I can see that $y={e}^{-x}$ and $z=\mathrm{ln}(1-y)$. And, so $z=\mathrm{ln}(1-{e}^{-x})$. But how to go further?

Meossi91
2022-08-11
Answered

If $y=1-x+\frac{{x}^{2}}{2!}-\frac{{x}^{3}}{3!}+\dots $ and $z=-y-\frac{{y}^{2}}{2}-\frac{{y}^{3}}{3}-\dots $ then $\mathrm{ln}(\frac{1}{1-{e}^{x}})$

I can see that $y={e}^{-x}$ and $z=\mathrm{ln}(1-y)$. And, so $z=\mathrm{ln}(1-{e}^{-x})$. But how to go further?

I can see that $y={e}^{-x}$ and $z=\mathrm{ln}(1-y)$. And, so $z=\mathrm{ln}(1-{e}^{-x})$. But how to go further?

You can still ask an expert for help

Ben Logan

Answered 2022-08-12
Author has **13** answers

Once you know that $z=\mathrm{log}(1-{e}^{-x})$, you immediately have that

$\mathrm{log}\left(\frac{1}{1-{e}^{z}}\right)=\mathrm{log}\left(\frac{1}{1-{e}^{\mathrm{log}(1-{e}^{-x})}}\right)=\mathrm{log}\left(\frac{1}{1-(1-{e}^{-x})}\right)=\mathrm{log}\left(\frac{1}{{e}^{-x}}\right)=\mathrm{log}{e}^{x}=x$

And you've done.

$\mathrm{log}\left(\frac{1}{1-{e}^{z}}\right)=\mathrm{log}\left(\frac{1}{1-{e}^{\mathrm{log}(1-{e}^{-x})}}\right)=\mathrm{log}\left(\frac{1}{1-(1-{e}^{-x})}\right)=\mathrm{log}\left(\frac{1}{{e}^{-x}}\right)=\mathrm{log}{e}^{x}=x$

And you've done.

asked 2022-04-27

If $x={\mathrm{log}}_{12}27$ ,then what is the the value of $\mathrm{log}}_{616$ ?

asked 2022-09-19

Upper bound of natural logarithm

I was playing looking for a good upper bound of natural logarithm and I found that

$\mathrm{ln}x\le {x}^{1/e}$

apparently works: Can someone give me a formal proof of this inequality?

I was playing looking for a good upper bound of natural logarithm and I found that

$\mathrm{ln}x\le {x}^{1/e}$

apparently works: Can someone give me a formal proof of this inequality?

asked 2022-09-18

Homework help to rearrange formula

Given the equation

${V}_{m}=u(\mathrm{ln}{m}_{0}-\mathrm{ln}{m}_{8})-g{t}_{f}$

I need to solve for ${m}_{0}$ Here is what I have but it looks messy and I feel like there is sometihng wrong or a better way

1st attempt

$\begin{array}{rl}& {V}_{m}=u(\mathrm{ln}{m}_{0}-\mathrm{ln}{m}_{8})-g{t}_{f}\\ & {V}_{m}=u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}\\ & {V}_{m}=u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}\\ & {V}_{m}=u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}\\ & u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}-{V}_{m}=0\\ & u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-{V}_{m}=g{t}_{f}\\ & u(\mathrm{ln}{m}_{0})=g{t}_{f}+u(\mathrm{ln}{m}_{8})+{V}_{m}\\ & \mathrm{ln}{m}_{0}=(g{t}_{f}+u(\mathrm{ln}{m}_{8})+{V}_{m})\xf7u\\ & {e}^{(g{t}_{f}+u(\mathrm{ln}{m}_{8})+{V}_{m})\xf7u}={m}_{0}\end{array}$

2nd attempt - think this looks a little better but still not there yet

$\begin{array}{rl}& {V}_{m}=u(\mathrm{ln}\frac{{m}_{0}}{{m}_{8}})-g{t}_{f}\\ & {V}_{m}+g{t}_{f}=u(\mathrm{ln}\frac{{m}_{0}}{{m}_{8}})\\ & \frac{{V}_{m}+g{t}_{f}}{u}=\mathrm{ln}\frac{{m}_{0}}{{m}_{8}}\\ & {e}^{\frac{{V}_{m}+g{t}_{f}}{u}}=\frac{{m}_{0}}{{m}_{8}}\\ & {m}_{8}{e}^{\frac{{V}_{m}+g{t}_{f}}{u}}={m}_{0}\end{array}$

