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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