How do you solve #Log 5x + log x^2 - log x = 5#?

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Answer 1 · 2015-11-21T03:59:50

Elementary

Lets recall a few rules;

#log a + log b = log ab #

#log a - log b = log (a/b )#

Okay

#Log 5x + log x^2 - log x = 5#
#Log 5x^3 - log x = 5#

#Log ((5x^3)/x) = 5#

#Log 5x^2 = 5#

Now this is a base 10 logarithm

so #10 ^ 5 = 5 x^2#

Solve and you shall get# x = +-100sqrt2#