- anonymous

Given the parent functions f(x) = 5x − 1 and g(x) = 3x − 9, what is g(x) − f(x)?

- katieb

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- UsukiDoll

given:
g(x) = 3x-9
f(x) = 5x-1
and we need to solve g(x) - f(x)

- 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

3x-9(-5x+1)?

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## More answers

- 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

oh the equation is 3^x

- anonymous

im sorry

- UsukiDoll

ahhhhhhhh!
ok no problem... just rewrite what f(x) and g(x) was supposed to be

- UsukiDoll

ok was g(x) = 3^x-9 ?

- anonymous

f(x)=5x-1 g(x)=3^x-9

- anonymous

- UsukiDoll

ok we do the same process again
g(x) -f(x)
3^x-9-(5x-1)

- UsukiDoll

actually we have already done the distribution of the negative,
3^x-9-5x+1
now combine like terms for -9+1

- anonymous

-8

- UsukiDoll

\[3^x-9+1-5x\]

- UsukiDoll

yup \[3^x-8-5x\]

- UsukiDoll

since we can't combine any more terms, that's the answer

- 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

let me type the equation correctly

- anonymous

\[f(x)=\log_{10} x \]
g(x)=5x-2

- anonymous

firt answer is log10 x^5-2
second is log10(5x-2)^x

- anonymous

- 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

\[\log10_x[5x-2]\] use distribution

- anonymous

what do you distribute

- UsukiDoll

distribute the log

- UsukiDoll

\[\log_{10}x[5x-2]\]

- anonymous

i dont have a graphing or scientific calc to solve

- UsukiDoll

we can just distribute the log
|dw:1435728843362:dw|

- anonymous

x(log10 5x-log10 2)?

- UsukiDoll

|dw:1435728950642:dw| now distribute the log with the negative -2

- UsukiDoll

distribution is multiplication what's that log times -2?

- anonymous

idk i dont have a calc

- UsukiDoll

we don't need a calculator this time

- 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

so its either c or d?

- anonymous

wouldnt it be d because its - and not +?

- UsukiDoll

I'm not sure.. @jim_thompson5910 can help out.. I need to go

- anonymous

- 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

yes and i corrected how it is right under neath that

- 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

so d?

- anonymous

its backwards but

- jim_thompson5910

can you post a screenshot of the full thing? some symbols didn't make it (at least I think?)

- anonymous

##### 1 Attachment

- jim_thompson5910

D is close but not quite 100%

- jim_thompson5910

there has to be a typo that they made

- anonymous

##### 1 Attachment

- anonymous

could it be c because when you multiply its equivalent to adding with exponets or something like that lol?

- 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

i still dont understand please help its one in the morning

- 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

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

oh the other one has a typo I think

- anonymous

which would you pick

- jim_thompson5910

D I guess since it's close just off by a sign

- anonymous

it was wrong i guess it was c

- anonymous

thanks thoough!

- jim_thompson5910

you'll have to tell your teacher about the typo

- 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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