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Question Number 206253 by mnjuly1970 last updated on 10/Apr/24

$$$$$$f\left(x\right)={log}_2(x+2\sqrt{x}+4)$$$$\Rightarrow f^{-1}(13-4\sqrt{3})=?$$$$-----$$$$$$

Answered by cortano21 last updated on 10/Apr/24

$$\cancel{\underset{―}{\underbrace{}}}{}^{-1}(13-4\sqrt{3})=s\Rightarrow f\left(s\right)=13-4\sqrt{3}$$$$\Rightarrow \log {}_2({\left(\sqrt{s}\right)}^2+2\sqrt{s}+4)=13-4\sqrt{3}$$$$\Rightarrow \log {}_2({(\sqrt{s}+1)}^2+3)=13-4\sqrt{3}$$$$\Rightarrow {(\sqrt{s}+1)}^2+3=2^{13-4\sqrt{3}}$$$$\Rightarrow \sqrt{s}+1=\sqrt{2^{13-4\sqrt{3}}-3}$$$$s={(\sqrt{2^{13-4\sqrt{3}}-3}-1)}^2$$$$$$

Answered by A5T last updated on 10/Apr/24

$${log}_2[{(-1+\sqrt{2^x-3})}^2+2(-1+\sqrt{2^x-3})+4]$$$$={log}_2\left(2^x\right)=x$$$$\Rightarrow f\left[{(-1+\sqrt{2^x-3})}^2\right]=x\Rightarrow f^{-1}\left(x\right)={(-1+\sqrt{2^x-3})}^2$$$$\Rightarrow f^{-1}(13-4\sqrt{3})={(-1+\sqrt{2^{13-4\sqrt{3}}-3})}^2$$$$=2^{13-4\sqrt{3}}-2-2\sqrt{2^{13-4\sqrt{3}}-3}$$

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