(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.
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.