5log-4-2-3-6-6log-8-3-2-

Question Number 59401 by Pranay last updated on 09/May/19

5 {l o g}_{4 \sqrt{2}} (3 - \sqrt{6}) - 6 {l o g}_{8} (\sqrt{3} - \sqrt{2})

Answered by prakash jain last updated on 10/May/19

4 \sqrt{2} = {(\sqrt{2})}^{5}

8 = {(\sqrt{2})}^{6}

{5 \log}_{{(\sqrt{2})}^{5}} (3 - \sqrt{6}) - {6 \log}_{{(\sqrt{2})}^{6}} (\sqrt{3} - \sqrt{2})

Using \log_{a^{n}} b = \frac{1}{n} \log_{a} b

5 \frac{1}{5} \log_{\sqrt{2}} (3 - \sqrt{6}) - 6 \frac{1}{6} \log_{\sqrt{2}} (\sqrt{3} - \sqrt{2})

= \log_{\sqrt{2}} \frac{3 - \sqrt{6}}{\sqrt{3} - \sqrt{2}}

= \log_{\sqrt{2}} (\frac{3 - \sqrt{6}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}})

= \log_{\sqrt{2}} (3 \sqrt{3} - \sqrt{18} + 3 \sqrt{2} - \sqrt{12})

= \log_{\sqrt{2}} (3 \sqrt{3} - 3 \sqrt{2} + 3 \sqrt{2} - 2 \sqrt{3})

= \log_{\sqrt{2}} \sqrt{3}

= \frac{1}{2} \log_{\sqrt{2}} 3

= \frac{1}{2} \times \frac{1}{1 / 2} \log_{2} 3

= \log_{2} 3