# लघुगणकीय सर्वसमिकाओं की सूची

यहाँ जाएँ: भ्रमण, खोज

गणित में लघुगणक से संबन्धित बहुत सी सर्वसमिकाएँ हैं-

## बीजगणितीय सर्वसमिकाएँ

### सरल संक्रियाएँ (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.

 $\log_b(xy) = \log_b(x) + \log_b(y) \!\,$ because $b^c \cdot b^d = b^{c + d} \!\,$ $\log_b\!\left(\begin{matrix}\frac{x}{y}\end{matrix}\right) = \log_b(x) - \log_b(y)$ because $b^{c-d} = \tfrac{b^c}{b^d}$ $\log_b(x^d) = d \log_b(x) \!\,$ because $(b^c)^d = b^{cd} \!\,$ $\log_b\!\left(\!\sqrt[y]{x}\right) = \begin{matrix}\frac{\log_b(x)}{y}\end{matrix}$ because $\sqrt[y]{x} = x^{1/y}$ $x^{\log_b(y)} = y^{\log_b(x)} \!\,$ because $x^{\log_b(y)} = b^{\log_b(x) \log_b(y)} = b^{\log_b(y) \log_b(x)} = y^{\log_b(x)} \!\,$ $c\log_b(x)+d\log_b(y) = \log_b(x^c y^d) \!\,$ because $\log_b(x^c y^d) = \log_b(x^c) + \log_b(y^d) \!\,$

Where $b$, $x$, and $y$ are positive real numbers and $b \ne 1$. Both $c$ and $d$ are real numbers.

### स्वयंस्पष्ट सर्वसमिकाएँ (Trivial identities)

 $\log_b(1) = 0 \!\,$ because $b^0 = 1\!\,$ $\log_b(b) = 1 \!\,$ because $b^1 = b\!\,$

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 $x=0$.

### घातों का कैंसिल होना (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).

$b^{\log_b(x)} = x\text{ because }\mathrm{antilog}_b(\log_b(x)) = x \,$
$\log_b(b^x) = x\text{ because }\log_b(\mathrm{antilog}_b(x)) = x \,$

### आधार परिवर्तन (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).

#### उपपत्ति

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:

$\log_b a = \frac {1} {\log_a b}$
$\log_{b^n} a = {{\log_b a} \over n}$
$b^{\log_a d} = d^{\log_a b}$
$- \log_b a = \log_b \left({1 \over a}\right) = \log_{1 \over b} a$

$\log_{b_1}a_1 \,\cdots\, \log_{b_n}a_n = \log_{b_{\pi(1)}}a_1\, \cdots\, \log_{b_{\pi(n)}}a_n, \,$

where $\scriptstyle\pi\,$ is any permutation of the subscripts 1, ..., n. For example

$\log_b w\cdot \log_a x\cdot \log_d c\cdot \log_d z = \log_d w\cdot \log_b x\cdot \log_a c\cdot \log_d z. \,$

### संकलन/ब्यकलन (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 $a_0,\ldots, a_N > 0$.

### घात (Exponents)

A useful identity involving exponents:

$x^{\frac{\log(\log(x))}{\log(x)}} = \log(x)$

## कैलकुलस की सर्वसमिकाएँ

### सीमा

$\lim_{x \to 0^+} \log_a x = -\infty \quad \mbox{if } a > 1$
$\lim_{x \to 0^+} \log_a x = \infty \quad \mbox{if } a < 1$
$\lim_{x \to \infty} \log_a x = \infty \quad \mbox{if } a > 1$
$\lim_{x \to \infty} \log_a x = -\infty \quad \mbox{if } a < 1$
$\lim_{x \to 0^+} x^b \log_a x = 0$
$\lim_{x \to \infty} {1 \over x^b} \log_a x = 0$

The last limit is often summarized as "logarithms grow more slowly than any power or root of x".

### लघुगणकीय फलनों के अवकलज (Derivatives)

${d \over dx} \ln x = {1 \over x },$
${d \over dx} \log_b x = {1 \over x \ln b},$

Where $x > 0$, $b > 0$, and $b \ne 1$.

### समाकलन परिभाषा (Integral definition)

$\ln x = \int_1^x \frac {1}{t} dt$

### लघुगणकीय फलनों के समाकल (Integrals)

$\int \log_a x \, dx = x(\log_a x - \log_a e) + C$

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,

$\frac {d}{dx} \, x^{\left [ n \right ]} = n \, x^{\left [ n-1 \right ]}$
$\int x^{\left [ n \right ]}\,dx = \frac {x^{\left [ n+1 \right ]}} {n+1} + C$

## बड़ी संख्याओं का सन्निकटीकरण (Approximating large numbers)

The identities of logarithms can be used to approximate large numbers. Note that logb(a) + logb(c) = logb(ac), where a, b, and c are arbitrary constants. Suppose that one wants to approximate the 44th Mersenne prime, 232,582,657 − 1. To get the base-10 logarithm, we would multiply 32,582,657 by log10(2), getting 9,808,357.09543 = 9,808,357 + 0.09543. We can then get 109,808,357 × 100.09543 ≈ 1.25 × 109,808,357.

Similarly, factorials can be approximated by summing the logarithms of the terms.

## समिश्र लघुगणकीय सर्वसमिकाएँ (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.

### परिभाषाएँ

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 (-π, π].
Arg(z) is the principal value of the arg function, its value is restricted to (-π, π]. It can be computed using Arg(x+iy)= atan2(y, x).
$\operatorname{Log}(z) = \ln(|z|) + i \operatorname{Arg}(z)$
$e^{\operatorname{Log}(z)} = z$

The multiple valued version of log(z) is a set but it is easier to write it without braces and using it in formulas follows obvious rules.

log(z) is the set of complex numbers v which satisfy ev = z
arg(z) is the set of possible values of the arg function applied to z.

When k is any integer:

$\log(z) = \ln(|z|) + i \arg(z)$
$\log(z) = \operatorname{Log}(z) + 2 \pi i k$
$e^{\log(z)} = z$

### नियतांक (Constants)

Principal value forms:

$\operatorname{Log}(1) = 0$
$\operatorname{Log}(e) = 1$

Multiple value forms, for any k an integer:

$\log(1) = 0 + 2 \pi i k$
$\log(e) = 1 + 2 \pi i k$

### संकलन

Principal value forms:

$\operatorname{Log}(z_1) + \operatorname{Log}(z_2) = \operatorname{Log}(z_1 z_2) \pmod {2 \pi i}$
$\operatorname{Log}(z_1) - \operatorname{Log}(z_2) = \operatorname{Log}(z_1 / z_2) \pmod {2 \pi i}$

Multiple value forms:

$\log(z_1) + \log(z_2) = \log(z_1 z_2)$
$\log(z_1) - \log(z_2) = \log(z_1 / z_2)$

### घात

A complex power of a complex number can have many possible values.

Principal value form:

${z_1}^{z_2} = e^{z_2 \operatorname{Log}(z_1)}$
$\operatorname{Log}{\left({z_1}^{z_2}\right)} = z_2 \operatorname{Log}(z_1) \pmod {2 \pi i}$

Multiple value forms:

${z_1}^{z_2} = e^{z_2 \log(z_1)}$

Where k1, k2 are any integers:

$\log{\left({z_1}^{z_2}\right)} = z_2 \log(z_1) + 2 \pi i k_2$
$\log{\left({z_1}^{z_2}\right)} = z_2 \operatorname{Log}(z_1) + z_2 2 \pi i k_1 + 2 \pi i k_2$

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