Question
find the zero(es) of the function f(x)=log3(x-1)+log3(2x+3)
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GPT 3.5
To find the zero(s) of the function f(x) = log3(x-1) + log3(2x+3), we need to find the value(s) of x for which f(x) equals zero.
First, we need to remember that the logarithm of a negative number or zero is undefined. Therefore, both x - 1 and 2x + 3 must be greater than zero for the logarithms to be defined.
Setting x - 1 > 0 and 2x + 3 > 0, we have the following inequalities:
x - 1 > 0 --> x > 1
2x + 3 > 0 --> 2x > -3 --> x > -3/2
The intersection of the two inequalities gives us the valid range for x: x > 1.
Next, let's simplify the function:
f(x) = log3(x-1) + log3(2x+3)
Using the properties of logarithms, we can combine the two logarithms into a single logarithm:
f(x) = log3((x-1)(2x+3))
Now we set f(x) equal to zero and solve for x:
log3((x-1)(2x+3)) = 0
Using the exponentiation property of logarithms, we can rewrite this equation as:
(x-1)(2x+3) = 3^0
Simplifying further:
(x-1)(2x+3) = 1
Expanding the brackets:
2x^2 - x + 3x - 3 = 1
Combining like terms:
2x^2 + 2x - 4 = 0
Dividing both sides of the equation by 2:
x^2 + x - 2 = 0
This is a quadratic equation, which can be factored as:
(x + 2)(x - 1) = 0
Setting each factor equal to zero:
x + 2 = 0 --> x = -2
x - 1 = 0 --> x = 1
Therefore, the zeros of the function f(x) = log3(x-1) + log3(2x+3) are x = -2 and x = 1, but since we already determined that the valid range for x is x > 1, the only zero in the valid range is x = 1.
First, we need to remember that the logarithm of a negative number or zero is undefined. Therefore, both x - 1 and 2x + 3 must be greater than zero for the logarithms to be defined.
Setting x - 1 > 0 and 2x + 3 > 0, we have the following inequalities:
x - 1 > 0 --> x > 1
2x + 3 > 0 --> 2x > -3 --> x > -3/2
The intersection of the two inequalities gives us the valid range for x: x > 1.
Next, let's simplify the function:
f(x) = log3(x-1) + log3(2x+3)
Using the properties of logarithms, we can combine the two logarithms into a single logarithm:
f(x) = log3((x-1)(2x+3))
Now we set f(x) equal to zero and solve for x:
log3((x-1)(2x+3)) = 0
Using the exponentiation property of logarithms, we can rewrite this equation as:
(x-1)(2x+3) = 3^0
Simplifying further:
(x-1)(2x+3) = 1
Expanding the brackets:
2x^2 - x + 3x - 3 = 1
Combining like terms:
2x^2 + 2x - 4 = 0
Dividing both sides of the equation by 2:
x^2 + x - 2 = 0
This is a quadratic equation, which can be factored as:
(x + 2)(x - 1) = 0
Setting each factor equal to zero:
x + 2 = 0 --> x = -2
x - 1 = 0 --> x = 1
Therefore, the zeros of the function f(x) = log3(x-1) + log3(2x+3) are x = -2 and x = 1, but since we already determined that the valid range for x is x > 1, the only zero in the valid range is x = 1.