Properties of logarithm and exponent. Natural logarithm and number e. Expressions through complex numbers

Not bad at all, right? While mathematicians search for words to give you a long, confusing definition, let's take a closer look at this simple and clear one.

The number e means growth

The number e means continuous growth. As we saw in the previous example, e x allows us to link interest and time: 3 years at 100% growth is the same as 1 year at 300%, assuming "compound interest".

You can substitute any percentage and time values (50% for 4 years), but it is better to set the percentage as 100% for convenience (it turns out 100% for 2 years). By moving to 100%, we can focus solely on the time component:

e x = e percent * time = e 1.0 * time = e time

Obviously e x means:

how much will my contribution grow after x units of time (assuming 100% continuous growth).
for example, after 3 time intervals I will receive e 3 = 20.08 times more “things”.

e x is a scaling factor that shows what level we will grow to in x amount of time.

Natural logarithm means time

The natural logarithm is the inverse of e, a fancy term for opposite. Speaking of quirks; in Latin it is called logarithmus naturali, hence the abbreviation ln.

And what does this inversion or opposite mean?

e x allows us to substitute time and get growth.
ln(x) allows us to take growth or income and find out the time it takes to generate it.

For example:

e 3 equals 20.08. After three periods of time, we will have 20.08 times more than what we started with.
ln(08/20) would be approximately 3. If you are interested in growth of 20.08 times, you will need 3 time periods (again, assuming 100% continuous growth).

Still reading? The natural logarithm shows the time required to reach the desired level.

This non-standard logarithmic counting

Have you gone through logarithms - they are strange creatures. How did they manage to turn multiplication into addition? What about division into subtraction? Let's get a look.

What is ln(1) equal to? Intuitively, the question is: how long should I wait to get 1x more than what I have?

Zero. Zero. Not at all. You already have it once. It doesn't take much time to go from level 1 to level 1.

log(1) = 0

Okay, what about the fractional value? How long will it take for us to have 1/2 of the available quantity left? We know that with 100% continuous growth, ln(2) means the time it takes to double. If we let's turn back time(i.e., wait a negative amount of time), then we will get half of what we have.

ln(1/2) = -ln(2) = -0.693

Logical, right? If we go back (time back) to 0.693 seconds, we will find half the amount available. In general, you can turn the fraction over and take a negative value: ln(1/3) = -ln(3) = -1.09. This means that if we go back in time to 1.09 times, we will only find a third of the current number.

Okay, what about the logarithm of a negative number? How long does it take to “grow” a colony of bacteria from 1 to -3?

This is impossible! You can't get a negative bacteria count, can you? You can get a maximum (er...minimum) of zero, but there's no way you can get a negative number from these little critters. A negative bacteria count simply doesn't make sense.

ln(negative number) = undefined

"Undefined" means that there is no amount of time that would have to wait to get a negative value.

Logarithmic multiplication is just hilarious

How long will it take to grow fourfold? Of course, you can just take ln(4). But this is too simple, we will go the other way.

You can think of quadruple growth as doubling (requiring ln(2) units of time) and then doubling again (requiring another ln(2) units of time):

Time to grow 4 times = ln(4) = Time to double and then double again = ln(2) + ln(2)

Interesting. Any growth rate, say 20, can be considered a doubling right after a 10x increase. Or growth by 4 times, and then by 5 times. Or tripling and then increasing by 6.666 times. See the pattern?

ln(a*b) = ln(a) + ln(b)

The logarithm of A times B is log(A) + log(B). This relationship immediately makes sense when viewed in terms of growth.

If you are interested in 30x growth, you can wait ln(30) in one sitting, or wait ln(3) for tripling, and then another ln(10) for 10x. The end result is the same, so of course the time must remain constant (and it does).

What about division? Specifically, ln(5/3) means: how long will it take to grow 5 times and then get 1/3 of that?

