Alessandra Carrillo
2022-04-11
Answered

Use the same-derivative argument, as was done to prove the Productand Power Rules for logarithms, to prove the Quotient Ruleproperty.

You can still ask an expert for help

Jax Burns

Answered 2022-04-12
Author has **13** answers

The objective is to prove the property $\mathrm{ln}\left(\frac{b}{x}\right)=\mathrm{ln}b-\mathrm{ln}x$

Differential $\mathrm{ln}\left(\frac{b}{x}\right)$

$\frac{d}{dx}\left(\mathrm{ln}\left(\frac{b}{x}\right)\right)=\frac{1}{\frac{d}{x}}\frac{d}{dx}\left(\frac{b}{x}\right)$

$=\frac{x}{b}(-\frac{b}{{x}^{2}})$

$=\frac{-1}{x}$

Integrate both the sides of the equation with respect to x as,

$\mathrm{ln}\frac{b}{x}=(-\mathrm{ln}1)+c$

$c=\mathrm{ln}b$

Substitute $c=\mathrm{ln}b$ in the equation $\mathrm{ln}\left(\frac{b}{x}\right)=(-\mathrm{ln}x)+c$ as,

$\mathrm{ln}\frac{b}{x}=(-\mathrm{ln}x)+\mathrm{ln}b$

$\mathrm{ln}\left(\frac{b}{x}\right)=\mathrm{ln}b-\mathrm{ln}x$

Hence, Proved

asked 2022-07-23

Can this log question be simplified?

${2}^{lo{g}_{3}5}-{5}^{lo{g}_{3}2}.$ I don't know any formula that can apply to it or is there a formula?? Even a hint will be helpful.

${2}^{lo{g}_{3}5}-{5}^{lo{g}_{3}2}.$ I don't know any formula that can apply to it or is there a formula?? Even a hint will be helpful.

asked 2022-08-11

help with logarithmic integration.

I've been googling some tutorials on integrating logarithms for my calc 2 class and I've found a lot of good stuff. Unfortunately nothing has answered how to handle a problem that I have. I've tried u-substitution and have still been getting it wrong or just lost. My problem is the $\int (\frac{9}{{x}^{3}}+\frac{2}{9x})dx$ (sorry, I'm a little new to the math section and don't know how to format that nicely). So far I've tried finding a common denominator which changes the equation to $\frac{81+2{x}^{2}}{9{x}^{3}}$. I tried doing $u=9{x}^{3}$ and $du=27{x}^{2}dx$ or $\frac{1}{27}du={x}^{2}dx$. Now the problem is kind of ugly and I'm not sure what to do with that next. We haven't covered integration by parts yet in this class and I've found that comes up, so I'm not sure if it's possible to evaluate without that technique. I really appreciate any help!

Thank you in advance

I've been googling some tutorials on integrating logarithms for my calc 2 class and I've found a lot of good stuff. Unfortunately nothing has answered how to handle a problem that I have. I've tried u-substitution and have still been getting it wrong or just lost. My problem is the $\int (\frac{9}{{x}^{3}}+\frac{2}{9x})dx$ (sorry, I'm a little new to the math section and don't know how to format that nicely). So far I've tried finding a common denominator which changes the equation to $\frac{81+2{x}^{2}}{9{x}^{3}}$. I tried doing $u=9{x}^{3}$ and $du=27{x}^{2}dx$ or $\frac{1}{27}du={x}^{2}dx$. Now the problem is kind of ugly and I'm not sure what to do with that next. We haven't covered integration by parts yet in this class and I've found that comes up, so I'm not sure if it's possible to evaluate without that technique. I really appreciate any help!

Thank you in advance

asked 2022-06-22

How do I solve this integral?

As stated the title, I get to a point which I can't do anything, and I'm sure I've made a mistake some where, here is my full working out:

$\int {e}^{ix}\mathrm{cos}(x)dx\phantom{\rule{0ex}{0ex}}u={e}^{ix}\text{|}{u}^{\prime}=i{e}^{ie}\phantom{\rule{0ex}{0ex}}v=\mathrm{sin}(x)\text{|}{v}^{\prime}=\mathrm{cos}(x)\phantom{\rule{0ex}{0ex}}{e}^{ix}\mathrm{sin}(x)-\int i{e}^{ix}\mathrm{sin}(x)dx+C\phantom{\rule{0ex}{0ex}}{e}^{ix}\mathrm{sin}(x)-i\int {e}^{ix}\mathrm{sin}(x)dx+C\phantom{\rule{0ex}{0ex}}u={e}^{ix}\text{|}{u}^{\prime}=i{e}^{ie}\phantom{\rule{0ex}{0ex}}v=-\mathrm{cos}(x)\text{|}{v}^{\prime}=\mathrm{sin}(x)\phantom{\rule{0ex}{0ex}}{e}^{ix}\mathrm{sin}(x)-i(-{e}^{ix}\mathrm{cos}(x)+i\int {e}^{ix}cos(x)dx)+C\phantom{\rule{0ex}{0ex}}Let\int {e}^{ix}\mathrm{cos}(x)dx=I\phantom{\rule{0ex}{0ex}}I={e}^{ix}(\mathrm{sin}(x)+i\mathrm{cos}(x))-{i}^{2}I+C$

But (${i}^{2}=-1$) so the equation should become:

$I={e}^{ix}(\mathrm{sin}(x)+i\mathrm{cos}(x))+I+C$

And this is where I'm stuck, I can't simply take $I$ away from both sides, that would make

${e}^{ix}(\mathrm{sin}(x)+i\mathrm{cos}(x))=0$

What have a messed up in the process? And just to make it clear someone in a previous question didn't under understand what ${v}^{\prime}$ was, it's the same as $\frac{dv}{dx}$, thank you in advance.

