\(\frac{1}{5}\)
\(\frac{1}{2}\)
\(\frac{-1}{2}\)
(\frac{log_23}{log_27}\)
Correct answer is A
log\(_2\)6 - log\(_2\)3 = log\(_2\) (\(\frac{6}{3}\))
= log\(_2\)2... A
log\(_2\)8 - 2log\(_2\)\(\frac{1}{2}\)
log\(_2\)8 - log\(_2\)\(\frac{1}{2}\)\(^2\)
log\(_2\)8 - log\(_2\)\(\frac{1}{4}\)
= log\(_2\) 32... B
\(\frac{A}{B}\) = \(\frac{log_22}{log_232}\)
N.B. log\(_2\)2 = 1
= \(\frac{1}{5}\)
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