Lesson and presentation on the topics: "Natural logarithms. Base of a natural logarithm. 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 the logarithm can be a set of numbers that are greater than 0. Today we will also consider the logarithm, which is based on the number e. Such a logarithm is usually called the natural logarithm. It has its own notation: $\ln(n)$ is the natural logarithm. This notation is equivalent to: $\log_e(n)=\ln(n)$.
The exponential and logarithmic functions are inverse, 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 slope of the tangent to the graph of the function $y=e^x$ at the point (0;1) is 45°. Then the slope of the tangent to the graph of the natural logarithm at the point (1; 0) will also be equal to 45°. Both of these tangents will be parallel to the line $y=x$. Let's sketch the tangents:

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

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

In the course of higher mathematics it is proved that the derivative of an inverse function is the reciprocal of the derivative of the given function.
It doesn't make much sense to delve 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 $x=e$.
Solution.
The equation of the tangent to the graph of the function, at the point $x=a$, we remember well.
$y=f(a)+f"(a)(x-a)$.
Let us 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.

Example.
Investigate the function for monotonicity and extrema: $y=x^6-6*ln(x)$.
Solution.
Domain of the function $D(y)=(0;+∞)$.
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 $х=1$. Find the intervals of increase and decrease:

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