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anonymous

  • 4 years ago

use the exponent laws to simplify the following until you have only a single power. then evaluate.

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  1. anonymous
    • 4 years ago
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    \[2^{4} \times(4^{5}\div4^{2}) \over (6 + 7^{-3})^{0}\]

  2. anonymous
    • 4 years ago
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    ok sarah, first, (4^5)/(4^2)=4^3 4^3=2^4, so we have (2^4)*(2^4) in the denominator, but i can't see the numenator clearly, can u type it again

  3. anonymous
    • 4 years ago
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    i dont even know what demoninator and numerator mean

  4. anonymous
    • 4 years ago
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    |dw:1327953048715:dw|

  5. anonymous
    • 4 years ago
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    |dw:1327953037368:dw|

  6. myininaya
    • 4 years ago
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    Here are some laws to know: \[\frac{x^a}{x^b}=x^{a-b} ; x^0=1 , x \neq 0\]

  7. anonymous
    • 4 years ago
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    i know the laws i just dont know how to apply them.

  8. myininaya
    • 4 years ago
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    \[\frac{2^4 \cdot (4^{5-2})}{1}=2^4 \cdot 4^3\]

  9. myininaya
    • 4 years ago
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    So that should help

  10. myininaya
    • 4 years ago
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    and 2^4=2(2)(2)(2)=16 and 4^3=4(4)(4)=64 find the product of 64 and 16 and you will win! :)

  11. anonymous
    • 4 years ago
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    apparently im im supost to make the 2 into another number

  12. anonymous
    • 4 years ago
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    ugh!!! this is so confusing! !!!!!

  13. myininaya
    • 4 years ago
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    Could you tell me where you are having the problem?

  14. anonymous
    • 4 years ago
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    where? like the whole problem is a problem i dont know where to start or what to do

  15. anonymous
    • 4 years ago
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    im homeschooled i dont have a teacher to teach me this stuff

  16. myininaya
    • 4 years ago
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    well right a way i seen the bottom was one since something^0=1 (of course when that something isn't 0)

  17. myininaya
    • 4 years ago
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    then the two guys inside that parenthesis had the same base and i was like we should apply this law \[\frac{x^a}{x^b}=x^{a-b}\] since we had the same base and we are dividing

  18. myininaya
    • 4 years ago
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    so that is how i got \[\frac{2^4 \cdot (4^{5-2})}{1}\]

  19. myininaya
    • 4 years ago
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    \[2^4 \cdot (4^3)\]

  20. myininaya
    • 4 years ago
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    since 5-2=3 and anything over 1 is that anything

  21. anonymous
    • 4 years ago
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    but what happens to the stuff inside the brackets?

  22. anonymous
    • 4 years ago
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    and apparently i have to change the 2 so its the same base or sumthin

  23. myininaya
    • 4 years ago
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    \[2^4 \cdot 4^3\] you don't really need those parathesis anymore

  24. anonymous
    • 4 years ago
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    ok i need to simplify.

  25. myininaya
    • 4 years ago
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    oh you want to write it as 2^something?

  26. anonymous
    • 4 years ago
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    i dont even know what you mean!

  27. myininaya
    • 4 years ago
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    2 to some power?

  28. myininaya
    • 4 years ago
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    \[2^{something}\] is that what you are saying what you want?

  29. anonymous
    • 4 years ago
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    omg this is hopeless

  30. myininaya
    • 4 years ago
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    i was just going to write 2^4 as 2(2)(2)(2)=16 and 4^3=4(4)(4)=64 and find the product of 16 and 64 since we have \[2^4 \cdot 4^3\] \[16 \cdot 64\]

  31. anonymous
    • 4 years ago
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    ok look at the original question

  32. anonymous
    • 4 years ago
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    im supost to simplify that whole thing

  33. myininaya
    • 4 years ago
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    so you wanted it as 2^something

  34. anonymous
    • 4 years ago
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    with the same bases

  35. myininaya
    • 4 years ago
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    \[2^4 \cdot (2^2)^3\] since 4 is 2(2)=2^2

  36. myininaya
    • 4 years ago
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    \[2^4 \cdot 2^6\] since 2(3)=6

  37. anonymous
    • 4 years ago
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    i dont even know what u mean..

  38. anonymous
    • 4 years ago
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    omggggggggg!!!!!!

  39. myininaya
    • 4 years ago
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    \[2^{4+6}\] using the law \[x^a \cdot x^b=x^{a+b}\]

  40. myininaya
    • 4 years ago
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    i'm sorry can you tell me what is troubling you?

  41. anonymous
    • 4 years ago
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    thanks for your help but i dont understand any math

  42. anonymous
    • 4 years ago
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    i dont understand what to do or anything

  43. anonymous
    • 4 years ago
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    i need like a step by step explination

  44. myininaya
    • 4 years ago
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    All I'm doing is applying the law of exponents That is what I'm giving you

  45. myininaya
    • 4 years ago
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    But if you don't understand the step let me know

  46. anonymous
    • 4 years ago
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    i dont

  47. myininaya
    • 4 years ago
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    ok which one?

