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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)
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 does 76 plus 54 plus 92 plus 88 plus 76 plus 88 plus 75 plus 93 plus 92 plus 68 plus 88 plus 76 plus 76 plus 88 plus 80 plus 70 plus 88 plus 72 equal?
76 plus 54 plus 92 plus 88 plus 76 plus 88 plus 75 plus 93 plus 92 plus 68 plus 88 plus 76 plus 76 plus 88 plus 80 plus 70 plus 88plus 72 equal 1,440
What is 39 plus 39 plus 39 plus 39 plus 39 plus 39 plus 19 plus 19 plus 19 plus 19 plus 19 plus 19 plus 19 plus 19 plus 19 plus 19 plus 19 plus 19 plus 35 plus 35 plus 35 plus 35 plus 35 plus 35 plus?
The answer is 672.