Given the equation

${V}_{m}=u(\mathrm{ln}{m}_{0}-\mathrm{ln}{m}_{8})-g{t}_{f}$

I need to solve for ${m}_{0}$ Here is what I have but it looks messy and I feel like there is sometihng wrong or a better way

1st attempt

$\begin{array}{rl}& {V}_{m}=u(\mathrm{ln}{m}_{0}-\mathrm{ln}{m}_{8})-g{t}_{f}\\ & {V}_{m}=u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}\\ & {V}_{m}=u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}\\ & {V}_{m}=u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}\\ & u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-g{t}_{f}-{V}_{m}=0\\ & u(\mathrm{ln}{m}_{0})-u(\mathrm{ln}{m}_{8})-{V}_{m}=g{t}_{f}\\ & u(\mathrm{ln}{m}_{0})=g{t}_{f}+u(\mathrm{ln}{m}_{8})+{V}_{m}\\ & \mathrm{ln}{m}_{0}=(g{t}_{f}+u(\mathrm{ln}{m}_{8})+{V}_{m})\xf7u\\ & {e}^{(g{t}_{f}+u(\mathrm{ln}{m}_{8})+{V}_{m})\xf7u}={m}_{0}\end{array}$

2nd attempt - think this looks a little better but still not there yet

$\begin{array}{rl}& {V}_{m}=u(\mathrm{ln}\frac{{m}_{0}}{{m}_{8}})-g{t}_{f}\\ & {V}_{m}+g{t}_{f}=u(\mathrm{ln}\frac{{m}_{0}}{{m}_{8}})\\ & \frac{{V}_{m}+g{t}_{f}}{u}=\mathrm{ln}\frac{{m}_{0}}{{m}_{8}}\\ & {e}^{\frac{{V}_{m}+g{t}_{f}}{u}}=\frac{{m}_{0}}{{m}_{8}}\\ & {m}_{8}{e}^{\frac{{V}_{m}+g{t}_{f}}{u}}={m}_{0}\end{array}$

asked 2022-03-22

Logistic regression - Transposing formulas

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

$y={b}_{0}+{b}_{1}x$

Our sigmoid function

$p=\frac{1}{1+{e}^{-y}}$

Our logistic function

$\mathrm{ln}\left(\frac{p}{1-p}\right)={b}_{0}+{b}_{1}x$

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

1.$p=\frac{1}{1+{e}^{-y}}$

2.$\frac{1}{p}=1+{e}^{-y}$

3.$\frac{1}{p}-1={e}^{-y}$

4.$\mathrm{ln}(\frac{1}{p}-1)=\mathrm{ln}\left({e}^{-y}\right)$

5.$\mathrm{ln}(\frac{1}{p}-1)=-y$

6.$-\mathrm{ln}(\frac{1}{p}-1)=y$

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

$\mathrm{ln}\left(\frac{p}{1-p}\right)$

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

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

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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Find x

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How to solve for negative numbers in logarithmic equations

I am trying to solve the equation

Taking

But I clearly missed out

How do I solve this equation algebraically?

asked 2022-07-15

Logarithm question-I donot know but this question may be solved by any other way also.

Let $({x}_{0},{y}_{0})$ be the solution of the following equations.

$(2x{)}^{\mathrm{ln}2}=(3y{)}^{\mathrm{ln}3}$

${3}^{\mathrm{ln}x}={2}^{\mathrm{ln}y}$

Then ${x}_{0}$ is

A) $\frac{1}{6}$

B) $\frac{1}{3}$

C) $\frac{1}{2}$

D) $6$

I have tried this problem by taking log on both sides of the two equations. But, finally I could not make up to get the values of $x$ and $y$

Let $({x}_{0},{y}_{0})$ be the solution of the following equations.

$(2x{)}^{\mathrm{ln}2}=(3y{)}^{\mathrm{ln}3}$

${3}^{\mathrm{ln}x}={2}^{\mathrm{ln}y}$

Then ${x}_{0}$ is

A) $\frac{1}{6}$

B) $\frac{1}{3}$

C) $\frac{1}{2}$

D) $6$

I have tried this problem by taking log on both sides of the two equations. But, finally I could not make up to get the values of $x$ and $y$