If $\sqrt{36}=6$ is true, then why is $\sqrt{36}=-6$ false in verification of radical equations?

Lymnmeatlypamgfm
2022-04-23
Answered

If $\sqrt{36}=6$ is true, then why is $\sqrt{36}=-6$ false in verification of radical equations?

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jeffster830gyz

Answered 2022-04-24
Author has **21** answers

If $36=(+6)(+6)=(-6)(-6)$

Then$\sqrt{36}=\sqrt{\left|36\right|}=\pm 6$

Usually 6 is answer.

But when you have an equation is needed to accept both results.

Then

Usually 6 is answer.

But when you have an equation is needed to accept both results.

Landyn Whitney

Answered 2022-04-25
Author has **19** answers

In general notation, $\sqrt{a}=$ the positive (+) square root and $-\sqrt{a}=$ the negative (-) square root. So, in the instant case, $\sqrt{36}=6\text{}\text{and}\text{}-\sqrt{36}=-6$ .

asked 2022-03-30

Prove that if $e<y<x$ then $x}^{y}<{y}^{x$

My try:

I tried to use Taylor's theorem:

${y}^{x}-{x}^{y}={e}^{x\mathrm{ln}y}-{e}^{y\mathrm{ln}x}=1+x\mathrm{ln}y+o\left(x\mathrm{ln}y\right)-1-y\mathrm{ln}x-o\left(y\mathrm{ln}x\right)=$

$={\mathrm{ln}y}^{x}-{\mathrm{ln}x}^{y}+o\left(x\mathrm{ln}y\right)-o\left(y\mathrm{ln}x\right)=\mathrm{ln}\frac{{y}^{x}}{{x}^{y}}+o\left(x\mathrm{ln}y\right)-o\left(y\mathrm{ln}x\right)$

Hovewer I have a problem to show that${y}^{x}-{x}^{y}>0$ because I can't say that $\mathrm{ln}\frac{{y}^{x}}{{x}^{y}}>0$ because then I use with theses.

Have you some idea?

My try:

I tried to use Taylor's theorem:

Hovewer I have a problem to show that

Have you some idea?

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Question about basic exponential/logarithm properties

Solve for k:

${e}^{k/2}=a$

Solution:

${e}^{2k}=a$

$k/2=\mathbf{l}\mathbf{n}a$

$k=2\mathbf{l}\mathbf{n}a$

$=\mathbf{l}\mathbf{n}{a}^{2}$

My question is: why does $2\mathbf{l}\mathbf{n}a=\mathbf{l}\mathbf{n}{a}^{2}$? Why can you transfer the 2 to be an exponent of a?

Solve for k:

${e}^{k/2}=a$

Solution:

${e}^{2k}=a$

$k/2=\mathbf{l}\mathbf{n}a$

$k=2\mathbf{l}\mathbf{n}a$

$=\mathbf{l}\mathbf{n}{a}^{2}$

My question is: why does $2\mathbf{l}\mathbf{n}a=\mathbf{l}\mathbf{n}{a}^{2}$? Why can you transfer the 2 to be an exponent of a?

asked 2022-07-17

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Saw this problem and I thought I'd take a shot at it:

Find all $a$ such that

$\underset{x\to \mathrm{\infty}}{lim}{\left(\frac{x+a}{x-a}\right)}^{x}=e.$

Saw this problem and I thought I'd take a shot at it:

Find all $a$ such that

$\underset{x\to \mathrm{\infty}}{lim}{\left(\frac{x+a}{x-a}\right)}^{x}=e.$

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Factor each polynomial completely. If the polynomial cannot be factored, say it is prime.

$8{x}^{2}+88x+80$

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Use the properties of the natural logarithms to simplify the following expressions:

$a)\mathrm{ln}\mathrm{sin}\theta -\mathrm{ln}\frac{\mathrm{sin}\theta}{5}$

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Show $\left(\mathrm{log}n\right)}^{\left(\mathrm{log}n\right)}={2}^{\left(\mathrm{log}n\right)\left(\mathrm{log}\left(\mathrm{log}n\right)\right)$

I am having difficulty understanding how this follows.

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

Which logarithmic identities are used to go through each equality?

e.g. how do you first go from

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

and then to

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

(The log base must be 2 or else this equality won't hold)

I am having difficulty understanding how this follows.

Which logarithmic identities are used to go through each equality?

e.g. how do you first go from

and then to

(The log base must be 2 or else this equality won't hold)