Precalc: evaluate e^(ln (√11))? and log5(5^8)?

　
e^(ln x) = x ------> e^(ln(√11)) = √11
logₐ(a^x) = x ----> log₅(5^8) = 8

Please read :

e^[ln (√11)] = √11 (Rule: a^(log_a (x)) = x).

log_5(5^8) = 8 (Rule: log_a (a^x) = x).

** e^ln ( V11) = V(11) because lne = e^ln = 1
** log5 (5^8) = 8 because log base 5 of a number 5 = 1