Question 250873
log(x-1) + log(5x) = 2


three basic rules of logarithms are:


log(a*b) = log(a) + log(b) (rule 1)
log(a/b) = log(a) - log(b) (rule 2)
log(a^b) = b*log(a) (rule 3)


your equation can be made into the following by invoking rule 1.


log(x-1) + log(5x) = 2 becomes:


log((x-1)*(5x)) = 2


this becomes:


log(5x^2 - 5x) = 2


the basic rule of logarithms states:


y = log(a,x) if and only if a^y = x


in your equation, a base of 10 is implied.


your equation of log(5x^2 - 5x) = 2 is really:


log(10,(5x^2-5x)) = 2 which means:


log of (5x^2-5x) to the base of 10 = 2


using the basic rule of logarithms, this equation becomes:


log(10,(5x^2-5x)) = 2 if and only if:


10^2 = 5x^2-5x


this becomes:


100 = 5x^2 - 5x


subtract 100 from both sides of this equation to get:


5x^2 - 5x - 100 = 0 which is a quadratic equation.


divide both sides of this equation by 5 to get:


x^2 - x - 20 = 0


this factors out to be:


(x-5)*(x+4) = 0


solve for x to get:


x = 5 or x = -4


substitute in original quadratic equation to confirm these answers are good.


5x^2 - 5x = 100 becomes:


80 + 20 = 100 when x = -4 and becomes:


125 - 25 = 100 when x = 5.


both solutions are good.


plug these solutions into your original equation that you started with to get:


log(x-1) + log(5x) = 2 becomes:


log(4) + log(25) = 2 when x = 5.   


solving this equations gets 2 = 2 confirming x = 5 is a good solution.


log(x-1) + log(5x) = 2 becomes:


log(-5) + log(-20) = 2 when x = -4


this solutions is not valid because you can't take the log of a negative number.


your only valid solution is:


x = 5