81-1-log-5-9-3-3-log-6-3-409-7-2-log-25-7-125-log-25-6-

Question Number 121387 by Ar Brandon last updated on 07/Nov/20

\frac{{(81)}^{1 / \log_{5} 9} + 3^{3 / \log_{\sqrt{6}} 3}}{409} [{(\sqrt{7})}^{2 / \log_{25} 7} - {(125)}^{\log_{25} 6}]

Answered by 675480065 last updated on 07/Nov/20

(\frac{9^{\log_{9} 25} + 3^{\log_{3} 6 \sqrt{6}}}{409}) [7^{\log_{7} 25} - 5^{\log_{5} 6 \sqrt{6}}]

= (\frac{25 + 6 \sqrt{6}}{409}) (25 - 6 \sqrt{6})

= (\frac{1}{409}) (25^{2} - 36 (6)) = \frac{409}{409} = 1

Answered by liberty last updated on 07/Nov/20

[{(9)}^{\log_{9} {(5)}^{2}} + 3^{\log_{3} {(\sqrt{6})}^{3}}] [{(7)}^{\log_{7} (25)} - {(6)}^{\log_{25} (125)}] \times \frac{1}{409}

= [25 + 6 \sqrt{6}] [25 - 6 \sqrt{6}] \times \frac{1}{409}

= (625 - 216) \times \frac{1}{409} = 1

625 - 216

409.0

Commented by Ar Brandon last updated on 07/Nov/20

Cool. Thanks Sir