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f-x-log-2-x-2-x-4-f-1-13-4-3-

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Question Number 206253 by mnjuly1970 last updated on 10/Apr/24
f(x)= log_( 2) ( x + 2(√x) +4 ) ⇒ f^( −1) ( 13 −4(√3) ) = ? −−−−−
f(x)=log2(x+2x+4)⇒f−1(13−43)=?−−−−−
Answered by cortano21 last updated on 10/Apr/24
^(−1) (13−4(√3) )= s ⇒f(s)= 13−4(√3) ⇒log _2 (((√s) )^2 +2(√s) + 4) = 13−4(√3) ⇒ log _2 (((√s) +1)^2 +3) = 13−4(√3) ⇒ ((√s) +1)^2 +3 = 2^(13−4(√3)) ⇒ (√s) +1 = (√(2^(13−4(√3)) −3)) s = ((√(2^(13−4(√3)) −3)) −1)^2
⏟―−1(13−43)=s⇒f(s)=13−43⇒log2((s)2+2s+4)=13−43⇒log2((s+1)2+3)=13−43⇒(s+1)2+3=213−43⇒s+1=213−43−3s=(213−43−3−1)2
Answered by A5T last updated on 10/Apr/24
log_2 [(−1+(√(2^x −3)))^2 +2(−1+(√(2^x −3)))+4] =log_2 (2^x )=x ⇒f[(−1+(√(2^x −3)))^2 ]=x⇒f^(−1) (x)=(−1+(√(2^x −3)))^2 ⇒f^(−1) (13−4(√3))=(−1+(√(2^(13−4(√3)) −3)))^2 =2^(13−4(√3)) −2−2(√(2^(13−4(√3)) −3))
log2[(−1+2x−3)2+2(−1+2x−3)+4]=log2(2x)=x⇒f[(−1+2x−3)2]=x⇒f−1(x)=(−1+2x−3)2⇒f−1(13−43)=(−1+213−43−3)2=213−43−2−2213−43−3

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