What are the number of solutions for real x, which satisfy the equation -Maths 9th

1 Answer

Answer :

log2​x>0 2log2​log2​x+log21​​log2​(22​x)=1 ⇒2log2​log2​x−log2​log2​(22​x)=1[∵loga1​​b=−loga​b] $$\Rightarrow \log _{ 2 }{ \left[ \dfrac { { \left( \log _{ 2 }{ x }  \right)  }^{ 2 } }{ \log_{ 2 }\left( 2\sqrt { 2 } x \right)  }  \right]  } =1 \quad [\because  m\log a = \log a^m   \quad \text& \quad \log a-\log b=\log\dfrac{a}{b}]$$ ⇒log2​(22​)+log2​x(log2​x)2​=21=2[∵log(ab)=loga+logb] Substitute log2​x=t to obtain t2−2t−2log2​(22​)=0 ⇒t2−2t−3=0[∵2log2​22​=log2​(22​)2=log2​8=log2​23=3] ⇒t=3,−1=log2​x ⇒x=2−1or 23 That is x=21​  or 8 Hence, the answer is 8.

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