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.
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