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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."

Q: The log of a quotient is the log of the numerator divided by the log of the denominator?

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No. Insert the word "minus" in place of the word "plus", and you'll have a true statement.

The negative log of a number is the log of the number's reciprocal ('1' divided by the number).

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" 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(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

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No. The log of a quotient is the log of a denominator subtracted from the log of the numerator.

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