જો $x = {\log _5}(1000)$ અને $y = {\log _7}(2058)$ તો
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a
(a) \(x = {\log _5}1000 = 3{\log _5}10 = 3 + 3{\log _5}2 = 3 + {\log _5}8\)
(a) \(x = {\log _5}1000 = 3{\log _5}10 = 3 + 3{\log _5}2 = 3 + {\log _5}8\)
\(y = {\log _7}2058 = {\log _7}({7^3}.6) = 3 + {\log _7}6\)
As \({\log _5}8 > {\log _5}5\) i.e., \({\log _5}8 > 1\). \(x > 4\)
And \({\log _7}6 < {\log _7}7\) i.e., \({\log _7}6 < 1\)
\(\therefore y < 4\);
\(\therefore x > y\).
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