Which are the solutions of the equation lgx+lg(x-3)=1?
3 Answers
hala718 | High School Teacher | (Level 1) Educator Emeritus
lgx+ lg(x-3)=1
We know that:
logx+lgy= lgxy
and, log10=1
Now substitute;
lgx(x-3)=lg10
==> x(x-3)=10
==> x^2-3x = 10
==> x^2 -3x -10=0
Now factorize:
==> (x-5)(x+2)=0
x1= 5
x2=-2 ( this solution is impossible because log-2 is undefined and x can not be a negative value)
Then the only solution is x1=5
To check:
lg5 + lg(5-3)= lg5+lg2 = lg10=1
neela | High School Teacher | (Level 3) Valedictorian
logx+log(x-3) = 1.
log(x(x-3) = 1, as loga+logb = log(ab). Taking antologarithms with respect to base 10, we get:
x(x-3) = 10^1.
x^x-3x -10 = 0. Or
x = [3+ srt(3^2+4*10)}/2 = (3+sqrt49)/2 = 5. Or
x= (3-7)/2 = -2
giorgiana1976 | College Teacher | (Level 3) Valedictorian
For the beginning, we'll use the product property of logarithms: the sum of logarithms is the logarithm of the product.
lgx+lg(x-3)=1
lg [x(x-3)] = 1
The base of logarithm being 10, we'll re-write the equation:
[x(x-3)] = 10^1
We'll open the brackets:
x^2 - 3x = 10
We'll add -10 both sides of the equation:
x^2 - 3x - 10 = 0
We'll solve the quadratic equation:
x1 = [3 + sqrt(9+40)]/2
x1 = (3+7)/2
x1 = 5
x2 = (3-7)/2
x2 = -4/2
x2 = -2
The second solution, namely -2, is not convenient, because lg(-2) is undefined.
We'll check the first solution into equation:
lg5+lg(5-3)=1
lg5 + lg2 = 1
lg(5*2) = 1
lg 10 = 1
10 = 10^1
10 = 10
So, the solution x=5 is the only solution of the equation.