Problem 68 - Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \log (x-90)+\log x=3 $$

Step-by-Step Solution

Answer

The solution is \( x = 100 \).

Step 1: Recognize and Apply the Product Rule

Notice the equation involves the sum of two logarithmic terms. We can use the product rule for logarithms here, which states that \( \log a + \log b = \log(ab) \). Therefore, combine the logarithms: \( \log((x-90) \cdot x) = 3 \).

Step 2: Rewrite the Logarithmic Equation

Since now we have \( \log((x-90) \cdot x) = 3 \), rewrite it in exponential form to eliminate the logarithm: \((x-90) \cdot x = 10^3 \). This comes from knowing that \( \log_{10} y = z \) implies \( y = 10^z \).

Step 3: Solve the Quadratic Equation

The equation becomes a quadratic: \( x^2 - 90x = 1000 \), simplifying to \( x^2 - 90x - 1000 = 0 \). Next, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -90 \), and \( c = -1000 \).

Step 4: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac = (-90)^2 - 4(1)(-1000) = 8100 + 4000 = 12100 \).

Step 5: Apply the Quadratic Formula

Substitute the values into the quadratic formula: \( x = \frac{90 \pm \sqrt{12100}}{2} \). You find \( \sqrt{12100} = 110 \). Thus, \( x = \frac{90 \pm 110}{2} \).

Step 6: Determine the Possible Values for x

Calculating the values gives \( x = \frac{90 + 110}{2} = 100 \) or \( x = \frac{90 - 110}{2} = -10 \). Since \( x \) must be greater than 90 (ensuring \( x-90 > 0 \)), only \( x = 100 \) is valid.

Step 7: Verify the Solution

To verify, substitute \( x = 100 \) back into the original equation: \( \log(100-90) + \log(100) = \log(10) + \log(100) = 1 + 2 = 3 \). The left side equals the right side, confirming the solution.

Key Concepts

Quadratic Formula

The quadratic formula is a crucial tool used to solve quadratic equations, which are polynomials of the form \( ax^2 + bx + c = 0 \). This formula allows us to find the values of \( x \) that satisfy the equation. The quadratic formula is derived from the process of completing the square and is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a \), \( b \), and \( c \) are coefficients of the quadratic equation, and \( \pm \) indicates that there will generally be two solutions. The "\( \pm \)" gives both the positive and negative square roots, providing two potential values for \( x \).
For example, in our exercise, the equation \( x^2 - 90x - 1000 = 0 \) can be solved using the quadratic formula with \( a = 1 \), \( b = -90 \), and \( c = -1000 \). By applying these values to the formula, you get the solutions that help you determine the possible values for \( x \).

Product Rule for Logarithms

The product rule for logarithms is a valuable property that simplifies the addition of logarithmic terms. This rule states that the logarithm of a product is equal to the sum of the logarithms of the factors:

• \( \log_a (b \times c) = \log_a b + \log_a c \)

In this form, \( a \) is the base of the logarithm, while \( b \) and \( c \) are positive numbers.
In the context of our given logarithmic equation, \( \log(x-90) + \log(x) = 3 \), we apply the product rule to combine these logarithms into one:

• \( \log((x-90) \cdot x) = 3 \)

This transformation makes it possible to switch from a logarithmic to an exponential equation, which is often easier to solve.

Discriminant

The discriminant is a component of the quadratic formula under the square root sign, represented as \( b^2 - 4ac \). It gives us important information about the nature and number of solutions of a quadratic equation:

• If the discriminant is positive, the equation has two distinct real solutions.
• If the discriminant is zero, there is exactly one real solution (a repeated root).
• If the discriminant is negative, the equation has no real solutions (the solutions are complex).

For the equation \( x^2 - 90x - 1000 = 0 \), we calculate the discriminant to be \( (-90)^2 - 4(1)(-1000) = 12100 \). Since 12100 is positive, it indicates that there are two distinct real solutions. This step is essential to understanding what types of solutions are possible before applying the quadratic formula.

Exponential Form of Logarithms

Rewriting logarithmic equations in their exponential form is a common method used to eliminate the logarithm and simplify solving the equation. The connection between logarithms and exponents can be expressed as:

• \( \log_b y = x \) implies \( y = b^x \)

This means any logarithmic equation can be rewritten as an exponential equation, which often makes finding the solution more straightforward.
In the problem \( \log((x-90) \cdot x) = 3 \), converting this to an exponential form gives \((x-90) \cdot x = 10^3\). This transformation effectively removes the logarithm, turning the problem into a straightforward equation to solve. By doing this, you can focus on handling the algebraic equation created from the original problem, leading up to the use of the quadratic formula we discussed earlier.