No. The log of a quotient is the log of a denominator subtracted from the log of the numerator.
A log-log scale is a set of axes where each axis is logarithmic in scale.
to log in friendster.com
The anti-log of what ??? If log(12) = 1.07918, then antilog(1.07918) = 12 Did you want the anti-log of 12 ? That's 1,000,000,000,000.
k=log4 91.8 4^k=91.8 -- b/c of log rules-- log 4^k=log 91.8 -- b/c of log rules-- k*log 4=log91.8 --> divide by log 4 k=log 91.8/log 4 k= 3.260
Bhai Log was created on 2011-08-31.
log(x6) = log(x) + log(6) = 0.7782*log(x) log(x6) = 6*log(x)
tom dunsdons dad and mum log log log log log log log in my buttt
Not quite. The log(x/y) = log(x) - log(y) In words, this reads "The log of a quotient is the difference of the log of the numerator and the log of the denominator."
No. The log of a quotient is the log of a denominator subtracted from the log of the numerator.
"Log" is not a normal variable, it stands for the logarithm function.log (a.b)=log a+log blog(a/b)=log a-log blog (a)^n= n log a
log on to
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log(x) - log(6) = log(15)Add log(6) to each side:log(x) = log(15) + log(6) = log(15 times 6)x = 15 times 6x = 90
No. log 20 is a positive number , so it you subtract it from log 5 you get less than log 5. However, log10 5 = 1 - log102 = 2- log1020 . or log 5 - log 20 = log 5 - log 4*5 = log 5 - (log 5 + log 4) = log 5 - log 5 - log 4 = - log 4 But we do not need to do all of these computations, because log 5 is different from log 5 - log 20 by the law of the equality that says two equals remain equal if and only if we subtract (in our case) the same thing from them.
You have to use logarithms (logs).Here are a few handy tools:If [ C = D ], then [ log(C) = log(D) ]log(AB) = log(A) + log(B)log(A/B) = log(A) - log(B)log(Np) = p times log(N)
log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4