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keketsu
 3 years ago
Simplify: /[2x(2x)^(x1) + (ln (2x))(2x)^x/]
I'm particularly interested in how to work with /[(2x)^(x1)/]. thanks.
keketsu
 3 years ago
Simplify: /[2x(2x)^(x1) + (ln (2x))(2x)^x/] I'm particularly interested in how to work with /[(2x)^(x1)/]. thanks.

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vengeance921
 3 years ago
Best ResponseYou've already chosen the best response.1\[2x(2x)^{x1}+(Ln(2x))(2x)^{x}\] Is this the question?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2\[a^{(b+c)}=a^b * a^c\]

vengeance921
 3 years ago
Best ResponseYou've already chosen the best response.1They have a natural logarithm there, right?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2yes, but it looks to just care about factoring out the 2x

keketsu
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, vengeance, sorry about the extra brackets. And yes, amistre64, you gave me the rule I was looking for. Thanks guys. How do I mark these equations so that the system recognizes them as math problems, btw?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2\[(2x)^1*(2x)^{(x1)}=(2x)^{(x1+1)}\]

vengeance921
 3 years ago
Best ResponseYou've already chosen the best response.1@Keketsu=Can you just post the real question please? Lol

SchoolSlacker
 3 years ago
Best ResponseYou've already chosen the best response.02x(2x)^{x1} sorry for asking my own question in your question but i just wanted to know with this would you do the factorisation with the outer number first or the indices with the brackets?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2a demonstration to type in latex would be futile since the system would parse it as latex and not a demonstration :)

keketsu
 3 years ago
Best ResponseYou've already chosen the best response.1@Vengeance921, what you posted is the equation, but it is dealing with the (2x)^(x1) that I was having trouble with. So that's (2x^x) * (2x^1) or (2x^x)/2x?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2School, there is no real rule for procedure; only that it makes mathical sense in the end :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2\[\frac{2x*2x^x}{2x}\]is one way to present it, yes

SchoolSlacker
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks i just saw it in the equation, and i had no idea :P

keketsu
 3 years ago
Best ResponseYou've already chosen the best response.1Wonderful. Thanks for the help.
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