OpenStudy (anonymous):

If f(x) = log3 (x + 1), what is f−1(2)?

3 years ago
OpenStudy (anonymous):

the question is a little vague.. please specify

3 years ago
OpenStudy (anonymous):

1, 8, 10, or 27

3 years ago
OpenStudy (anonymous):

The answers are what I just typed, they are the ones to pick from atleast... I ma just utterly stuck on this equation.

3 years ago
OpenStudy (anonymous):

its okay. are you sure its asking for f-1(2)?

3 years ago
OpenStudy (anonymous):

no sorry hold on it's f^-1 (2)

3 years ago
OpenStudy (anonymous):

hey i have that same problem

3 years ago
OpenStudy (anonymous):

.o. @naahhbass algebra two?

3 years ago
OpenStudy (anonymous):

yehh for my worksheet

3 years ago
OpenStudy (anonymous):

also is the 3 of log 3 the base or coefficient?

3 years ago
OpenStudy (anonymous):

base

3 years ago
OpenStudy (anonymous):

see that changes the problem. give me a sec

3 years ago
OpenStudy (anonymous):

okki dokki

3 years ago
OpenStudy (anonymous):

okay so the answer is 1

3 years ago
OpenStudy (anonymous):

but here is how you get it: calculating the log is not the problem. its how to to get to f^-1(2) that throws you off?

3 years ago
OpenStudy (anonymous):

yes it is

3 years ago
OpenStudy (anonymous):

its simple. so if you look at f^-1(2).. you can also write it as \[\frac{ 1 }{ f(2) }\]

3 years ago
OpenStudy (anonymous):

so you just calculate f(2) as it is: \[f(2)= \log_{3} (2+1)= \log_{3} 3 = 1 \]

3 years ago
OpenStudy (anonymous):

and 1/1 is 1 tadaaa c:

3 years ago
OpenStudy (anonymous):

Awesome thank you so very much!

3 years ago
OpenStudy (anonymous):

you are welcome

3 years ago
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