⇒log(x−3)=2log(x−7) ⇒(x−3)=(x−7)2 ⇒x−3=x2−14x+49 x2−15x+52=0 Discriminant = 152−4×52=225−208=17>0 Two distinct values of x exists However, since we have the log function involved here, we will have to verify the solutions. Log is defined only for positive terms So. if log(x - 3) and log(x - 7) are defined, x has to be greater than 7. For the equation x2−15x+52=0 one of the values
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(15−√17) This is less than 7. Since x cannot be less than 7 there is only one possible solution