express as a single logarithm: PLEASE HELP with a few questi

[murkr]

hello everyone, i am new here and plan to use this forum often, you guys seem very friendly. I am in college algebra right now and im having a huge problem with a few questions on my review packet for the exam coming up, the test is in a few days and these questions are going to be similar to the exam. i have been looking at my math book for the last 3-4 hours and i just cant find any similar questions, im stuck. very frustrated & stressed from this as well.

i scanned the math problem packet my teacher gave me to my pc, i mostly need help on "Express as a single logarithm" problems, they are the last 4 questions. but if you got extra time i would really appreciate it if you can help me on the top questions as well, but i know that might be asking to much so if anything just an answer to atleast one question from problems 11,12,13, or 14 would be great. thank you so much



if possible show your work to how you solved the problem, because i really would like to know how to do these. but if you cant i understand, it would be hard to type it out.

Thanks again everyone, this really is going to help me understand it a lot better.

 

[arthur ohlsten]

8) A is correct

9) D is correct
1-1/x^3= [x^3-1] /x^3
factor [x^3-1]
x^3-1=[x-1][x^2+x+1]

or log [ 1-1/x^3]=log{[x-1][x^2+x+1]/x^3}
log[x-1] log[x^2+x+1] -3 log x


10) A is correct


11) A is correct

12 A is correct

13) D is correct
=====================================
14)
x^2-6x-27=[x+3][x-9]

x^2-2x-15=[x-5][x+3]

x^2-18x+81=[x-9]^2
the inverse ln is
[x+3][x-9] [x+6][x-9]^2
--------------------------------
[x-5] [x-5][x+3]

[x-9]^3 [x+6]
-----------------
[x-5]^2

Answer C

please check my work for errors
Arthur

 

[murkr]

are you sure problem #8 is correct? because someone posted something else, but i dont know how to copy it to this forum, it wont let me, i thought #8 was C as well, but this other guy said it was A and showed some work

also from problem #13 i got "B" and you got "D"

 

[BigGlenntheHeavy]

$\displaystyle 8) \ Base \ is \ understood \ to \ be \ 17, \ then \ we \ have: \ log\frac{(5^{1/4})}{n^2m^2}$

$\displaystyle log\frac{(5^{1/4})}{n^2m^2} \ = \ log \ (5^{1/4})-log(nm)^2$

$\displaystyle = \ \frac{1}{4}log(5)-2log|nm| \ = \ \frac{1}{4}log(5)-2[log|n|+log|m|] \ = \ \frac{1}{4}log(5)-2log|n|-2log|m|$

$\displaystyle Hence, \ A \ is \ correct.$

$\displaystyle Note: \ When \ doing \ problems \ involving \ logs \ and \ the \ answer \ in \ the \ back \ of \ the \ book \ is \ different$

$\displaystyle \ than \ yours, \ it \ "aint \ necessarily \ so" \ that \ your \ answer \ is \ wrong \ as \ there \ are \ so \ many \ ways \ to$

$\displaystyle finalize \ a \ log \ problem. \ Trick: \ Plug \ a \ number \ (no \ 0 \ or \ 1) \ into \ your \ answer \ and \ the \ answer \ in$

$\displaystyle the \ back \ of \ the \ book. \ If \ they \ are \ the \ same, \ then \ you \ are \ correct, \ just \ with \ a \ different \ form \ of$

$\displaystyle \ the \ same \ problem.$

$\displaystyle For \ example. \ problem \ 9, \ D \ is \ correct.$

$\displaystyle log\bigg|1-\frac{1}{x^3}\bigg| \ = \ log\bigg|\frac{x^3-1}{x^3}\bigg| \ = \ log|x^3-1|-log|x^3|. \ I \ would \ stop \ here.$

$\displaystyle However, \ they \ continue, \ to \ wit: \ log||x-1||x^2+x+1||-3log|x| \ = \ log|x-1|+log|x^2+x+1|-3log|x|$

$\displaystyle Both \ answers \ are \ correct, \ just \ of \ different \ forms.$

 

[murkr]

$\displaystyle 8) \ Base \ is \ understood \ to \ be \ 17, \ then \ we \ have: \ log\frac{(5^{1/4})}{n^2m^2}$

