How do you solve #log x = log 5# and find any extraneous solutions?

1 Answer
George C.
Jun 1, 2016

#x=5#

Explanation:

The function #f(x) = 10^x# is strictly monotonically increasing on its (implicit) domain #(-oo, oo)# with range #(0, oo)#.

Its inverse #f^(-1)(x) = log(x)# is strictly monotonically increasing on its (implicit) domain #(0, oo)# with range #(-oo, oo)#.

For each of these functions, since they are strictly monotonically increasing, they are also one-one.

So if #log x = log 5# then #x = 5#.

graph{(y - 10^x)(x - 10^y) = 0 [-10, 10, -5, 5]}

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