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Question Number 167010 by mnjuly1970 last updated on 04/Mar/22

Answered by LowLevelLump last updated on 04/Mar/22

h(x)=f(x−ln 2)+a ln ((1/2)e^(2x) −e^x +1)=ln (e^(2(x−ln 2)) −e^(x−ln 2) +(1/2))+a ln ((1/2)e^(2x) −e^x +1)=ln (e^(2x−ln 4) −e^(x−ln 2) +(1/2))+a ln ((1/2)e^(2x) −e^x +1)=ln ((1/4)e^(2x) −(1/2)e^x +(1/2))+a ln ((1/2)e^(2x) −e^x +1)=ln ((1/4)e^(2x) −(1/2)e^x +(1/2))+ln e^a ln ((1/2)e^(2x) −e^x +1)=ln ((1/4)e^a e^(2x) −(1/2)e^a e^x +(1/2)e^a ) Let t=e^x , therefore ln ((1/2)t^2 −t+1)=ln ((1/4)e^a t^2 −(1/2)e^a t+(1/2)e^a ) (1/2)t^2 −t+1=(1/4)e^a t^2 −(1/2)e^a t+(1/2)e^a (1/2)t^2 −t+1=(1/2)e^a ((1/2)t^2 −t+1) 1=(1/2)e^a e^a =2 e^a =e^(ln 2) a=ln 2

h(x)=f(x−ln2)+aln(12e2x−ex+1)=ln(e2(x−ln2)−ex−ln2+12)+aln(12e2x−ex+1)=ln(e2x−ln4−ex−ln2+12)+aln(12e2x−ex+1)=ln(14e2x−12ex+12)+aln(12e2x−ex+1)=ln(14e2x−12ex+12)+lnealn(12e2x−ex+1)=ln(14eae2x−12eaex+12ea)Lett=ex,thereforeln(12t2−t+1)=ln(14eat2−12eat+12ea)12t2−t+1=14eat2−12eat+12ea12t2−t+1=12ea(12t2−t+1)1=12eaea=2ea=eln2a=ln2

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