Great, growth by 5 times is ln(5). An increase of 1/3 times will take -ln(3) units of time. So,

ln(5/3) = ln(5) – ln(3)

This means: let it grow 5 times, and then “go back in time” to the point where only a third of that amount remains, so you get 5/3 growth. In general it turns out

ln(a/b) = ln(a) – ln(b)

I hope that the strange arithmetic of logarithms is starting to make sense to you: multiplying growth rates becomes adding growth time units, and dividing becomes subtracting time units. No need to memorize the rules, try to understand them.

Using the natural logarithm for arbitrary growth

Well, of course,” you say, “this is all good if the growth is 100%, but what about the 5% that I get?”

No problem. The "time" we calculate with ln() is actually a combination of interest rate and time, the same X from the e x equation. We just decided to set the percentage to 100% for simplicity, but we are free to use any numbers.

Let's say we want to achieve 30x growth: take ln(30) and get 3.4 This means:

e x = height
e 3.4 = 30

Obviously, this equation means "100% return over 3.4 years gives 30x growth." We can write this equation as follows:

e x = e rate*time
e 100% * 3.4 years = 30

We can change the values of “bet” and “time”, as long as the bet * time remains 3.4. For example, if we are interested in 30x growth, how long will we have to wait at an interest rate of 5%?

ln(30) = 3.4
rate * time = 3.4
0.05 * time = 3.4
time = 3.4 / 0.05 = 68 years

I reason like this: "ln(30) = 3.4, so at 100% growth it will take 3.4 years. If I double the growth rate, the time required will be halved."

100% for 3.4 years = 1.0 * 3.4 = 3.4
200% in 1.7 years = 2.0 * 1.7 = 3.4
50% for 6.8 years = 0.5 * 6.8 = 3.4
5% over 68 years = .05 * 68 = 3.4.

Great, right? The natural logarithm can be used with any interest rate and time because their product remains constant. You can move variable values as much as you like.

Cool example: Rule of seventy-two

The Rule of Seventy-Two is a mathematical technique that allows you to estimate how long it will take for your money to double. Now we will deduce it (yes!), and moreover, we will try to understand its essence.

How long will it take to double your money at 100% interest compounded annually?

Oops. We used the natural logarithm for the case of continuous growth, and now you are talking about annual compounding? Wouldn't this formula become unsuitable for such a case? Yes, it will, but for real interest rates like 5%, 6% or even 15%, the difference between annual compounding and continuous growth will be small. So the rough estimate works, um, roughly, so we'll pretend that we have a completely continuous accrual.

Now the question is simple: How quickly can you double with 100% growth? ln(2) = 0.693. It takes 0.693 units of time (years in our case) to double our amount with a continuous increase of 100%.

So, what if the interest rate is not 100%, but say 5% or 10%?

Easily! Since bet * time = 0.693, we will double the amount:

rate * time = 0.693
time = 0.693 / bet

It turns out that if the growth is 10%, it will take 0.693 / 0.10 = 6.93 years to double.

To simplify the calculations, let's multiply both sides by 100, then we can say "10" rather than "0.10":

time to double = 69.3 / bet, where the bet is expressed as a percentage.

Now it’s time to double at a rate of 5%, 69.3 / 5 = 13.86 years. However, 69.3 is not the most convenient dividend. Let's choose a close number, 72, which is convenient to divide by 2, 3, 4, 6, 8 and other numbers.

time to double = 72 / bet

which is the rule of seventy-two. Everything is covered.

If you need to find the time to triple, you can use ln(3) ~ 109.8 and get

time to triple = 110 / bet

Which is another useful rule. The "Rule of 72" applies to growth in interest rates, population growth, bacterial cultures, and anything that grows exponentially.

What's next?

Hopefully the natural logarithm now makes sense to you - it shows the time it takes for any number to grow exponentially. I think it's called natural because e is a universal measure of growth, so ln can be considered a universal way of determining how long it takes to grow.

Every time you see ln(x), remember "the time it takes to grow X times". In an upcoming article I will describe e and ln in conjunction so that the fresh scent of mathematics will fill the air.

