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Log5 2 plus log5 10 - log5 4?
log5(2) + log5(10) - log5(4) = log5(20/4) = log5(5) = 1
How do you solve log5 20 plus log5 10 - 3log5 2?
log5 20 + log5 10 - 3log5 2 = log5 [(20*10)/(2^3)] = log5 25 = 2 (log5 5) = 2
How do you solve log1 plus log2 plus log3?
log1 + log2 + log3 = log(1*2*3) = log6
What is the value of log5 plus log 5 10 - log 5 4?
1.268293446
What is log base 3 of (x plus 1) log base 2 of (x-1)?
The browser which is used for posting questions is almost totally useless for mathematical questions since it blocks most symbols.I am assuming that your question is about log base 3 of (x plus 1) plus log base 2 of (x-1).{log[(x + 1)^log2} + {log[(x - 1)^log3}/log(3^log2) where all the logs are to the same base - whichever you want. The denominator can also be written as log(3^log2)This can be simplified (?) to log{[(x + 1)^log2*(x - 1)^log3}/log(3^log2).As mentioned above, the expression can be to any base and so the expression becomesin base 2: log{[(x + 1)*(x - 1)^log3}/log(3) andin base 3: log{[(x + 1)^log2*(x - 1)}/log(2)
What is log2 6 plus log2 7 as a single logarithm?
You can combine ( \log_2 6 ) and ( \log_2 7 ) using the property of logarithms that states ( \log_a b + \log_a c = \log_a (b \cdot c) ). Therefore, ( \log_2 6 + \log_2 7 = \log_2 (6 \cdot 7) = \log_2 42 ).
Condense 9lnx minus lnx plus 3 minus lnx minus 3?
i believe it is 7lnx, but don't quote me on it.
What is y when log₂(y-4) plus log₂y equals 2?
Original problem: log2(y-4) + log2y = 2The first step is applying a log property to the first term which states that logbn + logba = logb(na)So we can change the left side of the equation to log2[y(y-4)] or log2(y2-4y).Then we apply another log property which states logbn = a is equivalent to ba = nSo we can change log2(y2-4y) = 2 to the equation 22 = y2-4yor 4 = y2 - 4yNow we can swing the four around to make the quadratic y2 - 4y - 4 = 0Unfortunately, whoever assigned you this problem decided to make it more difficult because the expression y2 - 4y - 4 is unfactorable, meaning you have to use the quadratic formula. With it, you should be able to find the two values for y: 2 ± √32/2
What do means by 3 dB?
3 dB is a change in power by a factor of 2. If it is plus, i.e. +3dB, power is doubled. If it is minus, i.e. -3dB, power is halved. 6 dB, then is a factor of four, or quarter; 6 dB is a factor of eight, or eighth, etc. The actual equation is 3 log2 (POWER OUT / POWER IN).
Could you Condense the expression 5log4 2 plus 7 log4 x plus 4 log 4 y?
You use the identities: log(ab) = b log(a), and log(ab) = log a + log b. In this case, 5 log42 + 7 log4x + 4 log4y = log432 + log4x7 + log4y4 = log4 (32x7y4).
How do you solve this logarithmic equation 2logX equals log2 plus log3X-4?
assuming the bases are the same, you can factor the log out, so you get:x^2 = 2 + 3x-4 (note that 2logx can be rewritten as logx^2 )x^2 - 3x +2 = 0You can factor this out: (x-1)(x-2) = 0so x must be one or x is two --> X=1 V X=2
What is the connection between the bandwidth of a signal and the data rate that it represents?
The data rate (C) is equal to the bandwidth (B) times the logarithm base 2 of 1 plus the signal-to-noise ratio (S/N) (how much interference is introduced in the transmission of data)C = B x log2(1 + S/N)So your data rate is directly proportional to your bandwidth. If you increase your bandwidth, your data rate will also increase provided the signal-to-noise ratio isn't affected.
