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Mathematics 21 Online
OpenStudy (anonymous):

I'm supposed to be working on the laws of exponents. answers are supposed to be in form ae^b

OpenStudy (anonymous):

\[\sqrt[3]{-8e ^{6x}} \] and \[(e ^{2x}e ^{3x})/e ^{5}\]

terenzreignz (terenzreignz):

For the first one, recall that \[\huge \sqrt[a]{b}=b^{\frac{1}{a}}\]

OpenStudy (anonymous):

so with that just to make sure..would i pull the negative along when i pull out the 8?

OpenStudy (anonymous):

\[8^{ \frac{ 6x }{ 3}}\]

terenzreignz (terenzreignz):

Keep it there. It's the safest thing to do... Now... \[\huge \sqrt[3]{-8e^{6x}}=\left(-8e^{6x}\right)^{\frac{1}{3}}\]

OpenStudy (anonymous):

so then i would have 6x/3 for my exponent? this is the point where i usually start getting confused because of all the rules with exponents

terenzreignz (terenzreignz):

It's okay, that's why you have me :) We're taking things one at a time, here... So, remember this law of exponents \[\huge (ab)^m=a^mb^m\] ? Now would be a good time to apply it \[\huge \left(-8e^{6x}\right)^{\frac{1}{3}}=\left[\left(-8\right)\left(e^{6x}\right)\right]^{\frac{1}{3}}\]

OpenStudy (anonymous):

okay so then i would have (2) and the other part will be e^6/3x ?

OpenStudy (anonymous):

i mean -2

terenzreignz (terenzreignz):

Very good :) I was about to correct you, but you beat me to it. That's why I wanted to keep the negative sign :) So, yeah, you get \[\huge -2e^{\frac{6x}{3}}\] And now just simplify that fraction. Before anything else, that fractional exponent is a fraction, which can be reduced to lowest terms:)

OpenStudy (anonymous):

so then i would have \[-2e ^{2x}\]

terenzreignz (terenzreignz):

Precisely. Well done there :)

OpenStudy (anonymous):

wow. thank you so much!! I would not have figured that out on my own!

terenzreignz (terenzreignz):

In time, you'll get the hang of it, and it'll all be second nature :) Now, you need help with the second question?

OpenStudy (anonymous):

yes i do.

OpenStudy (anonymous):

i know with exponents and division there is some kind of subtraction, but I couldn't get it

terenzreignz (terenzreignz):

Well, this question is, in my opinion, simpler than the first one~ It doesn't involve radical signs :D Just remember these two rules. \[\huge a^ma^n=a^{m+n}\] \[\huge \frac{a^m}{a^n}=a^{m-n}=\frac{1}{a^{n-m}}\]

terenzreignz (terenzreignz):

So, let's begin :) You're given \[\huge \frac{e^{2x}e^{3x}}{e^{5}}\] Let's do things one at a time, here...

terenzreignz (terenzreignz):

In the numerator, you multiply two exponentials with the same base. Using the rule posted above, you can just add their exponents, right? Go ahead now...

OpenStudy (anonymous):

e^5x

terenzreignz (terenzreignz):

That's right, so you're left with \[\huge \frac{e^{5x}}{e^5}\] Careful here, there are so many ways to go wrong... Use the rule about division.

OpenStudy (anonymous):

1/e^x or would it be 1/e^5x-5 ?

terenzreignz (terenzreignz):

Neither. Now review this rule again. I've only put in once instance so you don't get confused :) \[\huge \frac{a^m}{a^n}=a^{m-n}\]

OpenStudy (anonymous):

e^5x-5 i would keep the e right? because that kind of represents the a and since the 5x and 5 are not the same i couldn't directly subtract them to get an actual number

terenzreignz (terenzreignz):

Yep. You're left with \[\huge e^{5x-5}\] I don't see a way to further simplify this...maybe \[\huge e^{5(x-1)}\] But in any case, good job :)

OpenStudy (anonymous):

yeah! thank you so much!! I really need to go look online and find all those exponent rules again. thank you so much for your help

terenzreignz (terenzreignz):

No problem :)

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