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December 3rd, 2010, 11:30 AM   #1
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Exponential and Logarithmic functions and equations Part 2

Hi it's me again. This is part 2 of the pretest. ty


[attachment=3:9mhh7y0t]11.PNG[/attachment:9mhh7y0t]
11.
(8 x log 32/log 4 + 8 X log 8/log 2) (log2 16)2
( 20 + 24 ) - (log2 16)2
44 (log 16/log2)2
44 (4)2
22 16
28


[attachment=2:9mhh7y0t]12.PNG[/attachment:9mhh7y0t]
12.
This I'm Totally lost on - how to do it?

[attachment=1:9mhh7y0t]13.PNG[/attachment:9mhh7y0t]
13.
I believe I know how to do this and will post it soon once I get time to wrap my mind around it. Feel free to solve it if you like. I'll do it myself also. ty


[attachment=0:9mhh7y0t]14.PNG[/attachment:9mhh7y0t]

Ok and this one is the big mess up for me. Please help me with detailed instructions on how to solve it and the answer please.
14.
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File Type: png 11.PNG (20.5 KB, 246 views)
File Type: png 12.PNG (54.4 KB, 246 views)
File Type: png 13.PNG (27.0 KB, 246 views)
File Type: png 14.PNG (88.0 KB, 246 views)
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December 3rd, 2010, 12:14 PM   #2
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Re: Exponential and Logarithmic functions and equations Part

hello there
u have amistake in question 11 the answer should be 16 notic that log 32 +log8 =log(32*=log 256=log4^4 since the base is 4 u get 4 so u have 8*4=32 now the other part is truly 16 so u have 32-16=16
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December 3rd, 2010, 12:23 PM   #3
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Re: Exponential and Logarithmic functions and equations Part

last question
a: true loga+logb= loga*b this is arule in logarithm( u should make areview on the rules of logarithm)
b:false: loga-logb=log(a/b) (this is another rule) so the answer should be log(x^2/y)
c:false mlogn=logn^m not the other way (this is another rule)
d:true
e:false loga)^m is not equal to 3logalogmn)^3=(logmn)^3 this is the simplest form
f:false the root should be inside the log meaning log root(c^2) :root i mean the third root
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December 3rd, 2010, 12:28 PM   #4
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Re: Exponential and Logarithmic functions and equations Part

Hi Bday,

It might be useful to post these things one at a time so multiple helpers can jump on board.

I'll pick 12A

Simplify:















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December 3rd, 2010, 01:07 PM   #5
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Re: Exponential and Logarithmic functions and equations Part

Quote:
Originally Posted by islam
hello there
u have amistake in question 11 the answer should be 16 notic that log 32 +log8 =log(32*=log 256=log4^4 since the base is 4 u get 4 so u have 8*4=32 now the other part is truly 16 so u have 32-16=16
The bases are different. Without using change of base formula.....

-(\log_216)^2=" />








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December 3rd, 2010, 01:19 PM   #6
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Re: Exponential and Logarithmic functions and equations Part

13A.













-6 is no good! x = 3.
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December 3rd, 2010, 01:34 PM   #7
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Re: Exponential and Logarithmic functions and equations Part


13B















Disregard
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December 3rd, 2010, 01:54 PM   #8
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Re: Exponential and Logarithmic functions and equations Part

For 12B, use:



so that becomes





If your book meant then we would have:

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December 3rd, 2010, 02:47 PM   #9
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Re: Exponential and Logarithmic functions and equations Part

Another look at 12A:











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December 3rd, 2010, 06:39 PM   #10
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Re: Exponential and Logarithmic functions and equations Part

Quote:
Originally Posted by masters
Quote:
Originally Posted by islam
hello there
u have amistake in question 11 the answer should be 16 notic that log 32 +log8 =log(32*=log 256=log4^4 since the base is 4 u get 4 so u have 8*4=32 now the other part is truly 16 so u have 32-16=16
The bases are different. Without using change of base formula.....

-(\log_216)^2=" />








yes right sorry my mistake
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