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Question-200821




Question Number 200821 by Calculusboy last updated on 24/Nov/23
Answered by shunmisaki007 last updated on 24/Nov/23
ln(10!)  =ln(10∙9∙8∙7∙6∙5∙4∙3∙2∙1)  =ln((2∙5)∙3^2 ∙2^3 ∙7∙(2∙3)∙5∙2^2 ∙3∙2∙1)  =ln(2^8 ∙3^4 ∙5^2 ∙7)  =8ln(2)+4ln(3)+2ln(5)+ln(7) ...★
$$\mathrm{ln}\left(\mathrm{10}!\right) \\ $$$$=\mathrm{ln}\left(\mathrm{10}\centerdot\mathrm{9}\centerdot\mathrm{8}\centerdot\mathrm{7}\centerdot\mathrm{6}\centerdot\mathrm{5}\centerdot\mathrm{4}\centerdot\mathrm{3}\centerdot\mathrm{2}\centerdot\mathrm{1}\right) \\ $$$$=\mathrm{ln}\left(\left(\mathrm{2}\centerdot\mathrm{5}\right)\centerdot\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{2}^{\mathrm{3}} \centerdot\mathrm{7}\centerdot\left(\mathrm{2}\centerdot\mathrm{3}\right)\centerdot\mathrm{5}\centerdot\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}\centerdot\mathrm{2}\centerdot\mathrm{1}\right) \\ $$$$=\mathrm{ln}\left(\mathrm{2}^{\mathrm{8}} \centerdot\mathrm{3}^{\mathrm{4}} \centerdot\mathrm{5}^{\mathrm{2}} \centerdot\mathrm{7}\right) \\ $$$$=\mathrm{8ln}\left(\mathrm{2}\right)+\mathrm{4ln}\left(\mathrm{3}\right)+\mathrm{2ln}\left(\mathrm{5}\right)+\mathrm{ln}\left(\mathrm{7}\right)\:…\bigstar \\ $$
Commented by Calculusboy last updated on 26/Nov/23
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Commented by Calculusboy last updated on 26/Nov/23
$$ \\ $$

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