BD Merit - Freelancing, BCS, Bank, Primary, Admission, NTRCA

Discuss Forum

1. $\log_{2} \sqrt{6} + \log_{2} \sqrt{\frac{2}{3}} =$

Answer: Option B

Explanation:

{\log_{3}}^{\sqrt{6}} + {\log_{2}}^{\sqrt{\frac{2}{3}}} = {\log_{2}}^{6^{\frac{1}{2}}} + {\log_{2}}^{{\frac{2}{3}}^{\frac{1}{2}}} = \frac{1}{2} ({\log_{2}}^{6} + {\log_{2}}^{\frac{2}{3}}) = \frac{1}{2} {\log_{2}}^{(6 \times \frac{2}{3})} = \frac{1}{2} {\log_{2}}^{4} = \frac{1}{2} \log_{2} {(2)}^{2} = \frac{1}{2} \times 2 {\log_{2}}^{2} = {\log_{2}}^{2} = 1

Post your comments here:

Name:

Mobile:

Email:(Optional)

» Your comments will be displayed only after manual approval.