### Jefferson's FINAL SCRIBE - ACTION ON EXPONENTS AND LOGARITHMS!!

HEY HO! Its scribe specialist and former scribe emperor, Jefferson presenting today's scribe post! Attention! The logarithm test will be on Friday, the wiki project will also be due on Friday.

Mr.K started off today's lesson with a quiz. For those who were absent here is a copy of the quiz:

Now lets go through the quiz together shall we?

(1) Find log

_{10}(0.001)

solution:

log10(0.001) = -3

(2) Find log

_{2}(64) + log

_{3}(9)

solution:

log2(64) + log3(9) = 6 + 2

log2(64) + log3(9) = 8

(3) 3

^{x}= 7; which is true?

(a) 7= log

_{x}(3) (b) x = log

_{7}(3) (c) 3 = log

_{x}(7) (d) 7 = log

_{3}(x) (e) x=log

_{3}(7)

solution:

REMEMBER a log is an exponent. "7" is the power so the log format would be:

log

_{base}power = exponent

so the answer would be (e) x=log

_{3}(7)

(4) if log

_{a}(25) = 4, what is a?

(a) 1/5 (b) sqr(5) (c) 5 (d) log2 (5) (e) none of these

Log

_{a}(25) = 4

a = sqr(5)

(5) find log

_{100}(1 000 000)

log100(1000000) = 3

(6) If log

_{10}2 ≈ 0.301, which of these is the closest to the log

_{10}(2000)?

(a) 0.6 (b) 3 (c ) 6 (d) 30 (e) 60 (f)6

log

_{10}(2000) = log

_{10}(2 x 1000)

log

_{10}(2000) = log10(2) + log

_{10}(1000)

log

_{10}(2000) = 0.301 + 3

log

_{10}(2000) = 3.301

so the answer is (b) 3

(7) If log

_{10}2 ≈ 0.301, which of these is the closest to log

_{10}(8)?

(a) 0.3 (b) 0.6 (c )0.9 (d) 1.2 (e) 2.4 (f) 6

log

_{10}8 = log

_{10}(2)

^{3}

log

_{10}8 = log

_{10}(2)

^{3}

log

_{10}8 = 3log

_{10}(2)

log

_{10}8 = 3(0.301)

log

_{10}8 = 0.903

and the answer is (c) 0.9

(8) If log

_{10}2 ≈ 0.301, which of these is the closest to log

_{2}(10)?

(a) 0.5 (b) 1 (c ) 3 (d)5 (e)20 (f)50

log2(10) = log2(2*5)

log2(10) = log2(2) + log2(5)

log2(10) = 1 + 2.32

log2(10) = 3.32

and the answer is (c) 3

(9) find 2

^{log2(17)}

solution:

17

(10) find 4

^{log2(3)}

4

^{log2(3)}= 9

(2

^{2})

^{log2(3)}= 9

2

^{2log2(3)}= 9

Log

_{2}(3)

^{2}= 9

2log

_{2}(9) = 9

*not sure if this is correct Mr k , or anyone else please correct help me correct this*

(11) if log

_{3}((10) = k, then log

_{9}(10) =

(a) 2K (b) k/3 (c )k/2 (d)k

^{2}(e)2.4 (f)sqr(K)

log

_{3}(10) = K

3

^{K }=10

*since we don't know what the log9(10) is we make it equal to any variable we name, lets call it "n" courtesy of Mr k.*

log

_{9}(10) = n

9

^{n}= 10

3

^{K}= 9

^{n}

3

^{K}= 3

^{2n}

K = 2n

k/2 = n

and the answer is k/2

After the quiz, we were put into groups and faced a difficult word problem. Again, for those who were absent, here is a copy of the hand out:

AND NOW LETS GO OVER THE QUESTIONS!!

(a) The intensity of thunder is 10

^{12}times the intensity of barely audible sound. What is the decibal level of thunder?

I = 10

^{12}* 10

^{-12}

I = 1

* I = intensity of thunder * barely audible sound *

D = 10log(1/10

^{-12})

D = 10log10

^{12}

D = 120

(b) If the decibal levels of a subway train entering a station and normal conversation are 100 dB and 60 dB, respectively, how many times as intense as normal conversation is the noise of the subway train?

D

_{t}= 100

100 = 10log(I * 10

^{12})

10 = logI + log10

^{12}

-12 + 10 = logI + logI

-2 = logI

10

^{-2}= I

_{t}

D

_{c}= 60

60 = 10log(10

^{12}* I)

6 = log(10)

^{12}+ logI

6 = 12 + logI

-12 + 6 = logI

-6 = logI

10

^{-6}= log I

_{n}

I

_{t}/I

_{n}= 10

^{-12}/10

^{-6}

= 10

^{4}

=10000 times more intense

*I had just resently obtain the last bit of the answers after class, unfortunately I do not understand what's going on. Hopefully Mr K will explain this solution when he sees this scribe*

WELL I AM DONE!! Took me way too long to do this scribe. Operation rebellion will commence by a new member of the anti scribe take over movement, namely me =). My first victim will be NONE OTHER THAN JANET!!

that is all, *runs away*

later days, Jefferson

_{}

## 2 Comments:

This is a good post Jefferson. Your pictures of the quiz and questions from class are too small. They do not magnify when clicked on. The thought was good when you added them into your post but unless you have special eyesight they are unreadable.

Mr. Harbeck

Sargent Park School

You're in The Hall Of Fame now! ;-)

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