Q: Are you a log

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log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4

how do i log in

log(36,200) = 4.558709 (rounded)log[log(36,200)] = 0.658842 (rounded)

False When logs are taken, division becomes subtraction, so the log of a quotient is the log of the numerator minus the log of the denominator.

Assuming you are asking about the natural logarithms (base e):log (-1) = i x pithereforelog (log -1) = log (i x pi) = log i + log pi = (pi/2)i + log pi which is approximately 1.14472989 + 1.57079633 i

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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."

For a quotient x/y , then its log is logx - log y . NOT log(x/y)

"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

1

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

log(9x) + log(x) = 4log(10)log(9) + log(x) + log(x) = 4log(10)2log(x) = 4log(10) - log(9)log(x2) = log(104) - log(9)log(x2) = log(104/9)x2 = 104/9x = 102/3x = 33 and 1/3