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How do you solve #logx+log5=1#?

1 Answer

Apr 15, 2016

#x = 2# (Assuming #Log# is #Log_ 10#)

Explanation:

#Log(x) + Log(5) = 1#

Applying #Log(a) + Log(b) = Log(ab)#
# -> Log(5x) = 1#
Assuming #Log# is #Log_10#
#-> 5x = 10^1#
#5x = 10#
#x = 2#

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