Evaluate log of x+ log of x+15=2

Solution:

Step 1: Combine the logarithmic terms.

Step 1.1: Utilize the logarithm rule that combines logs with the same base: $\log_b(x) + \log_b(y) = \log_b(xy)$.

$\log(x) + \log(x + 15) = 2$ simplifies to $\log(x(x + 15)) = 2$.

Step 1.2: Expand the product inside the logarithm.

$\log(xx + x \cdot 15) = 2$.

Step 1.3: Simplify the expanded expression.

Step 1.3.1: Multiply $x$ by itself.

$\log(x^2 + x \cdot 15) = 2$.

Step 1.3.2: Rearrange the terms inside the log.

$\log(x^2 + 15x) = 2$.

Step 2: Convert the logarithmic equation to its exponential form.

Using the definition of a logarithm, $\log_b(x) = y$ is equivalent to $b^y = x$.

$(10)^2 = x^2 + 15x$.

Step 3: Solve the quadratic equation for $x$.

Step 3.1: Express the equation as $x^2 + 15x = (10)^2$.

$x^2 + 15x = 100$.

Step 3.2: Calculate $10^2$.

$x^2 + 15x = 100$.

Step 3.3: Bring all terms to one side to set the equation to zero.

$x^2 + 15x - 100 = 0$.

Step 3.4: Factor the quadratic expression.

Step 3.4.1: Find two numbers that multiply to $-100$ and add up to $15$.

The numbers are $-5$ and $20$.

Step 3.4.2: Write the factors based on these numbers.

$(x - 5)(x + 20) = 0$.

Step 3.5: Apply the zero product property.

If a product equals zero, then at least one of the factors must be zero.

$x - 5 = 0$ or $x + 20 = 0$.

Step 3.6: Solve the first factor for $x$.

Step 3.6.1: Set $x - 5$ to zero.

$x - 5 = 0$.

Step 3.6.2: Isolate $x$.

$x = 5$.

Step 3.7: Solve the second factor for $x$.

Step 3.7.1: Set $x + 20$ to zero.

$x + 20 = 0$.

Step 3.7.2: Isolate $x$.

$x = -20$.

Step 3.8: Combine the solutions.

The potential solutions are $x = 5$ and $x = -20$.

Step 4: Determine the valid solution(s).

Since the logarithm is undefined for non-positive numbers, exclude any negative solutions.

$x = 5$ is the only valid solution.

Knowledge Notes:

The problem involves solving a logarithmic equation. Here are the relevant knowledge points and detailed explanations:

1. Logarithm Properties: Logarithms have several properties that are useful in solving equations:

   • Product Property: $\log_b(x) + \log_b(y) = \log_b(xy)$.

   • Exponential Form: A logarithmic equation $\log_b(x) = y$ can be rewritten in exponential form as $b^y = x$.

2. Quadratic Equations: A quadratic equation is an equation of the form $ax^2 + bx + c = 0$. There are various methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula.

3. Factoring: Factoring involves expressing a quadratic expression as a product of its factors. The AC method is one approach where you look for two numbers that multiply to give the product of $a$ and $c$ (the coefficients of $x^2$ and the constant term) and add up to $b$ (the coefficient of $x$).

4. Zero Product Property: If a product of factors equals zero, then at least one of the factors must be zero. This property allows us to set each factor equal to zero and solve for the variable.

5. Domain of Logarithmic Functions: The argument of a logarithmic function must be positive. Therefore, any solution that results in a non-positive argument for the logarithm is not valid.

6. Exponential Equations: When converting a logarithmic equation to its exponential form, it's important to remember that the base of the logarithm becomes the base of the exponent on one side of the equation, with the other side being the argument of the log.

In solving the given problem, these concepts are applied in sequence to simplify the logarithmic equation, convert it to a quadratic equation, factor and solve the quadratic, and finally determine the valid solution(s) based on the domain of the logarithmic function.