Subtraction of Logarithm

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In summary: G}(\theta,\lambda)\lambda}{\lambda-i2{\phi}'(\theta)})-log(1-\frac{\tilde{G_n}(\theta,\lambda)\lambda}{\lambda-i2{\phi}'(\theta)})=log(1-\frac{\lambda}{\lambda-i2{\phi}'(\theta)}\frac{\tilde{G_n}(\theta,\lambda)-\tilde{G}(\theta,\lambda)}{1-\tilde{G_n}(\theta,\lambda)\lambda(\lambda-2i\phi'(\theta))^{-1}}).In summary, using the properties of log and
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HACR
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Homework Statement


If [tex]\tilde{G_n}(\theta,\lambda)= \sum_{k=1}^{n} \tilde{g_(k,1)}(i\lambda)\frac\{{theta}^{k}}{k!}[/tex], show that [tex]log(1-\frac{\tilde{G}(\theta,\lambda)\lambda}{\lambda-i2{\phi}'(\theta)})-log(1-\frac{\tilde{G}_n(\theta,\lambda)\lambda}{\lambda-i2{\phi}'(\theta)})=log(1-\frac{\lambda}{\lambda-i2{\phi}'(\theta)}\frac{\tilde{G_n}(\theta,\lambda)-\tilde{G}(\theta,\lambda)}{1-\tilde{G_n}(\theta,\lambda)\lambda(\lambda-2i\phi'(\theta))^{-1}})[/tex]

Homework Equations



log(a/b)=log(a)-log(b)

The Attempt at a Solution



Using the property of log, I came up with [tex]log(\frac{\lambda-2i\{phi}'(\theta)-\tilde{G}(\theta,\lambda)\lambda}{\lambda-i2{\phi}'(\theta)-\tilde{G_n}(\theta,\lambda)\lambda})[/tex] but it's not really equal as you can see.
 
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Dear student,

Your attempt at a solution is on the right track, but there are a few errors. First, in the property of log that you used, the denominator should be b, not a. So it should be log(a/b) = log(a) - log(b).

Second, you are missing a set of parentheses in your solution. It should be log(1-\frac{\lambda}{\lambda-i2{\phi}'(\theta)}\frac{\tilde{G_n}(\theta,\lambda)-\tilde{G}(\theta,\lambda)}{1-\tilde{G_n}(\theta,\lambda)\lambda(\lambda-2i\phi'(\theta))^{-1}}).

Finally, to get to the desired solution, you need to use the property of log again. Specifically, log(1/x) = -log(x). Applying this, we get log(1-\frac{\lambda}{\lambda-i2{\phi}'(\theta)}) + log(\frac{1}{1-\tilde{G_n}(\theta,\lambda)\lambda(\lambda-2i\phi'(\theta))^{-1}}).

Substituting this into your previous attempt, we get log(\frac{\lambda-2i{\phi}'(\theta)-\tilde{G}(\theta,\lambda)\lambda}{\lambda-i2{\phi}'(\theta)-\tilde{G_n}(\theta,\lambda)\lambda}) + log(\frac{1}{1-\tilde{G_n}(\theta,\lambda)\lambda(\lambda-2i\phi'(\theta))^{-1}}).

Using log(a/b) = log(a) - log(b) again, we can simplify this to log(\frac{\lambda-2i{\phi}'(\theta)-\tilde{G}(\theta,\lambda)\lambda}{\lambda-i2{\phi}'(\theta)-\tilde{G_n}(\theta,\lambda)\lambda}\frac{1}{1-\tilde{G_n}(\theta,\lambda)\lambda(\lambda-2i\phi'(\theta))^{-1}}).

Finally, we can use the fact that \tilde{G_n}(\theta,\lambda)= \sum_{k=1}^{n} \tilde{g_(k,1)}(i\lambda)\frac{\theta^{k}}{k!} to
 

What is subtraction of logarithm?

Subtraction of logarithm is a mathematical operation that involves subtracting the logarithms of two numbers. This is done by using the properties of logarithms, specifically the property that states that the logarithm of a quotient is equal to the difference of the logarithms of the individual numbers.

What are the properties of logarithms that are used in subtraction?

The two main properties of logarithms used in subtraction are the quotient property and the power property. The quotient property states that the logarithm of a quotient is equal to the difference of the logarithms of the individual numbers. The power property states that the logarithm of a number raised to a power is equal to the product of the power and the logarithm of the number.

Can logarithms with different bases be subtracted?

Yes, logarithms with different bases can be subtracted. However, before subtracting, the logarithms must be rewritten with the same base. This can be done by using the change of base formula, which states that the logarithm of a number with base a can be rewritten as the logarithm of the same number with base b divided by the logarithm of a with base b.

What is the result of subtracting a larger logarithm from a smaller logarithm?

When subtracting a larger logarithm from a smaller logarithm, the result will be a negative number. This is because the logarithm of a number less than 1 is always negative, and subtracting a larger number from a smaller number results in a negative difference.

Can logarithms be subtracted in any order?

No, logarithms cannot be subtracted in any order. The order of subtraction matters, as it can affect the final result. It is important to follow the correct order of operations, which is to subtract the logarithms in the order given from left to right.

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