My Math Forum Exponential and Logarithmic functions and equations Part 2

 Algebra Pre-Algebra and Basic Algebra Math Forum

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]

14.
Attached Images
 11.PNG (20.5 KB, 246 views) 12.PNG (54.4 KB, 246 views) 13.PNG (27.0 KB, 246 views) 14.PNG (88.0 KB, 246 views)

 December 3rd, 2010, 12:14 PM #2 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 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
 December 3rd, 2010, 12:23 PM #3 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 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
 December 3rd, 2010, 12:28 PM #4 Member   Joined: Nov 2009 Posts: 67 Thanks: 0 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: $9 \log_aa\sqrt[3]{a}+3(\log_aab^2-2\log_ab)=$ $\log_a(a\sqrt[3]{a})^9+3\log_a(ab^2)-6\log_ab=$ $\log_a\left(a^{\frac{4}{3}}\right)^9+\log_a\left(a b^2\right)^3-\log_ab^6=$ $\log_aa^{12}+\log_aa^3b^6-\log_ab^6=$ $\log_aa^{12}+\log_aa^3+\log_ab^6-\log_ab^6=$ $\log_a(a^{12} \cdot a^3)+0=$ $\log_aa^{15}=\underline{{\color{red}15}}$
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.....

$8(\log_432+\log_2-(\log_216)^2=" />

$8\log_432+8\log_28-(\log_216)^2=$

$\log_432^8+\log_28^8-(4)^2=$

$\log_44^{20}+\log_22^{24}-16=$

$\underline{{\color{red}20+24-16=28}}$

 December 3rd, 2010, 01:19 PM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,948 Thanks: 1140 Math Focus: Elementary mathematics and beyond Re: Exponential and Logarithmic functions and equations Part 13A. $\log_2(x\,+\,5)\,+\,\log_2(x\,-\,2)\,=\,3$ $\log_2((x\,+\,5)(x\,-\,2))\,=\,3$ $\log_2(x^2\,+\,3x\,-\,10)\,=\,3$ $x^2\,+\,3x\,-\,10\,=\,2^3\,=\,8$ $x^2\,+\,3x\,-\,18\,=\,0$ $(x\,+\6)(x\,-\,3)\,=\,0\,\Rightarrow\,x\,=\,-6,\,x\,=\,3$ -6 is no good! x = 3.
 December 3rd, 2010, 01:34 PM #7 Member   Joined: Nov 2009 Posts: 67 Thanks: 0 Re: Exponential and Logarithmic functions and equations Part 13B $\ln(3x^2-4)-\ln x=0$ $\ln\left(\dfrac{3x^2-4}{x}\right)=0$ $e^0=\dfrac{3x^2-4}{x}=1$ $3x^2-4=x$ $3x^2-x-4=0$ $(3x-4)(x+1)=0$ $x=\dfrac{4}{3}\:\:\text{or}\:\:\ x=-1$ Disregard $x=-1$
 December 3rd, 2010, 01:54 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Exponential and Logarithmic functions and equations Part For 12B, use: $\log_{\frac{1}{a}}x=\frac{\log_a x}{\log_a\left(\frac{1}{a}\right)}=-\log_a x$ so that $\log_3\left(3x^2-3x-18\right)+\log_{\frac{1}{3}}\left(9x+18\right)$ becomes $\log_3\left(3x^2-3x-18\right)-\log_3\left(9x+18\right)=\log_3\left(\frac{3x^2-3x-18}{9x+18}\right)=$ $\log_3\left(\frac{x^2-x-6}{3(x+2)}\right)=\log_3\left(\frac{(x-3)(x+2)}{3(x+2)}\right)=\log_3\left(\frac{x-3}{3}\right)=\log_3(x-3)-\log_3 3$ If your book meant $z=\log_3(x-3)$ then we would have: $z-1$
 December 3rd, 2010, 02:47 PM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,948 Thanks: 1140 Math Focus: Elementary mathematics and beyond Re: Exponential and Logarithmic functions and equations Part Another look at 12A: $9\log_aa\sqrt[3]{a}\,+\,3(\log_aab^2\,-\,2\log_ab)$ $9\log_aa^{\frac43}\,+\,3(\log_aab^2\,-\,2\log_ab)$ $12\,+\,3(\log_aab^2\,-\,2\log_ab)$ $12\,+\,3(\log_aa\,+\,\log_ab^2\,-\,2\log_ab)$ $12\,+\,3(1\,+\,2\log_ab\,-\,2\log_ab)$ $12\,+\,3\,=\,15$
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.....

$8(\log_432+\log_2-(\log_216)^2=" />

$8\log_432+8\log_28-(\log_216)^2=$

$\log_432^8+\log_28^8-(4)^2=$

$\log_44^{20}+\log_22^{24}-16=$

$\underline{{\color{red}20+24-16=28}}$
yes right sorry my mistake

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