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

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Answer 1 · 2016-07-12T01:22:55

5 (Assuming #log = log_10#)

#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#