Log3(x2+2x+2)=0 what is the answer?
X^2+2x+2=1
X^2+2x+1=0
(X+1)(x+1)=0
X=-1
if log3 (a) = 0
then
3^[log3(a)]= a = 3^0 = 1
so
x^2 + 2 x + 2 = 1
x^2 + 2 x + 1 = 0
(x+1)(x+1) = 0
x = -1
To solve the equation log3(x^2 + 2x + 2) = 0, we need to isolate the variable x.
Step 1: Start by converting the logarithmic equation into exponential form.
In exponential form, the equation becomes: 3^0 = x^2 + 2x + 2.
Step 2: Simplify 3^0.
Since any number (except zero) raised to the power of zero is equal to 1, we have: 1 = x^2 + 2x + 2.
Step 3: Rearrange the equation in standard quadratic form.
By subtracting 1 from both sides of the equation: 0 = x^2 + 2x + 1.
Step 4: Factor the quadratic equation.
The equation can be factored into (x + 1)(x + 1) = 0.
Thus, x + 1 = 0.
Step 5: Solve for x.
By subtracting 1 from both sides of x + 1 = 0, we find that x = -1.
Therefore, the solution to the equation log3(x^2 + 2x + 2) = 0 is x = -1.
To find the answer to the equation Log3(x^2+2x+2) = 0, we need to first understand the properties of logarithms.
Logarithms are the inverse of exponentiation. In this case, Log3 represents a logarithm with base 3. The equation Log3(x^2+2x+2) = 0 can be rewritten in exponential form as 3^0 = x^2 + 2x + 2.
In exponential form, any base raised to the power of 0 is equal to 1. Therefore, 3^0 is equal to 1. Thus, the equation becomes 1 = x^2 + 2x + 2.
This is a quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = 2, and c = 2. We can now solve this quadratic equation to find the values of x.
To solve it, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
Substituting the values in, we have x = (-(2) ± √((2)^2 - 4(1)(2))) / (2(1)).
Simplifying further, x = (-2 ± √(4 - 8)) / 2.
x = (-2 ± √(-4)) / 2.
The square root of a negative number is not a real number, so we cannot simplify the equation any further. This means that the equation has no real solutions.
Hence, there is no answer to the equation Log3(x^2+2x+2) = 0.