As stated the title, I get to a point which I can't do anything, and I'm sure I've made a mistake some where, here is my full working out:

$\int {e}^{ix}\mathrm{cos}(x)dx\phantom{\rule{0ex}{0ex}}u={e}^{ix}\text{|}{u}^{\prime}=i{e}^{ie}\phantom{\rule{0ex}{0ex}}v=\mathrm{sin}(x)\text{|}{v}^{\prime}=\mathrm{cos}(x)\phantom{\rule{0ex}{0ex}}{e}^{ix}\mathrm{sin}(x)-\int i{e}^{ix}\mathrm{sin}(x)dx+C\phantom{\rule{0ex}{0ex}}{e}^{ix}\mathrm{sin}(x)-i\int {e}^{ix}\mathrm{sin}(x)dx+C\phantom{\rule{0ex}{0ex}}u={e}^{ix}\text{|}{u}^{\prime}=i{e}^{ie}\phantom{\rule{0ex}{0ex}}v=-\mathrm{cos}(x)\text{|}{v}^{\prime}=\mathrm{sin}(x)\phantom{\rule{0ex}{0ex}}{e}^{ix}\mathrm{sin}(x)-i(-{e}^{ix}\mathrm{cos}(x)+i\int {e}^{ix}cos(x)dx)+C\phantom{\rule{0ex}{0ex}}Let\int {e}^{ix}\mathrm{cos}(x)dx=I\phantom{\rule{0ex}{0ex}}I={e}^{ix}(\mathrm{sin}(x)+i\mathrm{cos}(x))-{i}^{2}I+C$

But (${i}^{2}=-1$) so the equation should become:

$I={e}^{ix}(\mathrm{sin}(x)+i\mathrm{cos}(x))+I+C$

And this is where I'm stuck, I can't simply take $I$ away from both sides, that would make

${e}^{ix}(\mathrm{sin}(x)+i\mathrm{cos}(x))=0$

What have a messed up in the process? And just to make it clear someone in a previous question didn't under understand what ${v}^{\prime}$ was, it's the same as $\frac{dv}{dx}$, thank you in advance.

asked 2022-06-13

Solving for x in a equation involving natural logarithms

How would I solve for x in this equation here:

$\mathrm{ln}(x)+\mathrm{ln}(1/x+1)=3$

I realize that the answer is ${e}^{3}-1$ , but I am not sure as to how to get it. Any input is appreciated.

How would I solve for x in this equation here:

$\mathrm{ln}(x)+\mathrm{ln}(1/x+1)=3$

I realize that the answer is ${e}^{3}-1$ , but I am not sure as to how to get it. Any input is appreciated.

asked 2022-07-12

Solve ${2}^{x}={x}^{2}$

I've been asked to solve this and I've tried a few things but I have trouble eliminating $x$. I first tried taking the natural log:

$x\mathrm{ln}\left(2\right)=2\mathrm{ln}\left(x\right)$

$\frac{\mathrm{ln}\left(2\right)}{2}}={\displaystyle \frac{\mathrm{ln}\left(x\right)}{x}$

I don't know what to do from here so I decided to try another method:

${2}^{x}={2}^{{\mathrm{log}}_{2}\left({x}^{2}\right)}$

$x={\mathrm{log}}_{2}\left({x}^{2}\right)$

And then I get stuck here, I'm all out of ideas. My guess is I've overlooked something simple…

I've been asked to solve this and I've tried a few things but I have trouble eliminating $x$. I first tried taking the natural log:

$x\mathrm{ln}\left(2\right)=2\mathrm{ln}\left(x\right)$

$\frac{\mathrm{ln}\left(2\right)}{2}}={\displaystyle \frac{\mathrm{ln}\left(x\right)}{x}$

I don't know what to do from here so I decided to try another method:

${2}^{x}={2}^{{\mathrm{log}}_{2}\left({x}^{2}\right)}$

$x={\mathrm{log}}_{2}\left({x}^{2}\right)$

And then I get stuck here, I'm all out of ideas. My guess is I've overlooked something simple…

asked 2022-03-24

Logarithmic equations and their related exponential equations

I am learning about logarithms and I'd like some examples of the following:

Examples where the logarithmic value is a different positive integer

Examples where the logarithmic value is a different negative integer

Examples where the logarithmic value is a different non-integer fractional value

I am learning about logarithms and I'd like some examples of the following:

Examples where the logarithmic value is a different positive integer

Examples where the logarithmic value is a different negative integer

Examples where the logarithmic value is a different non-integer fractional value

asked 2021-10-18

Use The Properties Of Logarithms To Simplify

$\mathrm{log}}_{23.2}-{\mathrm{log}}_{20.025$