3^(-2x + 2) = 81? log(3^(-2x + 2)) = log(81) (-2x+2)log(3) = log(81) -2x = log(81)/log(3) - 2 x = (-1/2)(log(81)/log(3)) + 1
Using the natural (base e) logs, written as "ln", 3 is eln(3) and 5 is eln(5). Or in base 10, 3=10log(3) and 5=10log(5). Check it out by taking log of both sides: log(3) = log(10log(3)) = log(3) x log(10) =log(3) x 1=log(3).
If log(x) = y then log(x3) = 3*log(x) = 3*y so that x3 = antilog(3*y) So, to find the cibe of x 1) find log x 2) multiply it by 3 3) take the antilog of the result.
log(3√243)=log(27√3)
It is not possible.
Good luck...
Good luck...
I think there are 16 promotional points, i am not sure.
1. Structured 2. Semi - structured 3. Unstructured
1
3^(-2x + 2) = 81? log(3^(-2x + 2)) = log(81) (-2x+2)log(3) = log(81) -2x = log(81)/log(3) - 2 x = (-1/2)(log(81)/log(3)) + 1
Level = 20 times log (6/2) = 20 times log (3) = 9.54 dB
log (3 x 66) = log 3 + log 66
[log2 (x - 3)](log2 5) = 2log2 10 log2 (x - 3) = 2log2 10/log2 5 log2 (x - 3) = 2(log 10/log 2)/(log5/log 2) log2 (x - 3) = 2(log 10/log 5) log2 (x - 3) = 2(1/log 5) log2 (x - 3) = 2/log 5 x - 3 = 22/log x = 3 + 22/log 5
Log (x^3) = 3 log(x) Log of x to the third power is three times log of x.
which line of a sitrep report should contain a brief summary of enemy activity
Using the natural (base e) logs, written as "ln", 3 is eln(3) and 5 is eln(5). Or in base 10, 3=10log(3) and 5=10log(5). Check it out by taking log of both sides: log(3) = log(10log(3)) = log(3) x log(10) =log(3) x 1=log(3).