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- anonymous

Simplify: /[2x(2x)^(x-1) + (ln (2x))(2x)^x/]
I'm particularly interested in how to work with /[(2x)^(x-1)/]. thanks.

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- anonymous

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- anonymous

\[2x(2x)^{x-1}+(Ln(2x))(2x)^{x}\]
Is this the question?

- amistre64

\[a^{(b+c)}=a^b * a^c\]

- anonymous

They have a natural logarithm there, right?

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- amistre64

yes, but it looks to just care about factoring out the 2x

- anonymous

Yes, 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

\[(2x)^1*(2x)^{(x-1)}=(2x)^{(x-1+1)}\]

- amistre64

latex i assume

- anonymous

@Keketsu=Can you just post the real question please? Lol

- anonymous

2x(2x)^{x-1}
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

a demonstration to type in latex would be futile since the system would parse it as latex and not a demonstration :)

- anonymous

@Vengeance921, what you posted is the equation, but it is dealing with the (2x)^(x-1) that I was having trouble with.
So that's (2x^x) * (2x^-1) or (2x^x)/2x?

- amistre64

School, there is no real rule for procedure; only that it makes mathical sense in the end :)

- amistre64

\[\frac{2x*2x^x}{2x}\]is one way to present it, yes

- anonymous

Thanks i just saw it in the equation, and i had no idea :P

- anonymous

Wonderful. Thanks for the help.

- amistre64

youre welcome

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