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Solve \(\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)\).

Further Mathematics
WAEC 2018

Solve \(\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)\).

  • A. 4.75
  • B. 4.00
  • C. 1.75
  • D. 1.00
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Correct Answer: Option B
Explanation

\(\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)\)

Recall, \( 1 = \log_{2}2\), so

\(\log_{2}(12x - 10) = \log_{2}2 + \log_{2}(4x + 3)\)

= \(\log_{2}(12x - 10) = \log_{2}(2(4x + 3))\)

\(\implies (12x - 10) = 2(4x + 3) \therefore 12x - 10 = 8x + 6\)

\(12x - 8x = 4x = 6 + 10 = 16 \implies x = 4\)


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WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC offline past questions - with all answers and explanations in one app - Download for free
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709