How do you solve $\mathrm{ln}x+\mathrm{ln}(x-1)=1$ ?

sunshine022uv
2021-12-14
Answered

How do you solve $\mathrm{ln}x+\mathrm{ln}(x-1)=1$ ?

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asked 2022-06-27

Taking a logarithmic derivative of a function

I have the following expression:

$\mathrm{log}(1-\frac{r}{{r}_{s}})$

which I would like to take the following derivative of (and where ${r}_{s}$ is a constant):

$\frac{d(\mathrm{log}(1-\frac{r}{{r}_{s}}))}{d\mathrm{log}(r)}$

What kind of strategies could I employ to find a solution?

I have the following expression:

$\mathrm{log}(1-\frac{r}{{r}_{s}})$

which I would like to take the following derivative of (and where ${r}_{s}$ is a constant):

$\frac{d(\mathrm{log}(1-\frac{r}{{r}_{s}}))}{d\mathrm{log}(r)}$

What kind of strategies could I employ to find a solution?

asked 2022-06-18

Prove that $\frac{n-a}{n}<\frac{n+1-a}{n+1}$?

I have this math question that I'm kind of stuck on.

Prove that $\frac{n-a}{n}<\frac{n+1-a}{n+1}$

So far I have that: $\frac{n-a}{n}<\frac{n+1-a}{n+1}=n-a<\frac{n(n+1-a)}{n(n+1)}=n-\frac{n(n+1-a)}{n(n+1)}<a$

However, I'm not sure where to go from here. Thanks

I have this math question that I'm kind of stuck on.

Prove that $\frac{n-a}{n}<\frac{n+1-a}{n+1}$

So far I have that: $\frac{n-a}{n}<\frac{n+1-a}{n+1}=n-a<\frac{n(n+1-a)}{n(n+1)}=n-\frac{n(n+1-a)}{n(n+1)}<a$

However, I'm not sure where to go from here. Thanks

asked 2022-07-10

What's the formula to solve summation of logarithms?

I'm studying summation. Everything I know so far is that:

$\sum _{i=1}^{n}\text{}k=\frac{n(n+1)}{2}\text{}$

$\sum _{i=1}^{n}\text{}{k}^{2}=\frac{n(n+1)(2n+1)}{6}\text{}$

$\sum _{i=1}^{n}\text{}{k}^{3}=\frac{{n}^{2}(n+1{)}^{2}}{4}\text{}$

Unfortunately, I can't find neither on my book nor on the internet what the result of:

$\sum _{i=1}^{n}\mathrm{log}i$

$\sum _{i=1}^{n}\mathrm{ln}i$

is.

Can you help me out?

I'm studying summation. Everything I know so far is that:

$\sum _{i=1}^{n}\text{}k=\frac{n(n+1)}{2}\text{}$

$\sum _{i=1}^{n}\text{}{k}^{2}=\frac{n(n+1)(2n+1)}{6}\text{}$

$\sum _{i=1}^{n}\text{}{k}^{3}=\frac{{n}^{2}(n+1{)}^{2}}{4}\text{}$

Unfortunately, I can't find neither on my book nor on the internet what the result of:

$\sum _{i=1}^{n}\mathrm{log}i$

$\sum _{i=1}^{n}\mathrm{ln}i$

is.

Can you help me out?

asked 2022-04-28

Simplify expression using the properties of logarithms.

$3{\mathrm{log}}_{4}{r}^{2}-2{\mathrm{log}}_{4}r$

asked 2022-04-04

Express in terms of logarithms of x, y, z, or w.

${\mathrm{log}}_{4}\left(xz\right)$

asked 2022-06-15

Proof of generalization of a particular limit converging to ${e}^{\frac{1}{(p-1{)}^{2}}}$

I was reading a very old and long article on logarithms in a library it has pages turned yellow and had one pages titled - Tricky problems I managed to solve 5 out of the 6 but I couldn't do this 6th one . Prove that for all $p\in \mathbb{N}$

$\underset{n\to \mathrm{\infty}}{lim}\frac{({1}^{{1}^{p}}\times {2}^{{2}^{p}}\times {3}^{{3}^{p}}\times \dots {n}^{{n}^{p}})}{{n}^{\frac{1}{p+1}}}={e}^{\frac{1}{(p-1{)}^{2}}}$

I tried to manupilate this with the properties of logarithms but failed.

I was reading a very old and long article on logarithms in a library it has pages turned yellow and had one pages titled - Tricky problems I managed to solve 5 out of the 6 but I couldn't do this 6th one . Prove that for all $p\in \mathbb{N}$

$\underset{n\to \mathrm{\infty}}{lim}\frac{({1}^{{1}^{p}}\times {2}^{{2}^{p}}\times {3}^{{3}^{p}}\times \dots {n}^{{n}^{p}})}{{n}^{\frac{1}{p+1}}}={e}^{\frac{1}{(p-1{)}^{2}}}$

I tried to manupilate this with the properties of logarithms but failed.

asked 2022-05-15

The inverse function of a logarithm equation

I've tried many things with this question, and just can't seem to get it quite right, can someone please show me how to answer this question? Thank you in advance.

$g(x)=\mathrm{ln}(5x+25)\phantom{\rule{2em}{0ex}}{g}^{-1}(x)=\frac{\overline{){\displaystyle \phantom{X}}}}{\overline{){\displaystyle \phantom{X}}}}{e}^{x}\phantom{\rule{thinmathspace}{0ex}}\overline{){\displaystyle \phantom{XXX}}}$

I've tried many things with this question, and just can't seem to get it quite right, can someone please show me how to answer this question? Thank you in advance.

$g(x)=\mathrm{ln}(5x+25)\phantom{\rule{2em}{0ex}}{g}^{-1}(x)=\frac{\overline{){\displaystyle \phantom{X}}}}{\overline{){\displaystyle \phantom{X}}}}{e}^{x}\phantom{\rule{thinmathspace}{0ex}}\overline{){\displaystyle \phantom{XXX}}}$