How do you solve #log2+logx=1#?

1 Answer
Jul 12, 2016

Answer:

5 (Assuming #log = log_10#)

Explanation:

#log_10 2 + log_10 x = 1#

Using: #log_n a + log_n b = log_n(ab)# we may rewrite the equation as:

#log_10 2x = 1#

Using: #log_n a = b -> a = n^b# we may simplify the equation as:

#2x = 10^1#

#2x = 10#

#x = 5#