Answer:
x < -2 or x > 4
Step-by-step explanation:
[tex]\sf log_3 (x-1)^2 > 2[/tex]
We know that [tex]\sf log_b(a) = c[/tex] is [tex]\sf a = b^c[/tex] in exponential form.
In this case, we have > instead of =.
Thus
[tex]\sf log_3 (x-1)^2 > 2 \rightarrow (x-1)^2 > 3^2[/tex]
Solving,
[tex]\implies \sf \sqrt{(x-1)^2} > \sqrt{3^2}[/tex]
[tex]\implies \sf |x-1|>3[/tex]
As per the Absolute Value Inequalities rule,
[tex]\sf If \ |a| > b, \ then \ a>b \ or \ a<-b[/tex]
Hence,
[tex]\sf x-1 > 3 \ or \ x -1 <-3[/tex]
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Solving x - 1 > 3, we get x > 4.
Solving x - 1 < -3, we get x < -2.
Therefore x < -2 or x > 4
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