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Clavis TMUA Set C Paper 2 · Q13

Paper 2Logic & proofPowers & logarithmsAlgebra

Consider the following attempt to solve the equation log⁡2x+log⁡2(x−2)=3\log_2x+\log_2(x-2)=3.

  • (I) Combining the logarithms gives log⁡2(x(x−2))=3\log_2\big(x(x-2)\big)=3.
  • (II) Hence x(x−2)=8x(x-2)=8.
  • (III) So x2−2x−8=0x^2-2x-8=0, which factorises as (x−4)(x+2)=0(x-4)(x+2)=0.
  • (IV) Therefore the solutions are x=4x=4 and x=−2x=-2.

Which one of the following is true?

AThe working is correct.
BOnly one of x=4x=4 and x=−2x=-2 is a solution of the original equation, and the error arises as a result of step (I).
COnly one of x=4x=4 and x=−2x=-2 is a solution of the original equation, and the error arises as a result of step (II).
DOnly one of x=4x=4 and x=−2x=-2 is a solution of the original equation, and the error arises as a result of step (III).
ENeither x=4x=4 nor x=−2x=-2 is a solution of the original equation.
FThere is a further solution of the original equation that this working has missed.

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