How to solve for $k$ when the area about the $x$ axis and under the graph of the $f(x)=\frac{1}{x}$ from interval $x=[2,k]$ is equal to $\mathrm{ln}(4)$?

caschaillo7
2022-10-17
Answered

How to solve for $k$ when the area about the $x$ axis and under the graph of the $f(x)=\frac{1}{x}$ from interval $x=[2,k]$ is equal to $\mathrm{ln}(4)$?

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Kaylee Evans

Answered 2022-10-18
Author has **20** answers

Notice

$\underset{2}{\overset{k}{\int}}\frac{1}{x}dx=\mathrm{ln}4=2\mathrm{ln}2$

This is given. But

$\underset{2}{\overset{k}{\int}}\frac{1}{x}dx=\mathrm{ln}k-\mathrm{ln}2$

Hence,

$\mathrm{ln}k-\mathrm{ln}2=2\mathrm{ln}2\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}k=3\mathrm{ln}2\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}k=\mathrm{ln}{2}^{3}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}k={2}^{3}=8$

$\underset{2}{\overset{k}{\int}}\frac{1}{x}dx=\mathrm{ln}4=2\mathrm{ln}2$

This is given. But

$\underset{2}{\overset{k}{\int}}\frac{1}{x}dx=\mathrm{ln}k-\mathrm{ln}2$

Hence,

$\mathrm{ln}k-\mathrm{ln}2=2\mathrm{ln}2\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}k=3\mathrm{ln}2\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{ln}k=\mathrm{ln}{2}^{3}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}k={2}^{3}=8$

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I am trying to determine how long it will take$\left(t\right)$ for the half life of 500 amount of substance to fall below 100 (to be $\le 99$ -- I am only concerned with integers) when it has a half life of 5.

I have gotten this far in the calculation:

$99=500{\left(\frac{1}{2}\right)}^{\frac{t}{5}}$

So I guess my questions are as follows:

1.Have I set this up right?

2.Is there a better way to do this?

3.Can somebody point me in the direction of how to solve from here?

I am trying to determine how long it will take

I have gotten this far in the calculation:

So I guess my questions are as follows:

1.Have I set this up right?

2.Is there a better way to do this?

3.Can somebody point me in the direction of how to solve from here?

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$\mathrm{log}}_{5}(2x+3)={\mathrm{log}}_{53$

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I try to solve the following equation:

${(N+1)}^{{\mathrm{log}}_{N}125}=216$

I know the answer is 5 here but how could I rewrite the equations so I can solve it?

I tried to take the log of both sides but that didn't help me because I got stuck. Could anyone please explain me how to do this?

Thanks!

I try to solve the following equation:

I know the answer is 5 here but how could I rewrite the equations so I can solve it?

I tried to take the log of both sides but that didn't help me because I got stuck. Could anyone please explain me how to do this?

Thanks!

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$a={b}^{x}+{c}^{x}$, How to solve for $x$?

If $a={b}^{x}$, then $x$ could be written in terms of $a$ and $b$; $x={\displaystyle \frac{\mathrm{log}(a)}{\mathrm{log}(b)}}$

What about $a={b}^{x}+{c}^{x}$?

Could $x$ be written in terms of $a,b$ and $c$? $x=$?

If $a={b}^{x}$, then $x$ could be written in terms of $a$ and $b$; $x={\displaystyle \frac{\mathrm{log}(a)}{\mathrm{log}(b)}}$

What about $a={b}^{x}+{c}^{x}$?

Could $x$ be written in terms of $a,b$ and $c$? $x=$?

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Showing an inequality with ln

I have to show that the following inequation is true:

$\frac{\mathrm{ln}(x)+\mathrm{ln}(y)}{2}\le \mathrm{ln}(\frac{x+y}{2})$

I transformed it into

$\frac{\mathrm{ln}(x\cdot y)}{2}\le \mathrm{ln}(x+y)-\mathrm{ln}(2)$

because I thought that I better can show the inequation here, but I don't know how to proceed.

How can I proceed or am I completely wrong?

I have to show that the following inequation is true:

$\frac{\mathrm{ln}(x)+\mathrm{ln}(y)}{2}\le \mathrm{ln}(\frac{x+y}{2})$

I transformed it into

$\frac{\mathrm{ln}(x\cdot y)}{2}\le \mathrm{ln}(x+y)-\mathrm{ln}(2)$

because I thought that I better can show the inequation here, but I don't know how to proceed.

How can I proceed or am I completely wrong?

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Write each expression as a single natural logarithm. $2\mathrm{ln}8-3\mathrm{ln}4$