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how do you do log problems?
such as this one:
solve:
log 125 = 3
b
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You need to use the law of logs that log[sub]x[/sub] y = log y / log x.
In your example, that means that log 125/log b = 3.
Multiply by log b: log 125 = 3log b
Convert the right-hand side using another log law: log 125 = log(b³)
Take exponentials: 125 = b³
Take cube roots: 5 = b.
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ok i get it...
like this:
Log1/3 N = -3
log1/3 / logN = -3
Log1/3 = -3logN
log1/3 log (N^-3)
1/3 = N^-3
ok here is where i am getting stuck....
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and for this problem also.
log1/3A=-4
log1/3 / logA = -4
log1/3 = -4LogA
log1/3 = log(a^-4)
1/3 = a^-4
stuck
How do i find the number ..if i have -4 or -3. ? ?:/
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The exponential form a^x = b can be written as,
log b = x.
a
The problem is log 125 = 3.
b
It can be written as b^3 = 125.
b^3 = 5^3.
b =5.
Last edited by Prakash Panneer (2006-10-13 04:49:15)
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basicly, to remove the logarithm, you take the exponential with base of logarithm to each side
Last edited by luca-deltodesco (2006-10-13 04:53:22)
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Thank You So Very Much For Explaining
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Note that there is a difference between "log" and "\log" in LaTex:
The spacing is a lot better with \log, but in these cases, it doesn't matter all that much.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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yeh i usually try to use the proper ones, like \sin vs sin, \sec vs sec etc.
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which math is this?
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what do you mean holla?
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which math is this?
If you mean log, sec, sin, etc. They are just mathematical abbreviations.
LaTeX is being discussed.
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what year i mean..btw, kpsd...
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ok here is another problem
log 64/27 = 3
a
logA / log64/27 = 3
log64/27 = 3logA
log64/27 = logA^3
64/27 = a^3
A = 64/27^3
now i get
a = 64/19683
right
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Sorry, your answer is incorrect.
your question is log 64/27 =3
a
This can be written as a ^ 3 = 64/27
a = (64/27)^(1/3)
a = 4/3.
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