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2019 WAEC Mathematics Theory (a) Without using mathematical tables or calculator, simplify: \(\frac{log_28 + \log_216 - 4 \log_22}{\log_416}\) (b)...

Mathematics
WAEC 2019

(a) Without using mathematical tables or calculator, simplify: \(\frac{log_28 + \log_216 - 4 \log_22}{\log_416}\)


(b) If 1342\(_{five}\) - 241\(_{five}\) = x\(_{ten}\), find the value of x.
 

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Explanation

(a) \(\frac{\log_28 + \log_216 - 4\log_22}{\log_416}\) = \(\frac{\log_22^3 + \log_2 2^4 - 4\log_22}{\log_44^2}\)

= \(\frac{3 \log_2 2 + 4\log_22 - 4 \log_2^2}{2 \log 4^4}\)

But log\(_2\) 2 = \(\log_44 = 1\)

Then, \(\frac{\log_28+\log_216 - 4 \log_2^2}{\log_416}\) = \(\frac{3 + 4 - 4}{2}\) = 1\(\frac{1}{2}\)

(b) Subtract 241\(_{five}\) to get 1101\(_{five}\). Thereafter, converting to base ten to find the value of x = (5\(^3\) x 1) + (5\(^2\) x 1 + (5\(^1\) x 0) + (5\(^o\) x 1) = 125 + 25 + 0 + 1 = 151

(b) They subtracted 241\(_{five}\) from 1342\(_{five}\) to get 1101\(_{five}\). Thereafter, converting to base ten to find the value of x = (5\(^3\) x 1) + (5\(^2\) x 1) + (5\(^1\) x 0) + (5°x 1) = 125 + 25 + + 1 = 151.


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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 Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
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
WAEC offline past questions - with all answers and explanations in one app - Download for free