How do you solve #log x + log (x-3) = 1#?

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1 Answer
sente
Apr 15, 2016

#x = 5#

Explanation:

Assuming #log# is referring to the base-#10# logarithm, we can apply the properties that #log(a)+log(b) = log(ab)# and #10^log(x)=x# to obtain

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

#=> log(x(x-3)) = 1#

#=> 10^log(x(x-3)) = 10^1#

#=> x(x-3) = 10#

#=> x^2-3x - 10 = 0#

#=> x = (3+-sqrt(9+4*10))/2 = 3/2+-7/2# (by the quadratic formula)

Normally we would be done, however originally we had #log(x-3)# in the equation, meaning we have the restriction #x>3#, as the logarithm function is undefined in the reals for #x<=0#.

As #3/2-7/2 < 3# it is not a solution.

As #3/2+7/2 = 5 > 3#, it is a solution.

Thus we have the solution #x = 5#

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