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Mathematics 14 Online
OpenStudy (anonymous):

Given the parent functions f(x) = log10 x and g(x) = 5x − 2, what is f(x) • g(x)?

OpenStudy (anonymous):

@Astrophysics

OpenStudy (astrophysics):

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

OpenStudy (astrophysics):

You're just multiplying the functions

OpenStudy (anonymous):

Thanks! Can I ask one more?

OpenStudy (anonymous):

If f(x) = log2 (x + 4), what is f−1(3)?

OpenStudy (astrophysics):

The \[f^{-1}\] represents inverse

OpenStudy (anonymous):

yes:)

OpenStudy (astrophysics):

To find the inverse: Replace f(x) with y Switch x's and y's, so put x where y is and x where y is. Solve for y Replace y with f^-1(x)

OpenStudy (astrophysics):

Once you find the inverse just plug in 3 into the function and evaluate :-)

OpenStudy (anonymous):

y=log2(x+4)

OpenStudy (astrophysics):

good keep going

OpenStudy (anonymous):

x=log2(y+4)

OpenStudy (astrophysics):

right

OpenStudy (anonymous):

im confused on how to solve for "Y"

OpenStudy (astrophysics):

Since it's \[\log_2\] as the base we will have to take the power of 2 to on both sides so the following \[\huge 2^x = 2^{\log_2(y+4)} \implies 2^x = y+4\]

OpenStudy (anonymous):

hmm then wouldnt we have to get y alone?

OpenStudy (anonymous):

Would the answer be 8?

OpenStudy (loser66):

f(a) =b , hence \(f^{-1} (b) =a\) ok?

OpenStudy (loser66):

we need find \(f^{-1} (3) \) of \(f(x) = log 2(x+4)\), right? That is just let log 2(x+4) =3, and solve for x.

OpenStudy (astrophysics):

You can do either way you should get same result

OpenStudy (anonymous):

2?

OpenStudy (astrophysics):

No, \[y=2^x-4 \implies f^{-1}(x) = 2^x-4 \implies f^{-1}(3) = 2^3-4\]

OpenStudy (anonymous):

oh! okay so then I got, 4

OpenStudy (loser66):

confirm: \(log (2(x+4))\) or (x+4)*log 2??

OpenStudy (astrophysics):

Yes, 4 sounds betters :)

OpenStudy (anonymous):

2x2x2=4x2=8 8-4=4 :)

OpenStudy (astrophysics):

I think the original question was \[\log_2(x+4)\] right?

OpenStudy (anonymous):

yes

OpenStudy (astrophysics):

Ok we're good then

OpenStudy (astrophysics):

Thanks @Loser66

OpenStudy (anonymous):

yay! Thanks guysss:)

OpenStudy (loser66):

ok, clear.

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