Problem 68 - Elementary and 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: Use logarithmic property to combine logs

Combine the two logarithms on the left using the property \(\log a + \log b = \log(ab)\). This gives: \(\log((x-90)x) = 3\).

Step 2: Convert logarithmic equation to exponential form

Convert the logarithmic equation to its exponential form. The equation \(\log((x-90)x) = 3\) means \((x-90)x = 10^3\).

Step 3: Solve the quadratic equation

This simplifies to the quadratic equation \(x^2 - 90x - 1000 = 0\). Solve for \(x\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -90\), and \(c = -1000\).

Step 4: Evaluate the discriminant

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

Step 5: Calculate the roots

Substitute the discriminant into the quadratic formula: \(x = \frac{90 \pm \sqrt{12100}}{2}\). The square root of 12100 is 110, so: \(x = \frac{90 \pm 110}{2}\).

Step 6: Simplify the results

Compute the two possible values of \(x\): \(x_1 = \frac{90 + 110}{2} = 100\) and \(x_2 = \frac{90 - 110}{2} = -10\).

Step 7: Determine appropriate solutions within domain

Since \(x\) must be greater than 90 to make \(\log(x-90)\) defined, only \(x = 100\) is valid. \(x = -10\) is not a valid solution because it results in a negative argument in the logarithm.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool that helps in solving quadratic equations of the form \(ax^2 + bx + c = 0\). This formula can find the solutions (or roots) of any quadratic equation, provided you know the coefficients \(a\), \(b\), and \(c\). The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] It involves:

• Extracting the coefficients \(a\), \(b\), and \(c\) from the equation.
• Calculating the discriminant, \(b^2 - 4ac\), which determines the nature of the roots.
• Using the formula to find the roots by plugging in the values of \(a\), \(b\), and \(c\).

To solve the equation \(x^2 - 90x - 1000 = 0\) from our exercise, we use \(a = 1\), \(b = -90\), \(c = -1000\). The quadratic formula allows precise determination of the roots. Remember, the roots identified might need to satisfy additional conditions, like being within a specific domain, to be considered valid solutions.

Exponential Form

Transforming a logarithmic equation into exponential form can simplify solving it. Essentially, if you have a logarithmic equation like \(\log_b(y) = x\), you can rewrite it in exponential form as \(b^x = y\). This conversion is crucial for equations involving logarithms, like our exercise's step: \(\log((x-90)x) = 3\). Here, the base \(b\) is 10 (common for logarithms), and the equation can be translated to \((x-90)x = 10^3\), thus making it easier to handle. This conversion clarifies the arithmetic and enhances understanding, especially since exponential forms are often simpler in algebraic manipulation. Through this transformation step, you can focus on solving a straightforward equation rather than dealing with logarithmic complexities.

Discriminant

The discriminant is a special component of the quadratic formula, given as \(b^2 - 4ac\). It provides valuable information about the nature of a quadratic equation's roots. Understanding the discriminant helps predict the number and type of solutions:

• If the discriminant is positive, you have two distinct real roots.
• If it equals zero, there is exactly one real root (a repeated root).
• If it is negative, the equation has no real roots, only complex ones.

In our problem, the discriminant calculated is 12100, which is positive. This indicates the equation \(x^2 - 90x - 1000 = 0\) will yield two real roots. Calculating the discriminant before using the quadratic formula can help anticipate how the solutions will behave.

Logarithmic Properties

Logarithmic properties allow simplifications and restructuring of logarithmic expressions for easier solving or manipulation. Two vital properties used in solving logarithmic equations include:

• Product Property: \(\log_a(b) + \log_a(c) = \log_a(bc)\). This combines two logs into one, as used in the exercise: \(\log(x-90) + \log(x) = \log((x-90)x)\).
• Exponential Property: Converting between logarithmic and exponential forms, \(\log_b(y) = x\) implies \(b^x = y\).

This understanding enables efficient transformation and resolution of equations featuring logarithms. For instance, by using these properties, our original logarithmic equation becomes a quadratic one that can be more easily solved.Recognizing and applying these properties not only aids in solving equations but strengthens algebraic intuition.