If x = {log _5}(1000) and y = {log _7}(2058) | vedclass

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Question

(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

Vedclass · Accepted Answer

$x > y$

Vedclass · Answer

(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., ${\log _7}6 < 1$ $ herefore y < 4$; $ herefore x > y$.

If $x = {\log _5}(1000)$ and $y = {\log _7}(2058)$ then

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