How do you solve #log_2 (x) + log_6 (x) = 3#?

1 Answer
Sep 2, 2015

# x = 2^((3(log_2 3+1))/(log_2 3 + 2)) #

Explanation:

# log_a b = log_c b/log_c a #
# log (ab) = log a + log b #
# log_a a = 1 #
# log_a b = c <=> a^c = b #

# log_2 x + log_6 x = 3 #
# log_2 x + log_2 x/log_2 6 = 3 #
# log_2 x(1+1/log_2 6) = 3 #
# log_2 x = (3log_2 6)/(log_2 6 + 1) #
# x = 2^((3log_2 6)/(log_2 6 + 1)) #
# x = 2^((3(log_2 3+1))/(log_2 3 + 2)) #