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Question Number 93077 by hmamarques1994@gmail.com last updated on 10/May/20

  Se  f((√x) −1) = x+6,  log[f(1)] = ?

$$\:\:\mathrm{Se}\:\:\mathrm{f}\left(\sqrt{\mathrm{x}}\:−\mathrm{1}\right)\:=\:\mathrm{x}+\mathrm{6},\:\:\mathrm{log}\left[\mathrm{f}\left(\mathrm{1}\right)\right]\:=\:? \\ $$

Commented by john santu last updated on 11/May/20

f((√x)−1) = x+6   set (√x) −1 = p ⇒x = (p+1)^2   so f(x)= (x+1)^2 +6  f(1) = 10   log _(10)  f(1) = log _(10) (10) = 1  if you meant log f(1)= ln f(1)= ln(10)

$$\mathrm{f}\left(\sqrt{{x}}−\mathrm{1}\right)\:=\:{x}+\mathrm{6}\: \\ $$$${set}\:\sqrt{{x}}\:−\mathrm{1}\:=\:{p}\:\Rightarrow{x}\:=\:\left({p}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${so}\:{f}\left({x}\right)=\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{6} \\ $$$${f}\left(\mathrm{1}\right)\:=\:\mathrm{10}\: \\ $$$$\mathrm{log}\:_{\mathrm{10}} \:{f}\left(\mathrm{1}\right)\:=\:\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{10}\right)\:=\:\mathrm{1} \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{meant}\:\mathrm{log}\:\mathrm{f}\left(\mathrm{1}\right)=\:\mathrm{ln}\:\mathrm{f}\left(\mathrm{1}\right)=\:\mathrm{ln}\left(\mathrm{10}\right)\: \\ $$$$ \\ $$

Answered by Kunal12588 last updated on 10/May/20

(√x)−1=1  ⇒x=4  log_(10) (f(1))=1

$$\sqrt{{x}}−\mathrm{1}=\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{4} \\ $$$${log}_{\mathrm{10}} \left({f}\left(\mathrm{1}\right)\right)=\mathrm{1} \\ $$

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