1. Xishem

$\log_{2} \sqrt[5]{32}=\log_{2}2=1$

2. anonymous

what?

3. anonymous

how did you get log base 2 then 2?

4. anonymous

the fifth root of 32 is 2, because $2^5=32$

5. anonymous

then shouldn't the answer be five ?

6. anonymous

so your job is to find $\log_2(2)$ which asks the question, "what power do you raise the number "2" to to get an answer of 2?" and the answer is clearly 1

7. anonymous

$\log_2(2^5)=5$ yes

8. anonymous

but you have $\log_2(2)$

9. anonymous

wait..what?!

10. anonymous

$2^{y} = 2^{5}$

11. anonymous

ok lets go slow

12. anonymous

if your question was $\log_2(2^5)$ the answer would certainly be 5 correct?

13. anonymous

i.e. $\log_2(32)=5$

14. anonymous

wait one sec let me soak this in

15. anonymous

ok

16. anonymous

so you're telling me that 2^5 does not equal fifth root of 32?

17. anonymous

yes i am. 2^5 = 32 not the fifth root of 32

18. anonymous

so what is it? and how would u find what it is ?

19. anonymous

i want to know the fifth rote of 32

20. anonymous

but since 2^5 = 32 then we know that $\sqrt[5]{32}=2$

21. anonymous

it is 32^ 1/5

22. anonymous

you can think of it that way, yes

23. anonymous

im not understanding what you're telling me

24. anonymous

which means $(32)^{\frac{1}{5}}=(2^5)^{\frac{1}{5}}=2^1=2$

25. anonymous

in english: two the the fifth power is 32, therefore the fifth root of 32 is 2

26. anonymous

i think i understand that. thank you very much. can you help me with a few more?

27. anonymous

28. anonymous

$3 \log _{7} \sqrt[6]{49}$

29. anonymous

ok lets do this the easy way. first of all $\log(x^n)=n\log(x)$

30. anonymous

yep. i understand tha : - )

31. anonymous

and we can rewrite $\sqrt[6]{49}$ as $49^{\frac{1}{6}}$ yes?

32. anonymous

ummm im kind of confused on those

33. anonymous

is that just a general rule ?

34. anonymous

you need to understand exponential notation for logs, because logs are exponents.

35. anonymous

yes for example $\sqrt{x}=x^{\frac{1}{2}}$ $\sqrt[3]{x}=x^{\frac{1}{3}}$ etc

36. anonymous

and in general $\sqrt[n]{x}=x^{\frac{1}{n}}$

37. anonymous

okay awesome. just wrote that on my rules sheet XD

38. anonymous

so we start with $3\log_7(49^{\frac{1}{6}})$ and then bring the 1/6 right out front as a multiplier (coefficient)

39. anonymous

$\frac{1}{6}\times 3\log_7(49)$ $\frac{1}{2}\log_7(49)$ and idea what to do next?

40. anonymous

i really have no idea :(

41. anonymous

how can you write 49 with an exponent?

42. anonymous

what would u do if you have a number in front of the log? like the 1/2

43. anonymous

hold on, lets just take care of the 49 first. we can do this problem a different way second

44. anonymous

we would write $49=7^2$

45. anonymous

and then you have $\frac{1}{2}\log_77^2)$ which should be easy now

46. anonymous

how would we incorporate the 1/2?

47. anonymous

since $\log_b(b^n)=n$

48. anonymous

but its 1/2 log

49. anonymous

we leave it for a moment. the one half out front is just a number, let it sit there

50. anonymous

51. anonymous

we get $\frac{1}{2}\log_7(7^2)=\frac{1}{2}\times 2$

52. anonymous

the answer to $\log_7(7^2)=2$ and then $\frac{1}{2}\times 2=1$

53. anonymous

oh okay!!!!! wow i get that!

54. anonymous

if you like we can do this problem a different way, but maybe one way is enough, you tell me

55. anonymous

56. anonymous

yes it is 1

57. anonymous

i like this way. do u mind helping me with more?

58. anonymous

59. anonymous

$50 \log 5 \sqrt{125}$

60. anonymous

so we would have

61. anonymous

5 ^ x = square route 125 ^ 50?

62. anonymous

oh yikes leave the 50 out front!

63. anonymous

i mean you are right, but leave it there to make your life easier. think of how to write 125 as a number raised to a power

64. anonymous

okay. lol so i have 50 log base 5^x = 5^ 3

65. anonymous

i believe that is wrong

66. anonymous

close, you just forget the square root

67. anonymous

but i don't have the square route anymore i got it al out of the square rote

68. anonymous

$125=5^3$ and so $\sqrt{125}=5^{\frac{3}{2}}$

69. anonymous

i dont understand that

70. anonymous

just a moment ago we found out that $\sqrt{x}=x^{\frac{1}{2}}$ right?

71. anonymous

$\sqrt{125}=125^{\frac{1}{2}}=(5^3)^{\frac{1}{2}}=5^{\frac{3}{2}}$

72. anonymous

yes. oh okay i understand

73. anonymous

so then what would we have

74. anonymous

so now we have $50\log_5(5^{\frac{3}{2}})$ $50\times \frac{3}{2}$ etc

75. anonymous

as soon as you say $\log_b(\text{whatever})$ you should think of ways to write "whatever" as a power of b

76. anonymous

*see

77. anonymous

oh okay!!! wow!

78. anonymous

can u give me an example to try?

79. anonymous

$\log_3(\sqrt[5]{27})$

80. anonymous

can u do one with a number in front of the log

81. anonymous

the number out front does not really make any difference, but if you like i can write $10\log_3(\sqrt[5]{27})$

82. anonymous

okay let me get my paper lol

83. anonymous

6?

84. anonymous

yup!

85. anonymous

OMG

86. anonymous

really? i got it right?

87. anonymous

don't be so shocked, you will get the hang of it

88. anonymous

can u help me with more? lol

89. anonymous

after a couple practice problems you will think "not much to this"

90. anonymous

ok we can do another before i go

91. anonymous

$\log _{3} 4 - 1/2 \log _{3} (6x-5)$

92. anonymous

holy moly what are the instruction, write as a single log?

93. anonymous

lol i have no idea what to do at all

94. anonymous

first you need to make sure what the instructions are. what does it say before the problem?

95. anonymous

all it says is : solve

96. anonymous

maybe u an help me with a different one instead?

97. anonymous

you cannot "solve' because there is not an equation here. maybe it says "write as one log"

98. anonymous

which we can do if you like

99. anonymous

sure lets write it as one log

100. anonymous

by what we know before $n\log(x)=\log(x^n)$ so $\frac{1}{2}\log(6x-5)=\log((6x-5)^{\frac{1}{2}})=\log(\sqrt{6x-5})$

101. anonymous

also $\log(a)-\log(b)=\log(\frac{a}{b})$ so you can write the whole thing as $\log_3(\frac{4}{\sqrt{6x-5}})$

102. anonymous

103. anonymous

yes, although the text did not seem to provide a question

104. anonymous

okay. thank you so so so much for all your help! maybe i will pass math because of you! :D thanks again and have a wonderful day!!

105. anonymous

you too, good luck