You need a few more assumptions on a and b and you need to use the powr operator instead of ^, since ^ is only for the n-th power where n is a natural number. powr on the other hand is for any non-negative real number raised to the power of any other real number. (or similarly for complex numbers)
lemma
fixes a b :: real
assumes "a > 0" "a ≠ 1" "b > 0"
shows "∃x. a powr x = b"
proof
from assms show "a powr (log a b) = b" by simp
qed
You need a few more assumptions on
a
andb
and you need to use thepowr
operator instead of^
, since^
is only for then
-th power wheren
is a natural number.powr
on the other hand is for any non-negative real number raised to the power of any other real number. (or similarly for complex numbers)