Evaluate Log\(_2\) 8√2
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Correct Answer: Option C
Explanation:
whereLog\(_2\) 8√2→Log\(_2\) √128
→Log\(_2\) 128\(^\frac{1}{2}\)
=\(\frac{1}{2}\) * (Log\(_2\) 128)→\(\frac{1}{2}\) * (Log\(_2\) 2\(^7\))
=7 *\(\frac{1}{2}\) * (Log\(_2\) 2)
where(Log\(_2\) 2) = 1
→7 *\(\frac{1}{2}\) * 1
=\(\frac{7}{2}\) or 3.5
whereLog\(_2\) 8√2→Log\(_2\) √128
→Log\(_2\) 128\(^\frac{1}{2}\)
=\(\frac{1}{2}\) * (Log\(_2\) 128)→\(\frac{1}{2}\) * (Log\(_2\) 2\(^7\))
=7 *\(\frac{1}{2}\) * (Log\(_2\) 2)
where(Log\(_2\) 2) = 1
→7 *\(\frac{1}{2}\) * 1
=\(\frac{7}{2}\) or 3.5