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# adding and subtracting natural logs

From Wikipedia

List of logarithmic identities

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 log10, but not for log2. To find log2(3), one must calculate log10(3) / log10(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,&nbsp;...,&nbsp;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 log-probabilities:

\log_b (a+c) = \log_b a + \log_b (1+b^{\log_b c - \log_b a})
\log_b (a-c) = \log_b a + \log_b (1-b^{\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 (a-c) = \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 [ n-1 \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 logb(a)&nbsp;+&nbsp;logb(c) =&nbsp;logb(ac), where a, b, and c are arbitrary constants. Suppose that one wants to approximate the 44th Mersenne prime, 232,582,657&nbsp;&minus;&nbsp;1. To get the base-10 logarithm, we would multiply 32,582,657 by log10(2), getting 9,808,357.09543 =&nbsp;9,808,357&nbsp;+&nbsp;0.09543. We can then get 109,808,357&nbsp;&times;&nbsp;100.09543â‰ˆ&nbsp;1.25&nbsp;&times;&nbsp;109,808,357.

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)=

Question:So I was bored the other day and started messing with programming in Java again. I decided to make a square root function for BigInteger and BigDecimal because I noticed there was none. I also noticed there was not a function for Log, Log10, or the assorted. So I was trying to program one for natural log, which can then be used for log10 and then much more after that as well. My problem is that I am not getting an accurate log answers. Here is what I am doing.. I am using the taylor series that can be seen here: http://en.wikipedia.org/wiki/Natural_logarithm#Derivative.2C_Taylor_series now this only works for value when abs(x) < 1 so you plug your value and solve for x into (1+x) / (1-x) = y you get a new x and run through a loop the series adding/subtracting each term. However, I am not getting the right answer just testing with a value of 2 i keep getting -1.0986.... Here is my code, if anyone has any suggestions of how to take a natural log in a program or has done this before I would greatly appreciate it. Thanks a lot public static BigDecimal bigLN(BigDecimal value) { BigDecimal sum = new BigDecimal("0"); BigDecimal neg1 = new BigDecimal("-1"); BigDecimal term1 = new BigDecimal("0"); BigDecimal term2 = new BigDecimal("0"); BigDecimal total = new BigDecimal("0"); BigDecimal answer = new BigDecimal("0"); BigDecimal xSub1 = value.subtract(new BigDecimal("1")).divide(value.add(new BigDecimal("1")),100,RoundingMode.HALF_EVEN).subtract(new BigDecimal("1")); for (int count = 1; count<=100;count++){ term1 = xSub1.pow(count); term2 = neg1.pow(count+1); total = term1.multiply(term2).divide(new BigDecimal(Integer.toString(count)),100,RoundingMode.HALF_EVEN); sum = sum.add(total); } return sum; }

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

Question:The 1st problem is Log3(x+13) - Log3(x) = 3 the 2nd problem is log3(x+3) + log3(2x-4)=4

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(2x-4)=4 log3 [(x+3)(2x-4)] = 4 3^4 = (x+3)(2x-4) I assume you know how to solve quadratics.

Question:How do you add natural log functions that have different powers ex. (1/4)ln(2) + (1/2)ln(3) please explain your steps

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)

Question:please help me solve these log (3m +7) - log (m+4) = 2 log 6 - 3 log 3 the base being 6 log (2x+8) - log (2x^2 + 21x +61) = -3 the base being 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 cross-multiply, 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. ( :