Problem 325 - Precalculus

Question

Use the definition of a logarithm to solve the equation. \(10-4 \ln (9-8 x)=6\)

Step-by-Step Solution

Answer

The solution to \(10 - 4 \ln(9 - 8x) = 6\) is approximately \(x \approx 0.78525\).

Step 1: Simplify the Equation

Start by isolating the logarithmic term. Move everything else to the other side of the equation:\[ 10 - 4 \ln (9 - 8x) = 6 \]Subtract 10 from both sides:\[ -4 \ln (9 - 8x) = 6 - 10 \]This simplifies to:\[ -4 \ln (9 - 8x) = -4 \]

Step 2: Divide by the Coefficient of Logarithm

The next step is to get rid of the coefficient of the logarithm. Divide the entire equation by -4:\[ \ln (9 - 8x) = \frac{-4}{-4} \]Which simplifies to:\[ \ln (9 - 8x) = 1 \]

Step 3: Exponentiate to Remove Logarithm

Exponentiate both sides to eliminate the natural logarithm. Recall that if \ln(y) = x, then \exp(x) = y:\[ 9 - 8x = e^1 \]\[ 9 - 8x = e \]

Step 4: Solve for x

To solve for x, rearrange the equation:\[ 9 - e = 8x \]Divide both sides by 8:\[ x = \frac{9 - e}{8} \]

Step 5: Final Result

Substitute the value of e (approximately 2.718) to get a numerical value for x:\[ x \approx \frac{9 - 2.718}{8} \]\[ x \approx \frac{6.282}{8} \]\[ x \approx 0.78525 \]

Key Concepts

Exponentiation

Exponentiation is a fundamental mathematical operation that involves raising a number, known as the base, to the power of an exponent. This operation is denoted by the use of a superscript. For example, in the expression \( a^b \), \( a \) is the base and \( b \) is the exponent. The result is equal to multiplying the base by itself \( b \) times.Key points about exponentiation:

• It's an important operation in algebra and is used in solving equations.
• Exponentiation can be easily reversed using logarithms.
• The inverse operation of exponentiation is known as finding the logarithm.

One notable property is that any base raised to the power of zero equals one: \( a^0 = 1 \).In solving equations, exponentiation helps to "undo" logarithms, allowing us to transition from a logarithmic expression back to its original form. For example, in the logarithmic equation \( \ln (y) = x \), exponentiation yields \( y = e^x \). This step is crucial for isolating variables, leading us toward the solution of the equation.

Natural Logarithms

Natural logarithms are a special type of logarithm that use the mathematical constant \( e \) (approximately 2.718) as their base. The notation for a natural logarithm is \( \ln(x) \). Natural logarithms are particularly useful in calculus and complex exponential growth models.Key aspects of natural logarithms include:

• The natural logarithm of \( e \) is 1, since \( \ln(e) = 1 \).
• They possess unique properties that simplify many algebraic expressions.
• Natural logarithms are the inverse function of exponentiation when the base is \( e \).

Natural logarithms come into play when dealing with exponential growth and decay problems. For instance, if you have an exponential equation like \( e^x = y \), taking the natural logarithm of both sides, \( x = \ln(y) \), helps in solving for \( x \). This property allows us to work efficiently with equations involving exponential functions.

Solving Equations

Solving equations involves finding the value of the unknown variable that makes the equation true. It is a fundamental skill in mathematics and involves various methods depending on the type of equation.For logarithmic equations, such as the one discussed in the exercise, the steps to solve are:

• First, simplify the equation by moving terms around to isolate the logarithmic expression.
• Next, if a coefficient is present with the logarithm, divide the entire equation by this coefficient to isolate the logarithmic term completely.
• Then, apply exponentiation to remove the natural logarithm. This step involves using the property \( \ln(y) = x \) implies \( y = e^x \).
• Finally, solve for the unknown variable by rearranging the equation into a standard form and performing any arithmetic operations necessary.

In essence, solving such equations requires a clear understanding of the relationship between exponentiation and logarithms. Mastering these concepts allows for a seamless transition through the various forms of the equation until the unknown variable is isolated and determined.