Question

Find the zeros of the function f(x)=log3(x-1)+log3(2x+3)

Answer 1

To find the zeros of the function f(x)=log3(x-1)+log3(2x+3), we combine the two logs into a single logarithm and set it equal to zero, yielding a quadratic equation. The roots of this quadratic are x = -2 and x = 1, but only x = 1 is valid in the original expression.

Step-by-step explanation:

To find the zeros of the function f(x)=log3(x-1)+log3(2x+3), we apply the properties of logarithms. Specifically, the property that states the logarithm of the product of two numbers is the sum of their logarithms. Thus, we can combine the two logs into a single log.

First, we can rewrite the function as:

f(x)=log3((x-1)(2x+3))

To find the zeros, we set f(x)=0:

0 = log3((x-1)(2x+3))

Using the definition of a logarithm, this means that the argument of the log must be 1:

(x-1)(2x+3) = 1

Expanding and moving all terms to one side gives us:

2x^2 + 2x - 4 = 0

This is a quadratic equation, which we can solve by factoring, completing the square, or using the quadratic formula. To keep the answer concise, I'll skip directly to the roots:

x = -2 and x = 1

However, since the domain of a logarithm can only be positive, we must exclude any root that doesn't satisfy the original log functions. Checking both proposed solutions, we find that x = -2 makes the term (2x+3) negative, which isn't allowed in the log function. Therefore, the only zero of the original function is:

x = 1

Answer 2

The zeros of the function f(x)=log3(x-1)+log3(2x+3) are
`\(x = (1)/(2)\) and \(x = -4\)`.

Step-by-step explanation:

To find the zeros of the function
`\(f(x) = \log_3(x-1) + \log_3(2x+3)\)`, we need to set the function equal to zero and solve for
`\(x\)`:

`\[ \log_3(x-1) + \log_3(2x+3) = 0 \]`

Combine the logarithms using the product rule of logarithms, which

states that
`\(\log_a(b) + \log_a(c) = \log_a(bc)\)`:

`\[ \log_3((x-1)(2x+3)) = 0 \]`

Now, we can rewrite the equation in exponential form:

`\[ 3^0 = (x-1)(2x+3) \]`

Since
`\(3^0 = 1\)`, we have:

`\[ 1 = (x-1)(2x+3) \]`

Expand the equation:

`\[ 1 = 2x^2 + x - 3 \]`

Move all terms to one side to set the quadratic equation to zero:

`\[ 2x^2 + x - 4 = 0 \]`

Now, factor or use the quadratic formula to find the solutions for \(x\). Factoring, we get:

`\[ (2x - 1)(x + 4) = 0 \]`

Setting each factor to zero gives:

`\[ 2x - 1 = 0 \quad \text{or} \quad x + 4 = 0 \]`

Solving these equations gives:

`\[ x = (1)/(2) \quad \text{or} \quad x = -4 \]`

Therefore, the zeros of the function are
`\(x = (1)/(2)\) and \(x = -4\)`.