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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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)?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

@jim_thompson5910

- 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

@jim_thompson5910

- 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

- 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.