Solve the following algebraically.
log3(x) + log3(x-2) = 1
Are the solutions 3 and -1?
log(ab)=loga+logb
log3x+log[3(x-2)]=1
log[(3x)(3x-6)]=1
(3x)(3x-6)=10^1
9x^2-18x-10=0
find x and see if you would arrive at that
clearly -1 is not a solution, since log(-1) is not real.
log3(x(x-2)) = 1
x^2-2x = 3
x^2-2x-3 = 0
(x-3)(x+1) = 0
so, yes, x=3 is a solution, but -1 is extraneous.
To solve the equation algebraically, we will use the properties of logarithms to combine the two logarithms on the left side of the equation.
First, we can use the product rule of logarithms, which states that log base a (m) + log base a (n) is equal to log base a (m * n).
Applying the product rule to the equation, we combine the two logarithms with the same base, 3.
log3(x) + log3(x-2) = log3(x * (x-2))
Now, the equation becomes:
log3(x * (x-2)) = 1
Next, we will use the definition of logarithms to rewrite the equation in exponential form. In general, log base a (b) = c is equivalent to a^c = b.
So, we have:
3^1 = x * (x - 2)
Now, the equation becomes a quadratic equation:
3 = x^2 - 2x
Rearrange the equation into standard quadratic form:
x^2 - 2x - 3 = 0
To solve this equation, we can factor it:
(x - 3)(x + 1) = 0
Setting each factor equal to zero, we have two possible solutions:
x - 3 = 0, which gives x = 3
x + 1 = 0, which gives x = -1
So, the possible solutions to the equation are x = 3 and x = -1.
To verify our solutions, we can substitute them back into the original equation:
For x = 3:
log3(3) + log3(3-2) = 1
log3(3) + log3(1) = 1
1 + 0 = 1
The left side equals the right side, so x = 3 is a valid solution.
For x = -1:
log3(-1) + log3(-1-2) = 1
The logarithm of a negative number is undefined, so x = -1 is not a valid solution.
Therefore, the only solution to the equation is x = 3. The solution x = -1 is extraneous.