If $f\left(x\right)={\mathrm{log}}_{a}x$ , show that $\frac{f(x+h)-f\left(x\right)}{h}={\mathrm{log}}_{a}{(1+\frac{h}{x})}^{\frac{1}{h}},h\ne c0$

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2021-10-19
Answered

If $f\left(x\right)={\mathrm{log}}_{a}x$ , show that $\frac{f(x+h)-f\left(x\right)}{h}={\mathrm{log}}_{a}{(1+\frac{h}{x})}^{\frac{1}{h}},h\ne c0$

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Arham Warner

Answered 2021-10-20
Author has **102** answers

Step 1

The difference quotient$\frac{f(x+h)-f\left(x\right)}{h}$ can be used to calculate the first derivative of the function by taking the limit $h\to 0$ . It will give f'(x) if it exists.

For the given problem, two properties of logarithms will be used. The first is the difference property, which is$\mathrm{log}}_{a}x-{\mathrm{log}}_{a}y={\mathrm{log}}_{a}\frac{x}{y$ . The second one is the exponent property of logarithms, which is $y{\mathrm{log}}_{a}x={\mathrm{log}}_{a}{x}^{y}$

Step 2

Substitute$f\left(x\right)={\mathrm{log}}_{a}x$ in the difference quotient $\frac{f(x+h)-f\left(x\right)}{h}$ and simplify using the properties of logarithms.

$\frac{f(x+h)-f\left(x\right)}{h}=\frac{{\mathrm{log}}_{a}(x+h)-{\mathrm{log}}_{a}x}{h}=$

$=\frac{{\mathrm{log}}_{a}\frac{x+h}{x}}{h}$

$=\frac{1}{h}{\mathrm{log}}_{a}(1+\frac{h}{x})$

$={\mathrm{log}}_{a}{(1+\frac{h}{x})}^{\frac{1}{h}}$

It is defined for$h\ne 0\text{}\text{as}\text{}\frac{1}{h}$ is not defined.

The difference quotient

For the given problem, two properties of logarithms will be used. The first is the difference property, which is

Step 2

Substitute

It is defined for

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I am trying to understand the math behind Logistic regression. I am confused about transposing one formula to another.

Here is what I have: Our regression formula

Our sigmoid function

Our logistic function

From what I understand, we have to solve for y using the sigmoid function. I did this:

1.

2.

3.

4.

5.

6.

6 is y for our logistic function, but I don't see how can we get to our logistic function y

Can anyone help me transpose this or point me in the right direction?

If anyone downvotes me, please explain why in a comment.

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