Why is the linearization of any natural log function always going to be x?

To understand why the linearization of any natural log function is always going to be x, let's start with the definition of linearization.

The linearization of a function f(x) at a point a is a linear function L(x) that approximates the behavior of f(x) near the point a. Mathematically, it can be represented as:

L(x) = f(a) + f'(a)(x - a)

Here, f'(a) denotes the derivative of the function f(x) evaluated at the point a.

Now, let's consider the natural logarithm function, denoted as ln(x). The derivative of ln(x) is given by:

d/dx [ln(x)] = 1/x

To find the linearization of ln(x), let's choose a specific point a. Let's say we choose a = 1. Substituting this value into the formula for the linearization, we get:

L(x) = ln(1) + (1/1)(x - 1)
= 0 + x - 1
= x - 1

Therefore, the linearization of ln(x) at the point a = 1 is L(x) = x - 1.

Now, let's generalize this result. Since the derivative of ln(x) is 1/x, at any point a, the linearization of ln(x) can be written as:

L(x) = ln(a) + (1/a)(x - a)

Notice that the constant term ln(a) depends on the choice of the point a, while the coefficient of x, (1/a), is always a constant. This means that the linearization of ln(x) is always in the form of a linear equation where the coefficient of x is 1. Hence, the linearization of any natural log function is always going to be x.

In summary, the linearization of any natural log function is x because its linear approximation can be represented by a linear equation with a coefficient of x equal to 1.