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

Question

anonymous · Answer

Do you mean
$$f(x) = \log_{2} (x+4)$$
$$f ^{-1}(x) = ?$$

anonymous · Answer

Correction...
$$f ^{-1}(3) = ?$$

jdoe0001 · Answer

$\bf f(x) =y= log_2(x + 4)\qquad inverse\implies x= log_2(y + 4)\ \quad \ \quad \
\textit{log cancellation rule of }a^{log_ax}=x\qquad thus\ \quad \
x= log_2(y + 4)\implies 2^x=2^{log_2(y + 4)}\implies 2^x=y+4
$

jdoe0001 · Answer

solve for "y" and plug in  "3" for $\bf f^{-1}$

anonymous · Answer

yes thats what i meant! thank you

anonymous · Answer

where would f^-1 be though

jdoe0001 · Answer

?

jdoe0001 · Answer

well.. what did you get for the $\bf f^{-1}\quad ?$

anonymous · Answer

so when you solve for "y" you get 2^x - 4 = y

isaiah.feynman · Answer

Yes!

anonymous · Answer

f^-1 would be 9? cause 3 times three?

jdoe0001 · Answer

f(9)  just means , make "x" equals to 9 for all intances of "x"
the same is true for $\bf f^{-1}(9)$

jdoe0001 · Answer

$\bf 2^x-4=y\qquad \qquad f^{-1}(\color{red}{9})=2^{\color{red}{9}}-4$

anonymous · Answer

so how would i get the answer from that?

isaiah.feynman · Answer

The answer is already looking at you, what is 2 to the 9? punch your calculator whatever you get subtract 4 from it and thats your answer!