15−y=25(54−2y)\;\frac{1}{5^{-y}} = 25(5^{4-2y})5−y1=25(54−2y)
⟹ 5y=(52)(54−2y)\implies 5^{y} = (5^{2})(5^{4-2y})⟹5y=(52)(54−2y)
5y=52+4−2y5^{y} = 5^{2+4-2y}5y=52+4−2y
Comparing bases, we have
y=6−2yy = 6 - 2yy=6−2y
3y=6 ⟹ y=23y = 6 \implies y = 23y=6⟹y=2