## anonymous 4 years ago use the exponent laws to simplify the following until you have only a single power. then evaluate.

1. anonymous

$2^{4} \times(4^{5}\div4^{2}) \over (6 + 7^{-3})^{0}$

2. anonymous

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

i dont even know what demoninator and numerator mean

4. anonymous

|dw:1327953048715:dw|

5. anonymous

|dw:1327953037368:dw|

6. myininaya

Here are some laws to know: $\frac{x^a}{x^b}=x^{a-b} ; x^0=1 , x \neq 0$

7. anonymous

i know the laws i just dont know how to apply them.

8. myininaya

$\frac{2^4 \cdot (4^{5-2})}{1}=2^4 \cdot 4^3$

9. myininaya

So that should help

10. myininaya

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

apparently im im supost to make the 2 into another number

12. anonymous

ugh!!! this is so confusing! !!!!!

13. myininaya

Could you tell me where you are having the problem?

14. anonymous

where? like the whole problem is a problem i dont know where to start or what to do

15. anonymous

im homeschooled i dont have a teacher to teach me this stuff

16. myininaya

well right a way i seen the bottom was one since something^0=1 (of course when that something isn't 0)

17. myininaya

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

so that is how i got $\frac{2^4 \cdot (4^{5-2})}{1}$

19. myininaya

$2^4 \cdot (4^3)$

20. myininaya

since 5-2=3 and anything over 1 is that anything

21. anonymous

but what happens to the stuff inside the brackets?

22. anonymous

and apparently i have to change the 2 so its the same base or sumthin

23. myininaya

$2^4 \cdot 4^3$ you don't really need those parathesis anymore

24. anonymous

ok i need to simplify.

25. myininaya

oh you want to write it as 2^something?

26. anonymous

i dont even know what you mean!

27. myininaya

2 to some power?

28. myininaya

$2^{something}$ is that what you are saying what you want?

29. anonymous

omg this is hopeless

30. myininaya

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

ok look at the original question

32. anonymous

im supost to simplify that whole thing

33. myininaya

so you wanted it as 2^something

34. anonymous

with the same bases

35. myininaya

$2^4 \cdot (2^2)^3$ since 4 is 2(2)=2^2

36. myininaya

$2^4 \cdot 2^6$ since 2(3)=6

37. anonymous

i dont even know what u mean..

38. anonymous

omggggggggg!!!!!!

39. myininaya

$2^{4+6}$ using the law $x^a \cdot x^b=x^{a+b}$

40. myininaya

i'm sorry can you tell me what is troubling you?

41. anonymous

thanks for your help but i dont understand any math

42. anonymous

i dont understand what to do or anything

43. anonymous

i need like a step by step explination

44. myininaya

All I'm doing is applying the law of exponents That is what I'm giving you

45. myininaya

But if you don't understand the step let me know

46. anonymous

i dont

47. myininaya

ok which one?

48. anonymous

all of it !

49. myininaya

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

$2^{4} \times(4^{5}\div4^{2}) \over (6 + 7^{-3})^{0}$

51. myininaya

Ok on the bottom do you see how we can apply law 2?

52. anonymous

your supost to do brackets first tho

53. myininaya

It doesn't matter about all that we know the inside is not zero so we can apply law 2

54. anonymous

so its 1? like for sure?

55. anonymous

how do you know?

56. myininaya

That is right ! Great Job! so we have $2^{4} \times(4^{5}\div4^{2}) \over 1$ By law 2

57. myininaya

$2^{4} \times(4^{5}\div4^{2})$

58. myininaya

So now this is what is left over

59. anonymous

but u have to change the 2

60. myininaya

By order of operations we are suppose to look inside the parathesis

61. myininaya

What law can we apply inside the parathesis?

62. myininaya

We have division and the same base so we can use law 1 is what i called it above

63. myininaya

$2^4 \times (4^{5-2})$ remember this by law 1

64. myininaya

since 5-2=3 we can write $2^4 \times 4^3$

65. myininaya

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

i replaced 4 with 2^2 since 4=2^2

67. anonymous

i get all of it up to thr last bit

68. myininaya

$2^2=4?$

69. anonymous

im at 2^(4) x 4^(3)

70. myininaya

ok what about 2^4 * 4^3?

71. anonymous

im at that step

72. myininaya

ok let me know if you have a question about that step

73. anonymous

i have a question, like whats the next step after it

74. anonymous

how do we change the 2 into a 4

75. myininaya

this is what i wrote $2^4 \times(2^2)^3$

76. myininaya

$\text{ remember } 2^2=4$

77. anonymous

whats the mini number

78. myininaya

you mean the exponent?

79. anonymous

sure i guess?

80. myininaya

|dw:1327954725191:dw|

81. myininaya

That is a two.

82. anonymous

so after my last step what do i write lol

83. myininaya

$2^4 \times (2^2)^3$ 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

what does the dot mean

85. myininaya

multiply

86. myininaya

Do you know what to do last?

87. anonymous

k ima draw out what i have and u tell me if its correct ok?

88. anonymous

|dw:1327955081274:dw|

89. anonymous

|dw:1327955246139:dw|

90. anonymous

|dw:1327955289855:dw|

91. anonymous

|dw:1327955357557:dw|

92. anonymous

|dw:1327955389263:dw|

93. anonymous

|dw:1327955419197:dw|

94. anonymous

|dw:1327955435698:dw|

95. anonymous

is that it?

96. myininaya

that is right

97. anonymous

thank you so much! omg ! :)