anonymous
  • anonymous
Given the parent functions f(x) = 5x − 1 and g(x) = 3x − 9, what is g(x) − f(x)?
Mathematics
katieb
  • katieb
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UsukiDoll
  • UsukiDoll
given: g(x) = 3x-9 f(x) = 5x-1 and we need to solve g(x) - f(x)
UsukiDoll
  • UsukiDoll
so we plug these functions into g(x)-f(x) 3x-9-(5x-1) now distribute the negative on the right hand side of the equation
anonymous
  • anonymous
3x-9(-5x+1)?

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UsukiDoll
  • UsukiDoll
ok... that's good 3x-9-5x+1 now we combine like terms.. we can also rearrange this equation to make it easier to solve so instead of 3x-9-5x+1 we can rewrite it as 3x-5x-9+1
anonymous
  • anonymous
oh the equation is 3^x
anonymous
  • anonymous
im sorry
UsukiDoll
  • UsukiDoll
ahhhhhhhh! ok no problem... just rewrite what f(x) and g(x) was supposed to be
UsukiDoll
  • UsukiDoll
ok was g(x) = 3^x-9 ?
anonymous
  • anonymous
f(x)=5x-1 g(x)=3^x-9
anonymous
  • anonymous
UsukiDoll
  • UsukiDoll
ok we do the same process again g(x) -f(x) 3^x-9-(5x-1)
UsukiDoll
  • UsukiDoll
actually we have already done the distribution of the negative, 3^x-9-5x+1 now combine like terms for -9+1
anonymous
  • anonymous
-8
UsukiDoll
  • UsukiDoll
\[3^x-9+1-5x\]
UsukiDoll
  • UsukiDoll
yup \[3^x-8-5x\]
UsukiDoll
  • UsukiDoll
since we can't combine any more terms, that's the answer
anonymous
  • anonymous
Given the parent functions f(x) = log10 x and g(x) = 5x − 2, what is f(x) • g(x)? f(x) • g(x) = log10 x5x − 2 f(x) • g(x) = log10 (5x − 2)x f(x) • g(x) = 5x log10 x + 2 log10 x f(x) • g(x) = 2 log10 x − 5x log10 x
anonymous
  • anonymous
let me type the equation correctly
anonymous
  • anonymous
\[f(x)=\log_{10} x \] g(x)=5x-2
anonymous
  • anonymous
firt answer is log10 x^5-2 second is log10(5x-2)^x
anonymous
  • anonymous
UsukiDoll
  • UsukiDoll
we are using multiplication \[f(x) \cdot g(x) \] once again we place our f(x) function and g(x) into the formula
UsukiDoll
  • UsukiDoll
\[\log10_x[5x-2]\] use distribution
anonymous
  • anonymous
what do you distribute
UsukiDoll
  • UsukiDoll
distribute the log
UsukiDoll
  • UsukiDoll
\[\log_{10}x[5x-2]\]
anonymous
  • anonymous
i dont have a graphing or scientific calc to solve
UsukiDoll
  • UsukiDoll
we can just distribute the log |dw:1435728843362:dw|
anonymous
  • anonymous
x(log10 5x-log10 2)?
UsukiDoll
  • UsukiDoll
|dw:1435728950642:dw| now distribute the log with the negative -2
UsukiDoll
  • UsukiDoll
distribution is multiplication what's that log times -2?
anonymous
  • anonymous
idk i dont have a calc
UsukiDoll
  • UsukiDoll
we don't need a calculator this time
UsukiDoll
  • UsukiDoll
\[\log_{10}x[5x-2] \rightarrow \log_{10}x[5x]-\log_{10}x[2]\] based on your choices, it doesn't require you to calculate the log portion
anonymous
  • anonymous
so its either c or d?
anonymous
  • anonymous
wouldnt it be d because its - and not +?
UsukiDoll
  • UsukiDoll
I'm not sure.. @jim_thompson5910 can help out.. I need to go
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
This question right? Given the parent functions f(x) = log10 x and g(x) = 5x − 2, what is f(x) • g(x)? f(x) • g(x) = log10 x5x − 2 f(x) • g(x) = log10 (5x − 2)x f(x) • g(x) = 5x log10 x + 2 log10 x f(x) • g(x) = 2 log10 x − 5x log10 x
anonymous
  • anonymous
yes and i corrected how it is right under neath that
jim_thompson5910
  • jim_thompson5910
\[\Large f(x) = \log_{10}(x)\] \[\Large g(x) = 5x-2\] You will distribute the "log(x)" portion through like so \[\Large f(x) \cdot g(x) = \log_{10}(x)\cdot(5x-2)\] \[\Large f(x) \cdot g(x) = \log_{10}(x)\cdot 5x + \log_{10}(x)\cdot(-2)\] \[\Large f(x) \cdot g(x) = 5x\log_{10}(x)-2\log_{10}(x)\]
anonymous
  • anonymous
so d?
anonymous
  • anonymous
its backwards but
jim_thompson5910
  • jim_thompson5910
can you post a screenshot of the full thing? some symbols didn't make it (at least I think?)
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
D is close but not quite 100%
jim_thompson5910
  • jim_thompson5910
there has to be a typo that they made
anonymous
  • anonymous
anonymous
  • anonymous
could it be c because when you multiply its equivalent to adding with exponets or something like that lol?
jim_thompson5910
  • jim_thompson5910
hints: * look at rule 2 at the link below http://www.purplemath.com/modules/logrules.htm * factor 5x-5 and you'll see something cancel
anonymous
  • anonymous
i still dont understand please help its one in the morning
jim_thompson5910
  • jim_thompson5910
\[\Large f(x) = \log_{3}(5x-5)\] \[\Large g(x) = \log_{3}(x-1)\] Use rule 2 from that link \[\Large f(x)-g(x) = \log_{3}(5x-5)-\log_{3}(x-1)\] \[\Large f(x)-g(x) = \log_{3}\left(\frac{5x-5}{x-1}\right)\] \[\Large f(x)-g(x) = \log_{3}\left(\frac{5(x-1)}{x-1}\right)\] \[\Large f(x)-g(x) = \log_{3}\left(\frac{5\cancel{(x-1)}}{\cancel{x-1}}\right)\] \[\Large f(x)-g(x) = \log_{3}\left(5\right)\]
anonymous
  • anonymous
i thought we were on the other one still lol but yeah it make since now that i know what were talking about
jim_thompson5910
  • jim_thompson5910
oh the other one has a typo I think
anonymous
  • anonymous
which would you pick
jim_thompson5910
  • jim_thompson5910
D I guess since it's close just off by a sign
anonymous
  • anonymous
it was wrong i guess it was c
anonymous
  • anonymous
thanks thoough!
jim_thompson5910
  • jim_thompson5910
you'll have to tell your teacher about the typo
UsukiDoll
  • UsukiDoll
I second on that typo comment. Even I was in ??????????? land when I found out that there was a sign switch on C and D. It's like we have the answer, but the choices they gave messed up on the signs.

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