Solve the logarithm equation
Log (x^2-2x+2)=0
3
To solve the logarithm equation, we need to rewrite it in exponential form.
The equation is:
log(x^2-2x+2) = 0
By definition, a logarithm equation can be rewritten in exponential form as:
x^2-2x+2 = 10^0
Since any number raised to the power of 0 is 1, the equation simplifies to:
x^2-2x+2 = 1
Rearranging the terms and setting the equation to zero gives:
x^2-2x+1 = 0
Now, we can solve this quadratic equation by factoring:
(x-1)(x-1) = 0
The factor theorem tells us that if the product of two factors is zero, then at least one of the factors must be zero. Therefore:
x-1 = 0
Solving for x gives:
x = 1
Therefore, the solution to the logarithm equation log(x^2-2x+2) = 0 is x = 1.
To solve the logarithmic equation, we can rewrite it using the definition of logarithms.
The logarithmic equation can be rewritten as:
x^2 - 2x + 2 = 10^0
Since any number raised to the power of zero is 1, we simplify the equation further:
x^2 - 2x + 2 = 1
Now, let's solve the quadratic equation:
x^2 - 2x + 1 = 0
To solve this equation, we can either factorize or use the quadratic formula.
Using the quadratic formula, the solutions are:
x = (-(-2) ± √((-2)^2 - 4(1)(1))) / (2(1))
x = (2 ± √(4 - 4)) / 2
x = (2 ± √(0)) / 2
x = (2 ± 0) / 2
Therefore, the only solution to the equation is x = 1.
To solve the logarithm equation log(x^2 - 2x + 2) = 0, we can use the property of logarithm:
If log base b of a equals c, then b raised to the power of c is equal to a.
In this case, the base is assumed to be 10 because there is no indicated base. So, we have:
10^0 = x^2 - 2x + 2
Simplifying the equation, we have:
1 = x^2 - 2x + 2
Rearranging the equation, we get:
x^2 - 2x + 1 = 0
Now we can solve this quadratic equation by factoring:
(x - 1)^2 = 0
Taking the square root of both sides, we get:
x - 1 = 0
Solving for x, we get:
x = 1
Therefore, the solution to the logarithm equation log(x^2 - 2x + 2) = 0 is x = 1.