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calculusxy

  • one year ago

Help with exponents! Question posted below... MEDAL!

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  1. calculusxy
    • one year ago
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    \[\huge (x^4)^{-3} \times 2x^4\]

  2. calculusxy
    • one year ago
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    @hartnn

  3. hartnn
    • one year ago
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    first term \(\huge (a^b)^c = a^{bc}\)

  4. hartnn
    • one year ago
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    you subtracted the exponents, the rule tell us to multiply them :3

  5. anonymous
    • one year ago
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    @hartnn agree

  6. calculusxy
    • one year ago
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    Oh my godd

  7. calculusxy
    • one year ago
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    I am running our of my mind :\

  8. calculusxy
    • one year ago
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    *out

  9. hartnn
    • one year ago
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    \(4 \times (-3) = - (4\times 3) = .. \)

  10. calculusxy
    • one year ago
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    \[\huge x^{4 \times (-3)} = x^{-12}\]

  11. hartnn
    • one year ago
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    yussss

  12. hartnn
    • one year ago
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    now \(\huge a^b \times a ^c = a^{b+c}\)

  13. hartnn
    • one year ago
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    2 \((x^{-12} \times x^4) = .. ?\)

  14. calculusxy
    • one year ago
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    So here's the thing, I don't why we have to combine them... Is it because 2x^4 has two terms: 2 and x^4?

  15. Jhannybean
    • one year ago
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    theres a common base, \(x\)

  16. hartnn
    • one year ago
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    2 is a constant, lets throw that out of the process of multiplying 'x's

  17. calculusxy
    • one year ago
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    So just to clarify, if we have like 45x^5 then i will just have to take out the 45 and work with the x being the common base with another power right?

  18. hartnn
    • one year ago
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    ....with another term of x. yes.

  19. calculusxy
    • one year ago
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    \[\large 2(x^{-8}) = 2x^{-8}\]

  20. calculusxy
    • one year ago
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    Is that correct ?

  21. hartnn
    • one year ago
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    \(2 x^3 \times 45 x^5 = 2\times 45 x^{3+5}\)

  22. hartnn
    • one year ago
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    yes, thats correct :)

  23. hartnn
    • one year ago
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    2/x^8 also "looks" nice

  24. calculusxy
    • one year ago
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    My answer is \[\huge \frac{ 2 }{ x^8 }\]

  25. hartnn
    • one year ago
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    \(\huge \checkmark \)

  26. calculusxy
    • one year ago
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    The reason for why I have to put the x^8 to the denominator is because of the exponent of -8 right? The negative exponent tells the base to get to its reciprocal and then the exponent becomes positive?

  27. hartnn
    • one year ago
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    correcto! \(\Large a^{-b} = \dfrac{1}{a^b}\)

  28. calculusxy
    • one year ago
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    Thank you!

  29. hartnn
    • one year ago
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    welcome ^_^

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