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adding and subtracting natural logs
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From Wikipedia
In mathematics, there are several logarithmic identities.
Algebraic identities or laws
Using simpler operations
Logarithms can be used to make calculations easier. For example, two numbers can be multiplied just by using a logarithm table and adding.
Where b, x, and y are positive real numbers and b \ne 1. Both c and d are real numbers.
Trivial identities
Note that \log_b(0) \!\, is undefined because there is no number x \!\, such that b^x = 0 \!\, . In fact, there is a vertical asymptote on the graph of \log_b(x) \!\, at x=0.
Canceling exponentials
Logarithms and exponentials (antilogarithms) with the same base cancel each other. This is true because logarithms and exponentials are inverse operations (just like multiplication and division or addition and subtraction).
 b^{\log_b(x)} = x\text{ because }\mathrm{antilog}_b(\log_b(x)) = x \,
 \log_b(b^x) = x\text{ because }\log_b(\mathrm{antilog}_b(x)) = x \,
Changing the base
 \log_b a = {\log_d a \over \log_d b}
This identity is needed to evaluate logarithms on calculators. For instance, most calculators have buttons for ln and for log_{10}, but not for log_{2}. To find log_{2}(3), one must calculate log_{10}(3) / log_{10}(2) (or ln(3)/ln(2), which yields the same result).
Proof
 Let c=\log_b a.
 Then b^c=a.
 Take \log_d on both sides: \log_d b^c=\log_d a
 Simplify and solve for c: c\log_d b=\log_d a
 c=\frac{\log_d a}{\log_d b}
 Since c=\log_b a, then \log_b a=\frac{\log_d a}{\log_d b}
This formula has several consequences:
 \log_b a = \frac {1} {\log_a b}
 \log_{b^n} a = \over n}
 b^{\log_a d} = d^{\log_a b}
  \log_b a = \log_b \left({1 \over a}\right) = \log_{1 \over b} a
 \log_{b_1}a_1 \,\cdots\, \log_{b_n}a_n
where \scriptstyle\pi\, is any permutation of the subscripts 1, ..., n. For example
 \log_b w\cdot \log_a x\cdot \log_d c\cdot \log_d z
Summation/subtraction
The following summation/subtraction rule is especially useful in probability theory when one is dealing with a sum of logprobabilities:
 \log_b (a+c) = \log_b a + \log_b (1+b^{\log_b c  \log_b a})
 \log_b (ac) = \log_b a + \log_b (1b^{\log_b c  \log_b a})
which gives the special cases:
 \log_b (a+c) = \log_b a + \log_b \left(1+\frac{c}{a}\right)
 \log_b (ac) = \log_b a + \log_b \left(1\frac{c}{a}\right)
Note that in practice a and c have to be switched on the right hand side of the equations if c>a. Also note that the subtraction identity is not defined if a=c since the logarithm of zero is not defined.
More generally:
 \log _b \sum\limits_{i=0}^N a_i = \log_b a_0 + \log_b \left( 1+\sum\limits_{i=1}^N \frac{a_i}{a_0} \right) = \log _b a_0 + \log_b \left( 1+\sum\limits_{i=1}^N b^{\left( \log_b a_i  \log _b a_0 \right)} \right)
where a_0,\ldots ,a_N > 0.
Calculus identities
Limits
 \lim_{x \to 0^+} \log_a x = \infty \quad \mbox{if } a > 1
 \lim_{x \to 0^+} \log_a x = \infty \quad \mbox{if } a < 1
 \lim_{x \to \infty} \log_a x = \infty \quad \mbox{if } a > 1
 \lim_{x \to \infty} \log_a x = \infty \quad \mbox{if } a < 1
 \lim_{x \to 0^+} x^b \log_a x = 0
 \lim_{x \to \infty} {1 \over x^b} \log_a x = 0
The last limit is often summarized as "logarithms grow more slowly than any power or root of x".
Derivatives of logarithmic functions
 {d \over dx} \ln x = {1 \over x },
 {d \over dx} \log_b x = {1 \over x \ln b},
Where x > 0, b > 0, and b \ne 1.