$\displaystyle log\frac{(5^{1/4})}{n^2m^2} \ = \ log \ (5^{1/4})-log(nm)^2$

$\displaystyle = \ \frac{1}{4}log(5)-2log|nm| \ = \ \frac{1}{4}log(5)-2[log|n|+log|m|] \ = \ \frac{1}{4}log(5)-2log|n|-2log|m|$

$\displaystyle Hence, \ A \ is \ correct.$

$\displaystyle Note: \ When \ doing \ problems \ involving \ logs \ and \ the \ answer \ in \ the \ back \ of \ the \ book \ is \ different$

$\displaystyle \ than \ yours, \ it \ "aint \ necessarily \ so" \ that \ your \ answer \ is \ wrong \ as \ there \ are \ so \ many \ ways \ to$

$\displaystyle finalize \ a \ log \ problem. \ Trick: \ Plug \ a \ number \ (no \ 0 \ or \ 1) \ into \ your \ answer \ and \ the \ answer \ in$

$\displaystyle the \ back \ of \ the \ book. \ If \ they \ are \ the \ same, \ then \ you \ are \ correct, \ just \ with \ a \ different \ form \ of$

$\displaystyle \ the \ same \ problem.$

$\displaystyle For \ example. \ problem \ 9, \ D \ is \ correct.$

$\displaystyle log\bigg|1-\frac{1}{x^3}\bigg| \ = \ log\bigg|\frac{x^3-1}{x^3}\bigg| \ = \ log|x^3-1|-log|x^3|. \ I \ would \ stop \ here.$

$\displaystyle However, \ they \ continue, \ to \ wit: \ log||x-1||x^2+x+1||-3log|x| \ = \ log|x-1|+log|x^2+x+1|-3log|x|$

$\displaystyle Both \ answers \ are \ correct, \ just \ of \ different \ forms.$

thank you so much, that really helped me understand much more. im just having such a hard time with this.

i see you just verified #9 was D thank you.

but what about #13, im not sure if that square root sign is in both fractions or just the top one, im questions B and D for question #13. i asked two diff people and both gave me different answers so its really making it hard for me to understand it. and i cant find any similar problem in my book.

 

[BigGlenntheHeavy]

$\displaystyle Again, \ we'll \ forego \ the \ base.$

$\displaystyle \frac{1}{2}[log|x-3|-log|x|] \ = \ \frac{1}{2}log\frac{|x-3|}{|x|}$

\(\displaystyle = log\bigg[\frac{|x-3|}{|x|}\bigg]^{1/2} \ = \ log\sqrt\bigg[\frac{|x-3|}{|x|}\bigg] \ = \ B, \ I \ think \ as \ I \ can't \ read \ D.\)

$\displaystyle Note: \ B \ should \ have \ absolute \ value \ signs, \ what \ if \ x \ = \ 2?$

 

[murkr]

$\displaystyle Again, \ we'll \ forego \ the \ base.$

$\displaystyle \frac{1}{2}[log|x-3|-log|x|] \ = \ \frac{1}{2}log\frac{|x-3|}{|x|}$

\(\displaystyle = log\bigg[\frac{|x-3|}{|x|}\bigg]^{1/2} \ = \ log\sqrt\bigg[\frac{|x-3|}{|x|}\bigg] \ = \ B, \ I \ think \ as \ I \ can't \ read \ D.\)

$\displaystyle Note: \ B \ should \ have \ absolute \ value \ signs, \ what \ if \ x \ = \ 2?$

ohh i see now thank you BigGlenntheHeavy, you've been a big help , so then #14 is C correct? thats what answer i got when i did it

 

[BigGlenntheHeavy]

$\displaystyle 14 \ = \ B, \ again \ where \ are \ the \ absolute \ value \ signs?$

 

[murkr]

$\displaystyle 14 \ = \ B, \ again \ where \ are \ the \ absolute \ value \ signs?$

i dont know, my college teacher gave us all this packet. are you sure there are suppose to be absolute value signs on these type of questions? maybe this is simpler versions of these type of problems? i really dont understand why they are not there...
would there be a diff answer since there is no absolute value signs???

and you got B for #14. look above at "arthur ohlsten" posts, did he get most of them wrong or something? are you positive with your answers? i see you understand that it is missing some things, could there be these types of problems without absolute value signs which would result in different answers thats why "arthur ohlsten" got different answers?