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MISCELLANEOUS EXERCISE 6 (II) · Q172

Q.Solve for xx, log⁡x(8x−3)−log⁡x4=2\log_x(8x-3)-\log_x4=2.

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log⁡x(8x−3)−log⁡x4=2  ⟹  log⁡x(8x−34)=2  ⟹  8x−34=x2  ⟹  8x−3=4x2  ⟹  4x2−8x+3=0\log_x(8x-3)-\log_x4=2 \implies \log_x\left(\dfrac{8x-3}{4}\right)=2 \implies \dfrac{8x-3}{4}=x^2 \implies 8x-3=4x^2 \implies 4x^2-8x+3=0.

By the quadratic formula: x=8±64−488=8±48x=\dfrac{8\pm\sqrt{64-48}}{8}=\dfrac{8\pm4}{8}, giving x=32x=\dfrac32 or x=12x=\dfrac12.

Domain check (base xx needs x>0, x≠1x>0,\ x\ne1; argument 8x−3>08x-3>0 needs x>3/8x>3/8): both x=12x=\tfrac12 and x=32x=\tfrac32 satisfy x>3/8x>3/8, x≠1x\ne1, x>0x>0. …

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