Let u = (log2x)^2 – 6(log2x) + 12, where x is a real number. Then the equation x^u = 256 has : -Maths 9th

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Answer :

⇒  u=(log2​x)2−log2​x+12 ⇒  We can take log2​x=y ⇒  Then equation becomes y2−6y+12=u ⇒  Given that xu=256 Applying log we get, ⇒  ulog2​x=8 ∴   u=log2​x8​=y8​So our equation becomes, ⇒  y2−6y+12=y8​⇒  y3−6y2+12y−8=0   Substitute y=2 (2)2−6(2)2+12(2)−8=0 8−24+24−8=0 0=0 ∴   We get y=2 ⇒  So, log2​x=2 ⇒  x=22 ∴  x=4 ∴  The given equation has exactly one solution for x

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