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Question Number 82078 by jagoll last updated on 18/Feb/20

f(10^x ) = (√x)   what is f^(−1) (x)=?

$${f}\left(\mathrm{10}^{{x}} \right)\:=\:\sqrt{{x}}\: \\ $$$${what}\:{is}\:{f}^{−\mathrm{1}} \left({x}\right)=? \\ $$

Commented by john santu last updated on 18/Feb/20

hahaha..i think you can solve this  problem

$${hahaha}..{i}\:{think}\:{you}\:{can}\:{solve}\:{this} \\ $$$${problem}\: \\ $$

Answered by Kunal12588 last updated on 18/Feb/20

10^x =t  ⇒x log 10 = log t  ⇒x= ((log t)/(log 10))=log_(10) t  f(t)=(√(log_(10) t))  ⇒t=10^((f(t))^2 )   f^(−1) (t)=10^t^2    ⇒f^(−1) (x)=10^x^2

$$\mathrm{10}^{{x}} ={t} \\ $$$$\Rightarrow{x}\:{log}\:\mathrm{10}\:=\:{log}\:{t} \\ $$$$\Rightarrow{x}=\:\frac{{log}\:{t}}{{log}\:\mathrm{10}}={log}_{\mathrm{10}} {t} \\ $$$${f}\left({t}\right)=\sqrt{{log}_{\mathrm{10}} {t}} \\ $$$$\Rightarrow{t}=\mathrm{10}^{\left({f}\left({t}\right)\right)^{\mathrm{2}} } \\ $$$${f}^{−\mathrm{1}} \left({t}\right)=\mathrm{10}^{{t}^{\mathrm{2}} } \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\mathrm{10}^{{x}^{\mathrm{2}} } \\ $$

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