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

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

Mathematics
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I really need help with understanding this.
Answer Choices: a. f−1(3) = 2 b. f−1(3) = 3 c. f−1(3) = 9 d. f−1(3) = 18
@zepdrix help please

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Other answers:

@SkaterBoyShawn can you help me?
yes
okay good
\[f^{-1}(3)=a \implies f(a)=3 \\ \text{ so you need to solve } \log_2(a+6)=3 \text{ for } a \]
@freckles in my notes it was saying how i had to find the inverse function or something like that
let me solve it 1sec
You can find the inverse function then plug in 3 or you can just solve the equation above for a
ugh im still confused
i think i got the answer
on what part?
like why \[f^{-1}(3)=a \implies f(a)=3 ? \\ \text{ or on solving } \log_2(a+6)=3 \text{ for } a?\]
how? @SkaterBoyShawn and @freckles everything like what do you do with log2?
you want to write in the equivalent exponential form
recall \[\log_b(x)=y \implies b^{y}=x\]
i dont get any of this at all. Math is my weakest subject
so you don't know how to compare \[\log_b(x)=y \text{ to } \log_2(a+6)=3 \\ \text{ then use that } \log_b(x)=y \implies b^y=x \\ \text{ to write } \log_2(a+6)=3 \text{ in exponential form }\]?
|dw:1433709916685:dw| what is in place of the b? in place of the x? in place of the y?
i dont know how to write that in exponential form.
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
I'm just asking you to identify b,x, and y in that comparison
|dw:1433710114097:dw| do you not see that b is 2? can you try to identify what is place of x and y now?
i see that now. x is 6 and y is 3.
well x is everything in that log thing so x is a+6 not just 6
oh okay
\[\log_b(x)=y \implies b^y=x \\ \text{ so we can use this to write } \log_2(a+6)=3 \implies 2^{3}=a+6\]
can you solve 2^3=a+6 for a ?
a=2
yep
y would be f⁻¹(x) = 2ˣ-6 not sure yet for b is
so since f(2)=3 then f^(-1)(3)=2
oh okay then
And I think @SkaterBoyShawn is trying to attempt to find the inverse function and then plug in there instead which is cool too
so for problems like that all you have to do is plug in the numbers and solve for a?
\[y=\log2(x+6) \\ 2^y=x+6 \\ x=2^y-6 \\ f^{-1}(x)=2^x-6\] his inverse function is right
correct @freckles
so if you have done it this way you can just replace the x with 3
thank you
do u guys understand the the inverse function
no
no u don't
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)
@freckles u understand
understand what?
oh okay I got it now i think @freckles
In general: Let's say you have the following problem: \[\text{ Find } f^{-1}(m) \text{ if } f(x)=\log_b(x) \\ \\ \text{ assume } f^{-1}(m)=a \text{ well this means } f(a)=m \\ \\ \text{ so you replace } x \text{ with } a \text{ like so } \\ f(a)=\log_b(a) \\ \text{ now remember } f(a)=m \\ \text{ so we can replace } f(a) \text{ with } m \\ m=\log_b(a) \\ \text{ and solve for } a \\ b^m=b^{\log_b(a)} \\ b^m=a \\ a=b^m \\ \text{ so remember } a=f^{-1}(m) \\ \text{ and that is what we were aiming to find } \\ f^{-1}(m)=a=b^m \\ f^{-1}(m)=b^m\]
notice the is the inverse function of f(x) evaluated at x=m

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