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

Question 250873: log(x-1)+log(5x)=2
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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

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