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keketsu

  • 2 years ago

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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  1. vengeance921
    • 2 years ago
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    \[2x(2x)^{x-1}+(Ln(2x))(2x)^{x}\] Is this the question?

  2. amistre64
    • 2 years ago
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    \[a^{(b+c)}=a^b * a^c\]

  3. vengeance921
    • 2 years ago
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    They have a natural logarithm there, right?

  4. amistre64
    • 2 years ago
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    yes, but it looks to just care about factoring out the 2x

  5. keketsu
    • 2 years ago
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    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?

  6. amistre64
    • 2 years ago
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    \[(2x)^1*(2x)^{(x-1)}=(2x)^{(x-1+1)}\]

  7. amistre64
    • 2 years ago
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    latex i assume

  8. vengeance921
    • 2 years ago
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    @Keketsu=Can you just post the real question please? Lol

  9. SchoolSlacker
    • 2 years ago
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    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?

  10. amistre64
    • 2 years ago
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    a demonstration to type in latex would be futile since the system would parse it as latex and not a demonstration :)

  11. keketsu
    • 2 years ago
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    @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?

  12. amistre64
    • 2 years ago
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    School, there is no real rule for procedure; only that it makes mathical sense in the end :)

  13. amistre64
    • 2 years ago
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    \[\frac{2x*2x^x}{2x}\]is one way to present it, yes

  14. SchoolSlacker
    • 2 years ago
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    Thanks i just saw it in the equation, and i had no idea :P

  15. keketsu
    • 2 years ago
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    Wonderful. Thanks for the help.

  16. amistre64
    • 2 years ago
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    youre welcome

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