Question Number 104624 by dw last updated on 22/Jul/20
Commented by dw last updated on 22/Jul/20
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Commented by Dwaipayan Shikari last updated on 22/Jul/20
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Commented by Dwaipayan Shikari last updated on 22/Jul/20
$$\mathrm{2019}!\:{log}_{\mathrm{1001}} \mathrm{501}=\mathrm{0}.\mathrm{8998}\:\:.\mathrm{2019}!=\left({a}−\frac{\mathrm{27}}{\mathrm{100}}\right)\mathrm{2019}!\left({approximately}\right) \\ $$$${log}_{\mathrm{72}} \mathrm{144}=\mathrm{1}.\mathrm{16}={a} \\ $$$$ \\ $$