all the basis number is 3
2007-02-06 20:37:36 · 12 answers · asked by Moron 1 in Science & Mathematics ➔ Mathematics
all the basis number is 3
edit : i mean 3log 6 + 3log 21 - 3log 2 - 3log 7?
2007-02-06 20:49:28 · update #1
log(6) + log(21) - log(2) - log(7) (all logs to base 3)
Two of the laws of logarithms say :
For addition : log(M) + log(N) = log(M*N)
For subtraction : log(M) - log(N) = log(M/N)
The expression is thus : log[6*21/(2*7)]
2 into 6 goes 3 and 7 into 21 goes 3.
Expression is : log(3*3) = log(3^2)
Another law of logs states that : log(M^N) = Nlog(M)
Expression is now : 2log(3)
The definition of a logarithm is :
If p = log(N) to the base b, then N = b^p.
So if p = log(3) to the base 3, then 3 = 3^p,
which means p = 1, that is, log(3) to the base 3 = 1.
Thus, the expression = 2*1 = 2
Final answer is 2.
If you really meant to multiply each term by 3 (which I
really think you didn't mean to do), then the answer is 6.
2007-02-07 00:31:45 · answer #1 · answered by falzoon 7 · 1⤊ 0⤋
(log 6 - log 2) + (log 21 - log 7)
= log 6/2 + log 21/7 (using laws of logarithms)
= log 3 + log3
= 1 + 1
= 2
You have just posted an edit to say that a 3 should be included in working. In this case put a 3 in front of each set of brackets and answer = 6
2007-02-06 21:20:45 · answer #2 · answered by Como 7 · 1⤊ 0⤋
All logs are to the base three. Solve:
3log6 + 3log21 - 3log2 - 3log7
= 3{log(6*21) - log(2*7)} = 3{log(126) - log(14)}
= 3log(126/14) = 3log(9) = 3log(3²) = 3*2 = 6
2007-02-06 21:42:06 · answer #3 · answered by Northstar 7 · 0⤊ 1⤋
log 6 + log 21 - log 2 - log 7
= log (2*3) + log(7*3) - log 2 - log 7
= log 2 + log 3 + log 7 + log 3 - log 2 -log 7
= 2*log 3
= 2*1 [since the base is 3]
= 2
2007-02-06 20:53:02 · answer #4 · answered by Kristada 2 · 1⤊ 0⤋
log 6=log (2*3)=log2 +log 3
log 21=log(7*3)=log7+log 3
log 2+log 3+log 7+log 3-log 2-log 7 =2 log 3
2007-02-06 20:44:20 · answer #5 · answered by dheeru 2 · 0⤊ 1⤋
log 6 + log 21 = log(6*21) = log 126
log 2 + log 7 = log (2*7) = log 14
so your sum = log 126 - log 14 = log(126/14) = log 9
because it is base 3 so value = 2
2007-02-06 20:50:48 · answer #6 · answered by Mein Hoon Na 7 · 2⤊ 1⤋
Using the log properties:
log(a) + log(b) = log(ab), and
log(a) - log(b) = log(a/b),
log(6) + log(21) - log(2) - log(7) =
log(6*21) - log(2) - log(7) =
log(126) - log(2) - log(7) =
log(126/2) - log(7) =
log(63) - log(7) =
log(63 / 7) = log(9)
2007-02-06 20:41:37 · answer #7 · answered by Puggy 7 · 0⤊ 1⤋
log (3*2) + log(3*7) - log(2) - log(7)
= log(3) + log(2) + log(3)+log(7) - log(2) - log(7)
= log(3) * log(3)
= log(9)
2007-02-06 20:41:48 · answer #8 · answered by Rhul s 2 · 0⤊ 1⤋
log(tan a million)+log(tan 2)+log(tan 3)+......+log(tan 89) =log(tan1*tan2*tan3*.....tan89) =log{tan1*tan2*....tan44*tan45*tan(ninety-4... =log(tan1*tan2*....tan44*tan45*cot44...... =log(tan1*cot1*tan2*cot2*.....tan44*cot... =log(a million*a million*....a million*a million) =log1 =0
2016-12-17 04:22:33 · answer #9 · answered by Anonymous · 0⤊ 0⤋
it's easy:
log(6)+log(21)-log(2)-log(7)=
log(6)-log(2)+log(21)-log(7)=
log(6/2)+log(21/7)=
log(3)+log(3)=
1+1=2
2007-02-06 21:05:06 · answer #10 · answered by Michaela R 2 · 1⤊ 0⤋