How do you solve #36^(2p)=216^(p-1)#?

Before solving this exponential equation we can recognize that
#" "#
#color(red)36" and " color(red)216 " "# can be written as powers of base #" "color(red)6#
#" "#
#color(red)(36 = 6^2#
#" "#
#color(red)(216 = 6^3#
#" "#
#36^(2p) = 216^(p-1)#
#" "#
#rArrcolor(red)((6^2))^(2p) = color(red)((6^3))^(p-1)#
#" "#
#rArr6^(4p) = 6^(3(p-1))#
#" "#
Therefore,
#" "#
#4p = 3(p-1)#
#" "#
#rArr 4p = 3p - 3#
#" "#
#rArr 4p - 3p = -3#
#" "#
#rArr p = -3#