Problem 90 - A Graphical Approach to Precalculus with Limits

Question

The given equations are quadratic in form. Solve each and give exact solutions. $$(\log x)^{2}-6 \log x=7$$

Step-by-Step Solution

Answer

The solutions are \(x = 10^7\) and \(x = 0.1\).

Step 1: Recognize the Substitution

Observe that the equation \((\log x)^2 - 6 \log x = 7\) is quadratic in form. Let \(y = \log x\), transforming the equation into a standard quadratic form \(y^2 - 6y - 7 = 0\).

Step 2: Solve the Quadratic Equation

Now solve the quadratic equation \(y^2 - 6y - 7 = 0\). To factorize, find two numbers that multiply to \(-7\) and add to \(-6\). These numbers are \(-7\) and \(1\), so the equation factors as \((y - 7)(y + 1) = 0\).

Step 3: Find the Solutions for y

Set each factor equal to zero to find the solutions for \(y\): 1. \(y - 7 = 0\) gives \(y = 7\).2. \(y + 1 = 0\) gives \(y = -1\).

Step 4: Solve for x

Since \(y = \log x\), substitute back to solve for \(x\): 1. If \(y = 7\), then \(\log x = 7\) which implies \(x = 10^7\).2. If \(y = -1\), then \(\log x = -1\) which implies \(x = 10^{-1}\).

Key Concepts

Logarithmic Functions

Logarithmic functions are mathematical expressions that inverse exponential functions. For instance, if you have an equation like \( \log x = y \,\), it signifies that \( x = 10^y \.\) The logarithm measures the power to which the base (10 in this case) must be raised to obtain a certain number. So, if \( \log x = 7 \,\), \( x \) is the result of raising 10 to the seventh power, resulting in \( x = 10^7 \.\)

• Logarithmic functions help in solving equations where the variable exits as an exponent.
• It simplifies calculations, especially when dealing with large ranges of numbers in the context of exponential growth or decay.
• Logarithms with different bases operate under the same principles, where the base indicates the number being repeatedly multiplied.

Recognizing logarithmic forms and transforming them eases the solving of complex equations.

Factorization

Factorization is the process of breaking down an equation into a product of its simpler factors. It is a vital skill when dealing with quadratic equations.
The standard quadratic form is \( ax^2 + bx + c = 0 \,\) and the factorization strategy involves finding two numbers that multiply to \( c \,\) while also summing up to \( b \.\)

• Consider the equation \( y^2 - 6y - 7 = 0 \.\)
• Identify two factors of \( -7 \) that also add up to \(-6 \) which are \( -7 \) and \( 1 \.\)
• This leads to factorization as \((y - 7)(y + 1) = 0 \.\)

Efficient factorization simplifies the process of finding the roots of a quadratic equation.
The roots are values that satisfy the equation by making it equal to zero.

Substitution Method

The substitution method simplifies solving equations by replacing variables with simpler expressions. In complex equations, recognizing parts of the equation that can be substituted with single variables can ease difficulty.
Consider the equation \((\log x)^2 - 6 \, \log x = 7 \.\) By letting \( y = \log x \,\) the equation turns into a recognizable quadratic form: \(y^2 - 6y - 7 = 0 \.\)

• This step makes it easier to apply quadratic-solving techniques like factoring or using the quadratic formula.
• After solving \( y \), reverse the substitution to find \( x \).
• For instance, if \( y = 7 \,\) this means \( \log x = 7 \,\) implying \( x = 10^7 \.\)

Using substitution breaks down the complexity of logarithmic-related quadratic equations and provides a smoother solving process.