 10^2.
b) log base 2 problem: log 2 = 1 since 2=2^1. log 8 = 3 since 8 = 2^3. So log 2 + log 8 = 1+3=4
c)bases 5 and 2 problem: 5=5^1, 25=5^2, 8=2^3 so the result is 1+2 - 3 = 0 assuming you meant to have a + instead of an = in the problem.
d) 64 = 2^6 and 125 = 5^3 so the answer is (1/2)*6 + (1/3)*3 = 3+1 = 4
b) log base 2 problem: log 2 = 1 since 2=2^1. log 8 = 3 since 8 = 2^3. So log 2 + log 8 = 1+3=4
c)bases 5 and 2 problem: 5=5^1, 25=5^2, 8=2^3 so the result is 1+2 - 3 = 0 assuming you meant to have a + instead of an = in the problem.
d) 64 = 2^6 and 125 = 5^3 so the answer is (1/2)*6 + (1/3)*3 = 3+1 = 4
2010-01-16 3:20 pm
(a)
formula is
log a-log b = log(a/b)
so ans is
log(200/2)
log 100
value of log 100 is 2
so ans is == 2
-----------------------------
thereis use of two formulas
log a+log b= log a*b
and
alogb = log b^a
so 2log2 2 +log2 8
= log 2 2^2 +log 2 8
= log2 4 +log2 8
= log 2 (4*8)
= log 2 32
= log 2 2^5
= 5
note formula is
loga a^b = b
---------------------------
(c)
=log5 25 - log2 8
= log5 5^2 - log 2 2^3
= 2- 3
=1
= log5 5
-----------------------
(d)
1/2 log2 64+1/3log 5 125
log2 64^1/2 +log5 125^1/3
log2 8 +log5 5
log 2 2^3 + log 5 5
3 +1
=4
formula is
log a-log b = log(a/b)
so ans is
log(200/2)
log 100
value of log 100 is 2
so ans is == 2
-----------------------------
thereis use of two formulas
log a+log b= log a*b
and
alogb = log b^a
so 2log2 2 +log2 8
= log 2 2^2 +log 2 8
= log2 4 +log2 8
= log 2 (4*8)
= log 2 32
= log 2 2^5
= 5
note formula is
loga a^b = b
---------------------------
(c)
=log5 25 - log2 8
= log5 5^2 - log 2 2^3
= 2- 3
=1
= log5 5
-----------------------
(d)
1/2 log2 64+1/3log 5 125
log2 64^1/2 +log5 125^1/3
log2 8 +log5 5
log 2 2^3 + log 5 5
3 +1
=4
2010-01-16 3:15 pm
a) log 200 - log 2 = log (200/2) = log 100 = 2
b) 2 log_2 2 + log_2 8 = log_2 4 + log_2 8 = log_2 32 = 5
c) log_5 5 + log_5 25 - log_2 8 = 1 + 2 - 3 = 0
d) (1/2) log_2 64 + (1/3) log_5 125 = (1/2) (6) + (1/3) (3) = 3 + 1 = 4
b) 2 log_2 2 + log_2 8 = log_2 4 + log_2 8 = log_2 32 = 5
c) log_5 5 + log_5 25 - log_2 8 = 1 + 2 - 3 = 0
d) (1/2) log_2 64 + (1/3) log_5 125 = (1/2) (6) + (1/3) (3) = 3 + 1 = 4
參考: I have taught math for over 40 yr.
2010-01-16 3:13 pm
a)2
b)5
c)1≠ -1
d)4
b)5
c)1≠ -1
d)4
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