  48. anonymous
    • 4 years ago
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    all of it !

  49. myininaya
    • 4 years ago
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    ok so I'm going to write down all the law of exponents we will need: Law 1:\[\frac{x^a}{x^b}=x^{a-b}\] Law 2:\[x \neq 0 , x^0=1\] Law 3:\[x^a \cdot x^b=x^{a+b}\] Law 4:\[(x^a)^b=x^{ a \cdot b}\] These are the laws I'm going to use

  50. myininaya
    • 4 years ago
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    \[2^{4} \times(4^{5}\div4^{2}) \over (6 + 7^{-3})^{0} \]

  51. myininaya
    • 4 years ago
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    Ok on the bottom do you see how we can apply law 2?

  52. anonymous
    • 4 years ago
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    your supost to do brackets first tho

  53. myininaya
    • 4 years ago
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    It doesn't matter about all that we know the inside is not zero so we can apply law 2

  54. anonymous
    • 4 years ago
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    so its 1? like for sure?

  55. anonymous
    • 4 years ago
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    how do you know?

  56. myininaya
    • 4 years ago
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    That is right ! Great Job! so we have \[2^{4} \times(4^{5}\div4^{2}) \over 1\] By law 2

  57. myininaya
    • 4 years ago
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    \[2^{4} \times(4^{5}\div4^{2}) \]

  58. myininaya
    • 4 years ago
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    So now this is what is left over

  59. anonymous
    • 4 years ago
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    but u have to change the 2

  60. myininaya
    • 4 years ago
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    By order of operations we are suppose to look inside the parathesis

  61. myininaya
    • 4 years ago
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    What law can we apply inside the parathesis?

  62. myininaya
    • 4 years ago
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    We have division and the same base so we can use law 1 is what i called it above

  63. myininaya
    • 4 years ago
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    \[2^4 \times (4^{5-2})\] remember this by law 1

  64. myininaya
    • 4 years ago
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    since 5-2=3 we can write \[2^4 \times 4^3\]

  65. myininaya
    • 4 years ago
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    now since we want the same base we need find out how to write 4 such that it equals 2 to some power \[4=2^2 \] so we can write \[2^4 \times(2^2)^3\]

  66. myininaya
    • 4 years ago
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    i replaced 4 with 2^2 since 4=2^2

  67. anonymous
    • 4 years ago
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    i get all of it up to thr last bit

  68. myininaya
    • 4 years ago
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    \[2^2=4?\]

  69. anonymous
    • 4 years ago
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    im at 2^(4) x 4^(3)

  70. myininaya
    • 4 years ago
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    ok what about 2^4 * 4^3?

  71. anonymous
    • 4 years ago
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    im at that step

  72. myininaya
    • 4 years ago
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    ok let me know if you have a question about that step

  73. anonymous
    • 4 years ago
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    i have a question, like whats the next step after it

  74. anonymous
    • 4 years ago
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    how do we change the 2 into a 4

  75. myininaya
    • 4 years ago
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    this is what i wrote \[2^4 \times(2^2)^3 \]

  76. myininaya
    • 4 years ago
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    \[\text{ remember } 2^2=4\]

  77. anonymous
    • 4 years ago
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    whats the mini number

  78. myininaya
    • 4 years ago
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    you mean the exponent?

  79. anonymous
    • 4 years ago
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    sure i guess?

  80. myininaya
    • 4 years ago
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    |dw:1327954725191:dw|

  81. myininaya
    • 4 years ago
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    That is a two.

  82. anonymous
    • 4 years ago
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    so after my last step what do i write lol

  83. myininaya
    • 4 years ago
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    \[2^4 \times (2^2)^3\] We can rewrite that second part as \[2^{2 \cdot 3}=2^6\] by use of one of the other laws so now we are left with \[2^4 \cdot 2^6\]

  84. anonymous
    • 4 years ago
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    what does the dot mean

  85. myininaya
    • 4 years ago
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    multiply

  86. myininaya
    • 4 years ago
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    Do you know what to do last?

  87. anonymous
    • 4 years ago
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    k ima draw out what i have and u tell me if its correct ok?

  88. anonymous
    • 4 years ago
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    |dw:1327955081274:dw|

  89. anonymous
    • 4 years ago
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    |dw:1327955246139:dw|

  90. anonymous
    • 4 years ago
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    |dw:1327955289855:dw|

  91. anonymous
    • 4 years ago
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    |dw:1327955357557:dw|

  92. anonymous
    • 4 years ago
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    |dw:1327955389263:dw|

  93. anonymous
    • 4 years ago
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    |dw:1327955419197:dw|

  94. anonymous
    • 4 years ago
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    |dw:1327955435698:dw|

  95. anonymous
    • 4 years ago
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    is that it?

  96. myininaya
    • 4 years ago
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    that is right

  97. anonymous
    • 4 years ago
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    thank you so much! omg ! :)

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