anonymous
  • anonymous
Simplify: /[2x(2x)^(x-1) + (ln (2x))(2x)^x/] I'm particularly interested in how to work with /[(2x)^(x-1)/]. thanks.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[2x(2x)^{x-1}+(Ln(2x))(2x)^{x}\] Is this the question?
amistre64
  • amistre64
\[a^{(b+c)}=a^b * a^c\]
anonymous
  • anonymous
They have a natural logarithm there, right?

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amistre64
  • amistre64
yes, but it looks to just care about factoring out the 2x
anonymous
  • 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
  • amistre64
\[(2x)^1*(2x)^{(x-1)}=(2x)^{(x-1+1)}\]
amistre64
  • amistre64
latex i assume
anonymous
  • anonymous
@Keketsu=Can you just post the real question please? Lol
anonymous
  • 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
  • amistre64
a demonstration to type in latex would be futile since the system would parse it as latex and not a demonstration :)
anonymous
  • 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
  • amistre64
School, there is no real rule for procedure; only that it makes mathical sense in the end :)
amistre64
  • amistre64
\[\frac{2x*2x^x}{2x}\]is one way to present it, yes
anonymous
  • anonymous
Thanks i just saw it in the equation, and i had no idea :P
anonymous
  • anonymous
Wonderful. Thanks for the help.
amistre64
  • amistre64
youre welcome

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