Given the function f(x) = log2(x + 6), find the value of f^−1(3).

- anonymous

Given the function f(x) = log2(x + 6), find the value of f^−1(3).

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- schrodinger

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- anonymous

I really need help with understanding this.

- anonymous

Answer Choices:
a. f−1(3) = 2
b. f−1(3) = 3
c. f−1(3) = 9
d. f−1(3) = 18

- anonymous

@zepdrix help please

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## More answers

- anonymous

@SkaterBoyShawn can you help me?

- anonymous

yes

- anonymous

okay good

- freckles

\[f^{-1}(3)=a \implies f(a)=3 \\ \text{ so you need to solve } \log_2(a+6)=3 \text{ for } a \]

- anonymous

@freckles in my notes it was saying how i had to find the inverse function or something like that

- anonymous

let me solve it 1sec

- anonymous

ok @SkaterBoyShawn

- freckles

You can find the inverse function then plug in 3
or you can just solve the equation above for a

- anonymous

ugh im still confused

- anonymous

i think i got the answer

- freckles

on what part?

- freckles

like why \[f^{-1}(3)=a \implies f(a)=3 ? \\ \text{ or on solving } \log_2(a+6)=3 \text{ for } a?\]

- anonymous

how? @SkaterBoyShawn and @freckles everything like what do you do with log2?

- freckles

you want to write in the equivalent exponential form

- freckles

recall
\[\log_b(x)=y \implies b^{y}=x\]

- anonymous

i dont get any of this at all. Math is my weakest subject

- freckles

so you don't know how to compare
\[\log_b(x)=y \text{ to } \log_2(a+6)=3 \\ \text{ then use that } \log_b(x)=y \implies b^y=x \\ \text{ to write } \log_2(a+6)=3 \text{ in exponential form }\]?

- freckles

|dw:1433709916685:dw|
what is in place of the b?
in place of the x?
in place of the y?

- anonymous

i dont know how to write that in exponential form.

- anonymous

f⁻¹(3) = 2³-6 = 2 if u in verse it it i u would get 2^y = x+6
x = 2^y-6
f⁻¹(x) = 2ˣ-6

- freckles

I'm just asking you to identify b,x, and y in that comparison

- freckles

|dw:1433710114097:dw|
do you not see that b is 2?
can you try to identify what is place of x and y now?

- anonymous

i see that now. x is 6 and y is 3.

- freckles

well x is everything in that log thing
so x is a+6 not just 6

- anonymous

oh okay

- freckles

\[\log_b(x)=y \implies b^y=x \\ \text{ so we can use this to write } \log_2(a+6)=3 \implies 2^{3}=a+6\]

- freckles

can you solve 2^3=a+6 for a ?

- anonymous

a=2

- freckles

yep

- anonymous

y would be
f⁻¹(x) = 2ˣ-6 not sure yet for b is

- freckles

so since f(2)=3
then f^(-1)(3)=2

- anonymous

oh okay then

- freckles

And I think @SkaterBoyShawn is trying to attempt to find the inverse function and then plug in there instead which is cool too

- anonymous

so for problems like that all you have to do is plug in the numbers and solve for a?

- freckles

\[y=\log2(x+6) \\ 2^y=x+6 \\ x=2^y-6 \\ f^{-1}(x)=2^x-6\]
his inverse function is right

- anonymous

correct @freckles

- freckles

so if you have done it this way you can just replace the x with 3

- anonymous

thank you

- anonymous

do u guys understand the the inverse function

- anonymous

no

- anonymous

no u don't

- freckles

yeah
the function f(x)=log_b(x) is one to one so it does have an inverse function
to find something like this:
\[f^{-1}(3) \]
where f^(-1) is the inverse of f
you could say well I know the following
\[f^{-1}(3)=a \implies f(a)=3\]
so yes in the original you can replace x with a (or just leave the x there if you are feeling lazy) and then solve for a (or x if you left a x instead)

- anonymous

@freckles u understand

- freckles

understand what?

- anonymous

oh okay I got it now i think @freckles

- freckles

In general:
Let's say you have the following problem:
\[\text{ Find } f^{-1}(m) \text{ if } f(x)=\log_b(x) \\ \\ \text{ assume } f^{-1}(m)=a \text{ well this means } f(a)=m \\ \\ \text{ so you replace } x \text{ with } a \text{ like so } \\ f(a)=\log_b(a) \\ \text{ now remember } f(a)=m \\ \text{ so we can replace } f(a) \text{ with } m \\ m=\log_b(a) \\ \text{ and solve for } a \\ b^m=b^{\log_b(a)} \\ b^m=a \\ a=b^m \\ \text{ so remember } a=f^{-1}(m) \\ \text{ and that is what we were aiming to find } \\ f^{-1}(m)=a=b^m \\ f^{-1}(m)=b^m\]

- freckles

notice the is the inverse function of f(x) evaluated at x=m

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