Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems 17-24, solve for } x . \text { Hint: } \log _{a} b=c \Leftrightarrow a^{c}=b \text {. } $$ $$ 2 \log _{9}\left(\frac{x}{3}\right)=1 $$

Step-by-Step Solution

Answer

The value of \( x \) is 9.

Step 1: Simplify the Equation

Start by simplifying the given equation: \( 2 \log_9\left(\frac{x}{3}\right) = 1 \). We can divide both sides of the equation by 2 to isolate the logarithm term. So it becomes: \( \log_9\left(\frac{x}{3}\right) = \frac{1}{2} \).

Step 2: Apply the Logarithm Properties

According to the hint, \( \log_a b = c \) implies that \( a^c = b \). Therefore, here we have \( \log_9\left(\frac{x}{3}\right) = \frac{1}{2} \), which implies that \( 9^{\frac{1}{2}} = \frac{x}{3} \).

Step 3: Simplify the Exponential Expression

Calculate \( 9^{\frac{1}{2}} \). This expression represents the square root of 9, which simplifies to 3. Therefore, we have \( 3 = \frac{x}{3} \).

Step 4: Solve for x

Multiply both sides of the equation \( 3 = \frac{x}{3} \) by 3 to solve for \( x \). This gives \( x = 3 \times 3 \), resulting in \( x = 9 \).

Key Concepts

Properties of Logarithms

The properties of logarithms are essential tools that help us work with and simplify logarithmic expressions. Understanding these properties can make solving logarithmic equations easier. One fundamental property is that the logarithm of a quantity can be multiplied by a coefficient. For example, when you have an equation like \( 2 \log_9\left(\frac{x}{3}\right) = 1 \), you can use the property to rewrite it as \( \log_9\left(\frac{x}{3}\right)^2 = 1 \). This step simplifies the expression and aids in solving the equation.

Another important property is the logarithm rule: \( \log_a b = c \) means that \( a^c = b \). This is particularly useful in converting a logarithmic expression into an exponential form, which can then be solved using basic algebra. In our example, after simplifying the expression to \( \log_9\left(\frac{x}{3}\right) = \frac{1}{2} \), this property allows us to transform this into an equation of the form \( 9^{\frac{1}{2}} = \frac{x}{3} \).

Through these properties, we can untangle complex logarithmic equations and solve them systematically.

Exponential Expressions

Exponential expressions involve a base raised to a power or an exponent. They are linked to logarithms because logarithms are basically the inverse operations of exponentiation. After you have used logarithmic properties to simplify an equation, the next step often involves converting that logarithmic expression into an exponential form to solve for the variable.

In the equation \( \log_9\left(\frac{x}{3}\right) = \frac{1}{2} \), we apply the property \( \log_a b = c \) implies \( a^c = b \), to express it exponentially as \( 9^{\frac{1}{2}} = \frac{x}{3} \).
Calculating \( 9^{\frac{1}{2}} \) is crucial here. Since the exponent \( \frac{1}{2} \) indicates a square root, we find that \( 9^{\frac{1}{2}} = 3 \), because 3 is the square root of 9. This simplification transforms the equation into one that is much easier to handle algebraically.

Solving Equations

Solving equations often requires a series of logical steps that simplify the equation until the variable can be isolated. After converting a logarithmic equation to its exponential form, as in \( 9^{\frac{1}{2}} = \frac{x}{3} \), the next task is to solve for the variable, which is \( x \) in this case.

The key is to perform the same operation on both sides of the equation to maintain equality. Here, we are trying to solve for \( x \) in the expression \( 3 = \frac{x}{3} \). By multiplying both sides by 3, we eliminate the fraction and isolate \( x \). This yields \( x = 3 \times 3 \), resulting in \( x = 9 \).
These steps confirm that careful manipulation of equations and consistent application of algebraic rules will guide you to the correct solution. Understanding these methods is crucial for solving not just logarithmic equations, but also a broader range of algebraic problems.