anonymous
  • anonymous
Given the function f(x) = log2(x + 6), find the value of f^−1(3).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I really need help with understanding this.
anonymous
  • anonymous
Answer Choices: a. f−1(3) = 2 b. f−1(3) = 3 c. f−1(3) = 9 d. f−1(3) = 18
anonymous
  • anonymous
@zepdrix help please

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

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