For a quotient x/y , then its log is logx - log y . NOT log(x/y)
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)
A log-log scale is a set of axes where each axis is logarithmic in scale.
to log in friendster.com
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
log 100 base e = log 100 base 10 / log e base 10 log 100 base 10 = 10g 10^2 base 10 = 2 log 10 base 10 = 2 log e base 10 = 0.434294 (calculator) log 100 base e = 2/0.434294 = 4.605175
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."
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.
No. Insert the word "minus" in place of the word "plus", and you'll have a true statement.
True. For example: 4 X 104/2 X 108 = 2 X 10-4
It has no specific name. For example f(x) = sin(x)/log(x) where x not equal to 1
4 divided by 84 in log div = 0.047619047619047616
The negative log of a number is the log of the number's reciprocal ('1' divided by the number).
The derivative of a log is as follows: 1 divided by xlnb Where x is the number beside the log Where b is the base of the log and ln is just the natural log.
0.55
VK= RT/ZF * log [I+]out/[I+]inAccording to this equation, the equilibrium potential for potassium (VK) is equal to the product of the gas constant (R) and the temperature in degrees Kelvin (T) divided by the product of the valence of potassium (Z) and the Faraday constant (F) multiplied by the natural log of the quotient derived from the external and internal concentrations of potassium. Thus,
To calculate the number of decibels that power-level-'A' is greater than power-level-'B',-- Divide 'A' by 'B'-- Take the 'log' of the quotient-- Multiply the 'log' by 10 .If the result is negative, then 'A' is that many decibels lower than 'B'.
log y (3445.51/2400) / log 1.075 = x