Addendum: Natural logarithm of e

Quick quiz: what is ln(e)?

a math robot will say: since they are defined as the inverse of one another, it is obvious that ln(e) = 1.
understanding person: ln(e) is the number of times it takes to grow "e" times (about 2.718). However, the number e itself is a measure of growth by a factor of 1, so ln(e) = 1.

Think clearly.

September 9, 2013

Lesson and presentation on the topics: "Natural logarithms. The base of the natural logarithm. The logarithm of a natural number"

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What is natural logarithm

Guys, in the last lesson we learned a new, special number - e. Today we will continue to work with this number.
We have studied logarithms and we know that the base of a logarithm can be many numbers that are greater than 0. Today we will also look at a logarithm whose base is the number e. Such a logarithm is usually called the natural logarithm. It has its own notation: $\ln(n)$ is the natural logarithm. This entry is equivalent to the entry: $\log_e(n)=\ln(n)$.
Exponential and logarithmic functions are inverses, then the natural logarithm is the inverse of the function: $y=e^x$.
Inverse functions are symmetric with respect to the straight line $y=x$.
Let's plot the natural logarithm by plotting the exponential function with respect to the straight line $y=x$.

It is worth noting that the angle of inclination of the tangent to the graph of the function $y=e^x$ at point (0;1) is 45°. Then the angle of inclination of the tangent to the graph of the natural logarithm at point (1;0) will also be equal to 45°. Both of these tangents will be parallel to the line $y=x$. Let's diagram the tangents:

Properties of the function $y=\ln(x)$

1. $D(f)=(0;+∞)$.
2. Is neither even nor odd.
3. Increases throughout the entire domain of definition.
4. Not limited from above, not limited from below.
5. There is no greatest value, no minimum value.
6. Continuous.
7. $E(f)=(-∞; +∞)$.
8. Convex upward.
9. Differentiable everywhere.

In the course of higher mathematics it is proven that the derivative of an inverse function is the inverse of the derivative of a given function.
There is not much point in going into the proof, let's just write the formula: $y"=(\ln(x))"=\frac(1)(x)$.

Example.
Calculate the value of the derivative of the function: $y=\ln(2x-7)$ at the point $x=4$.
Solution.
In general, our function is represented by the function $y=f(kx+m)$; we can calculate the derivatives of such functions.
$y"=(\ln((2x-7)))"=\frac(2)((2x-7))$.
Let's calculate the value of the derivative at the required point: $y"(4)=\frac(2)((2*4-7))=2$.
Answer: 2.

Example.
Draw a tangent to the graph of the function $y=ln(x)$ at the point $х=е$.
Solution.
We remember well the equation of the tangent to the graph of a function at the point $x=a$.
$y=f(a)+f"(a)(x-a)$.
We sequentially calculate the required values.
$a=e$.
$f(a)=f(e)=\ln(e)=1$.
$f"(a)=\frac(1)(a)=\frac(1)(e)$.
$y=1+\frac(1)(e)(x-e)=1+\frac(x)(e)-\frac(e)(e)=\frac(x)(e)$.
The tangent equation at the point $x=e$ is the function $y=\frac(x)(e)$.
Let's plot the natural logarithm and the tangent line.

Example.
Examine the function for monotonicity and extrema: $y=x^6-6*ln(x)$.
Solution.
The domain of definition of the function $D(y)=(0;+∞)$.
Let's find the derivative of the given function:
$y"=6*x^5-\frac(6)(x)$.
The derivative exists for all x from the domain of definition, then there are no critical points. Let's find stationary points:
$6*x^5-\frac(6)(x)=0$.
$\frac(6*x^6-6)(x)=0$.
$6*x^6-6=0$.
$x^6-1=0$.
$x^6=1$.
$x=±1$.
The point $х=-1$ does not belong to the domain of definition. Then we have one stationary point $x=1$. Let's find the intervals of increasing and decreasing:

Point $x=1$ is the minimum point, then $y_min=1-6*\ln(1)=1$.
Answer: The function decreases on the segment (0;1], the function increases on the ray $)