(a) Without using mathematical tables or calculator, simplify: \(\frac{log_28 + \log_216 - 4 \log_22}{\log_416}\) (b) If 1342\(_{five}\) - 241\(_{five}\) = ...

Assessment: WAEC SSCE - General Mathematics - 2019 (Essay) Subject: General Mathematics

Question 1 Report

(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.
 

Answer Details

(a) Evaluate each logarithm to base 2.

\(\log_{2}8=\log_{2}2^{3}=3\), \(\log_{2}16=\log_{2}2^{4}=4\), \(4\log_{2}2=4(1)=4\).

Numerator \(=3+4-4=3\).

Denominator: \(\log_{4}16=\log_{4}4^{2}=2\).

\[\frac{\log_{2}8+\log_{2}16-4\log_{2}2}{\log_{4}16}=\frac{3}{2}\]

\(=\dfrac{3}{2}\)

(b) Convert each base-five number to base ten.

\[1342_{five}=1(5^{3})+3(5^{2})+4(5)+2=125+75+20+2=222\]\[241_{five}=2(5^{2})+4(5)+1=50+20+1=71\]

Therefore \(x=222-71=151\).

\(x=151\)

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