MCQ
If $x = {\log _5}(1000)$ and $y = {\log _7}(2058)$ then
- ✓$x > y$
- B$x < y$
- C$x = y$
- DNone of these
$y = {\log _7}2058 = {\log _7}({7^3}.6) = 3 + {\log _7}6$
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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$(A)$ $a=2, L=\frac{e^{4 \pi}-1}{e^\pi-1}$ $(B)$ $a=2, L=\frac{e^{4 \pi}+1}{e^\pi+1}$
$(C)$ $a=4, L=\frac{e^{4 \pi}-1}{e^\pi-1}$ $(D)$ $a=4, L=\frac{e^{4 \pi}+1}{e^\pi+1}$