Integral definition
 \ln x = \int_1^x \frac {1}{t} dt
Integrals of logarithmic functions
 \int \log_a x \, dx = x(\log_a x  \log_a e) + C
To remember higher integrals, it's convenient to define:
 x^{\left [n \right]} = x^{n}(\log(x)  H_n)
 x^{\left [ 0 \right ]} = \log x
 x^{\left [ 1 \right ]} = x \log(x)  x
 x^{\left [ 2 \right ]} = x^2 \log(x)  \begin{matrix} \frac{3}{2} \end{matrix} \, x^2
 x^{\left [ 3 \right ]} = x^3 \log(x)  \begin{matrix} \frac{11}{6} \end{matrix} \, x^3
Then,
 \frac {d}{dx} \, x^{\left [ n \right ]} = n \, x^{\left [ n1 \right ]}
 \int x^{\left [ n \right ]}\,dx = \frac {x^{\left [ n+1 \right ]}} {n+1} + C
Approximating large numbers
The identities of logarithms can be used to approximate large numbers. Note that log_{b}(a) + log_{b}(c) = log_{b}(ac), where a, b, and c are arbitrary constants. Suppose that one wants to approximate the 44th Mersenne prime, 2^{32,582,657} − 1. To get the base10 logarithm, we would multiply 32,582,657 by log_{10}(2), getting 9,808,357.09543 = 9,808,357 + 0.09543. We can then get 10^{9,808,357} × 10^{0.09543}â‰ˆ 1.25 × 10^{9,808,357}.
Similarly, factorials can be approximated by summing the logarithms of the terms.
Complex logarithm identities
The complex logarithm is the complex number analogue of the logarithm function. No single valued function on the complex plane can satisfy the normal rules for logarithms. However a multivalued function can be defined which satisfies most of the identities. It is usual to consider this as a function defined on a Riemann surface. A single valued version called the principal value of the logarithm can be defined which is discontinuous on the negative x axis and equals the multivalued version on a single branch cut.
Definitions
The convention will be used here that a capital first letter is used for the principal value of functions and the lower case version refers to the multivalued function. The single valued version of definitions and identities is always given first followed by a separate section for the multiple valued versions.
 ln(r) is the standard natural logarithm of the real number r.
 Log(z) is the principal value of the complex logarithm function and has imaginary part in the range (Ï€, Ï€].
 Arg(z) is the principal value of the arg function, its value is restricted to (Ï€, Ï€]. It can be computed using Arg(x+iy)=
From Yahoo Answers
Answers:A bit further down the page on Wikipedia that you link there is an additional formula to increase the speed of convergence: http://en.wikipedia.org/wiki/Natural_logarithm#Numerical_value However, still further down, it is mentioned that the Taylor series does not converge fast enough, and that the best is to calculate. There are additional formulas on this page: http://en.wikipedia.org/wiki/Logarithm#Calculation Finally, the following links show how to compute log in BASIC  the listings are not included but the method is described in more detail: http://www.atarimagazines.com/creative/v9n12/288_High_precision_functions_.php http://www.atarimagazines.com/creative/v10n1/205_High_precision_functions_.php
Answers:Subtracting Rule: loga (n)  loga (m) = loga (n/m) Addition Rule: loga (n) + loga (m) = loga (nm) log3 [(x+13)/x] = 3 3^3 = (x+13)/x 9x = x + 13 8x = 13 x = 13/8 log3(x+3) + log3(2x4)=4 log3 [(x+3)(2x4)] = 4 3^4 = (x+3)(2x4) I assume you know how to solve quadratics.
Answers:Easiest way to look at this is to combine the powers into the log again ln(2^1/4) + ln(3^1/2) Log rule ln (2^1/4 * 3^1/2)
Answers:When logs are subtracted, their arguments are divided. When logs are added, their arguments are multiplied. Coefficients of logs are exponents of their arguments. Log(3m+7)  Log(m+4) = Log (3m+7)/(m+4) 2Log6  3Log3 = Log36  Log27 = Log 4/3 So now, since each side of the equation is entirely one log, you can drop the log. (3m+7)/(m+4) = 4/3 So if you crossmultiply, you get: 3(3m+7) = 4(m+4) and solve for M. The second one: Log(2x+8)/(2x^2 + 21x + 61) = 3 So to break it out of the logarithm... 2^(3) = (2x+8)/(2x^2 + 21x + 61) And solve for x. ( :
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