What are logarithms?

Question

I have heard of logarithms, and done very little research at all. From that little bit of research I found out its in algebra 2. Sadly to say, I'm going into 9th grade, but yet I'm learning [calculus!?] and I don't know what a logarithm is! I find it in many places now. I deem it important to know what a logarithm is even though I'm jumping the gun in a sense. My understanding of concepts, is just like that of programming. In the mean time, you know its there, and your ITCHING SO HARD to find out what that is, but nope! For now we use it, tomorrow we learn what it does.

I just know that to identify a logarithm at my level, I just look for a log. :P

Answer 1

If you know what a power function is:

$$a^b=c$$

you can choose to solve for $a$ or $b$. If you want $a$, take $b$-th root on both sides:

$$a=\sqrt[b]{c}$$

Imagine $b=2$.

However, if you want to get $b$, you take the logarithm:

$$b=\log_{a}c$$

Here, I used the logarithm with a base $a$. Logarithms of different bases are related: they are simple multiples of each other. Common logarithms are $\log_{10}$ (the base is usually skipped), and the natural logarithm ($\log_e x=\ln x$) which is a very nicely-behaved function when you go further in calculus.

The numerical meaning of logarithm can be roughly understood as this: the whole part of the value of $\log_{10} x$ counts the number of digits in $x$. For instance

$$\log 1=0$$ $$\log 10=1$$ $$\log 100=2$$ $$\log 1000=3$$

and so on. Of course, you can evaluate something like $$\log 500=2.69...$$ $$\log 0.05=-1.30...$$

The rest of the properties follow from the definition that it inverts $a^b=c$. For instance, logarithm of a product can be split into sum of logarithms:

$$\log{ab}=\log a + \log b$$

Ultimately, it's just another elementary function, like roots, polynomials and so on.

Answer 2

Logarithms are the inverse to exponentiation. Simply put, $\log_a(k)$ is the solution to the equation

$$a^x=k$$

Here $a$ is called the "base" of the logarithm.

Obviously, $k\gt0$ and therefore, the logarithms of negative numbers are not defined, as long as we are dealing with the set of real numbers. Also $1^x=k$ will have no solutions for $k\ne1$, therefore $a$ must be not equal to $1$.

$(-1)^x$(or any other negative base) isn't well defined for non-integral $x$ and therefore $a\gt0$.

This sets some constraints on the domain of $\log$

• $a\ne1$
• $k\gt0$
• $a\gt0$

$\log_e(x)$ is called the natural logarithm, sometimes abbreviated as $\ln(x)$ or simply $\log(x)$

Answer 3

• a ++ $=$ ++ a, so incrementation has a single inverse operation, namely decrementation.

• Repeated incrementation is addition.

• $a+b=b+a$, so addition has a single inverse operation, namely subtraction.

• Repeated addition is multiplication.

• $a\cdot b=b\cdot a$, so multiplication has a single inverse operation, namely division.

• Repeated multiplication is exponentiation.

• But $a^b\neq b^a$ $\big($usually$\big)$, so exponentiation has not just one, but two inverse operations: root extraction $\big($when the exponent is known, and we want to find out the value of the base$\big)$, and logarithms $\big($for the opposite case: namely when the base is known, and we want to find out the value of the exponent$\big)$.

Answer 4

The function whose values at $x$ are $c\log{x}$ where $c$ is a constant is the most general function that satisfies, for all positive $x$ and $y$, the identity

$$h(x)+h(y)=h(xy)$$

Let $x\to e^x$ and $y\to e^y$, then we have $h(e^x)+h(e^y)=h(e^xe^y)=h(e^{x+y})$

Let $H(u)=h(e^u)$, then we have $H(x)+H(y)=H(x+y)$.

Let $x=y$, then we have $2H(x)=H(2x)\implies H(x)=\frac{1}{2}H'(2x)$.

Differentiating gives $H'(x)=H'(2x)$.

Repeated differentiation gives the general formula, $H^{(n)}(x)=2^{n-1}H^{(n)}(2x)$ for $n\ge0$.

Let $x=0$, then we have $(2^{n-1}-1)H^{(n)}(0)=0$

So $H^{(n)}(0)=0$ for all integers $n\ne 1$.

By considering the Maclaurin series for $f(x)$, we find that the most general function satisfying $H(x)+H(y)=H(x+y)$ is $cx$ where $c$ is a constant.

So $h(e^x)=cx$. Letting $x\to \log{x}$ we find the most general function satisfying the original function equation is $c\log{x}$.

Answer 5

It's funny how things come full circle...

Logarithms arise as a natural order within the subset of all numbers as the inverse of exponentiation. If exponentiation is repeated multiplication, then logarithms are repeated division. When this type of number first appeared, it went by the name of logarithmus, literally "ratio-number", and was invented by John Napier and Henry Briggs in the late 16th to early 17th century AD.

From a natural perspective, logarithms describe a pattern of growth in general, or can describe a scale such as pH.

Finally, from the perspective of the representations of quantities themselves, logarithms directly describe the nature of a particular representation of quantities of a given base. Each number can be represented in any base as the sum of unique powers of bases. Base two, the base commonly used to represent quantities in computers, is the sum of $n$ digits which correspond to the sum, $2^0 + 2^1 + 2^2 +...+ 2^n = \sum_{k=0}^{n}{2^n}$, for example; however, this can be generalized to all representations of numbers in any base with base $b$ and digital length $n$ as

$$\sum_{k=0}^{n}{b^k}$$