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anonymous

  • one year ago

How do you solve this problem? Please explain!

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  1. anonymous
    • one year ago
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    \[(4a ^{5/3})^{3/2}\]

  2. anonymous
    • one year ago
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    Solve for what?

  3. anonymous
    • one year ago
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    It says to simplify in this problem

  4. IrishBoy123
    • one year ago
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    basic rule is \(\large (b^n)^m = b^{n \times m}\) because it's \(b^n \times b^n \times b^n ......\) m times. so \(\large (a. \ b^n)^m = a^m \ . \ b^{n \times m}\) But you can go at this in a number of ways; and you might wish to try this approach: \(\large (4a^{5/3})^{3/2} = \sqrt{4a^{5/3} \ * 4a^{5/3} \ * 4a^{5/3} }\) it's not the immediate solution, you will get that by looking up the exponent rules as posted above, but it might help get you get comfortable with these kind of questions.....maybe under the \(^{3/2}\) umbrella you are cubing and square rooting....

  5. anonymous
    • one year ago
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    How would I multiply the numbers inside the square root?

  6. IrishBoy123
    • one year ago
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    well, imagine the square root is not there. jut multiply those out as you normally would: \(4 a^{5/3} \times 4 a^{5/3} \times4 a^{5/3} \) we'll worry about the square root sign after we have done that.

  7. anonymous
    • one year ago
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    Okay. When you are multiplying exponents you have to add them right?

  8. phi
    • one year ago
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    when multiplying the same base , you add the exponents.

  9. anonymous
    • one year ago
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    So without the square root it equals 4a^5?

  10. phi
    • one year ago
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    if you remember that x*x*x is x^3 but x is x^1 and x*x is x^2 you get the example x^1 * x^2 = x^3 (that is how I remember this rule)

  11. phi
    • one year ago
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    yes, but you also have 4*4*4 to do

  12. IrishBoy123
    • one year ago
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    nah you also have 4 x 4 x 4!!

  13. anonymous
    • one year ago
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    64a^5?

  14. IrishBoy123
    • one year ago
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    fine i'll leave it to @phi you are in great hands!! good luck

  15. anonymous
    • one year ago
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    then it would equal 8a.....how would I square root the 5?

  16. phi
    • one year ago
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    a^5 is a*a*a*a * a we can do square root of a^4 (it is a*a or a^2) square root of the "extra" a is either sqr(a) or a^1/2

  17. phi
    • one year ago
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    \[ \sqrt{64 a^4 } \sqrt{a} \]

  18. anonymous
    • one year ago
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    So the answer is \[8a \sqrt{a}\]?

  19. phi
    • one year ago
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    8a^2 sqr(a)

  20. anonymous
    • one year ago
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    So that would be simplified?

  21. phi
    • one year ago
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    they probably want you to use exponents, so write sqr(a) as a^(1/2)

  22. anonymous
    • one year ago
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    \[8a ^{2}a ^{1/2}\]

  23. anonymous
    • one year ago
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    Like that?

  24. phi
    • one year ago
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    yes, and though it looks a bit weird, we are multiplying the same base. (the a) so we add the exponents. you get 2.5 but generally people write it as an improper fraction

  25. anonymous
    • one year ago
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    Oh okay. Thanks!

  26. phi
    • one year ago
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    in other words, 2+ 1/2 is 4/2 + 1/2 = 5/2 and people write \[ 8 a^\frac{5}{2} \]

  27. anonymous
    • one year ago
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    Ya I got it. :)

  28. phi
    • one year ago
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    if you are doing a lot of this, I would use the "fast way" \[ (b^n)^m = b^{n \cdot m} \] for example \[ (4^1 \cdot a ^\frac{5}{3})^\frac{3}{2} \] becomes \[ 4^\frac{3}{2} \cdot a^{\frac{5}{3}\cdot \frac{3}{2} }\]

  29. phi
    • one year ago
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    notice the expoinet of a is 5*3/(3*2) which simplifies to 5/2 to find what 4^3/2 is, we have to do either \[ \left( 4^\frac{1}{2}\right)^3 = 2^3 = 8 \] or \[ \left( 4^3\right)^\frac{1}{2}= 64^\frac{1}{2} = 8 \]

  30. anonymous
    • one year ago
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    Oh that makes sense.

  31. phi
    • one year ago
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    if it doesn't , doing a dozen problems will be tedious.

  32. anonymous
    • one year ago
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    Ya that's true.

  33. anonymous
    • one year ago
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    Do you happen to know how to do this problem? How do you make both numbers have the same base? \[(5a ^{2/3})(4a ^{3/2})\]

  34. phi
    • one year ago
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    you are multiplying 4 "things" 5* a^(2/3) * 4 * a^(3/2) and you can switch the order: 5*4 * a^(2/3) * a^(3/2) does that help to know how to tackle this ?

  35. anonymous
    • one year ago
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    I think so. I got 20 as my final answer.

  36. phi
    • one year ago
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    yes 4*5 is 20 but you can't ignore the a's (but you can multiply them using "same base, add exponents" rule)

  37. anonymous
    • one year ago
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    Oh ya the a is \[a ^{1}\] So then the final answer would actually be 20a.

  38. phi
    • one year ago
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    when you *multiply* (like here) , and you have the same base, you *add* the exponents. in other words you get a^ new exponent and the new exponent is 2/3 + 3/2

  39. phi
    • one year ago
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    use a common denominator of 6 to get 4/6 + 9/6

  40. anonymous
    • one year ago
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    Oh....I was multiplying them again. So that would equal 13/6.

  41. phi
    • one year ago
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    yes. this stuff is confusing but try to remember the x*x = x^2 or (putting in the exponents) \[ x^1 \cdot x^1 = x^2\] as an example of adding the exponents

  42. anonymous
    • one year ago
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    Oh okay. So then it would equal \[20a ^{13/6}\]?

  43. phi
    • one year ago
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    yes

  44. anonymous
    • one year ago
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    Okay thanks again!

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