How do you solve #log (x – 2) + log x = log 3#?

1 Answer
Apr 6, 2016

Use the rules #log_an + log_am = log_a(n xx m)# and #log_an - log_am = log_a(n/m)#

Explanation:

Put all the logs on one side of the equation:

#log(x - 2) + logx - log3 = 0#

#log(((x + 2)x)/3) = 0#

#log((x^2 + 2x)/3) = 0#

Now, convert to exponential form. You are in base 10, since nothing is noted in subscript by the log.

#(x^2 + 2x)/3 = 10^0#

#(x^2 + 2x)/3 = 1#

#x^2 + 2x = 3#

#x^2 + 2x - 3 = 0#

#(x + 3)(x - 1) = 0#

#x = -3 and 1#

It is always vital to check your solutions after solving logarithmic equations. Neither solutions work, because the log of a negative number is undefined in the real number system. Therefore, this equation has no solution #(O/)#.

Hopefully this helps!