Problem 59 - Algebra 2

Question

Solve each equation. Check your solutions. $$ \log _{16} c-2 \log _{16} 3=\log _{16} 4 $$

Step-by-Step Solution

Answer

The solution is \( c = 36 \).

Step 1: Apply Logarithmic Identity

The given equation is \( \log_{16} c - 2 \log_{16} 3 = \log_{16} 4 \). Using the logarithmic identity \( a \log_b x = \log_b x^a \), rewrite \( 2 \log_{16} 3 \) as \( \log_{16} 3^2 \). This gives us the equation \( \log_{16} c - \log_{16} 9 = \log_{16} 4 \).

Step 2: Use the Quotient Rule

Apply the quotient rule of logarithms: \( \log_b x - \log_b y = \log_b \frac{x}{y} \). Rewrite the equation as \( \log_{16} \frac{c}{9} = \log_{16} 4 \).

Step 3: Exponential Form of Logarithms

Since the logarithmic expressions are equal, their arguments are equal as well. Set \( \frac{c}{9} = 4 \).

Step 4: Solve for c

To solve for \( c \), multiply both sides of the equation \( \frac{c}{9} = 4 \) by 9, resulting in \( c = 36 \).

Step 5: Check the Solution

Substitute \( c = 36 \) back into the original equation to verify: Calculate \( \log_{16} 36 - 2 \log_{16} 3 = \log_{16} 36 - \log_{16} 9 \). Using the quotient rule again gives \( \log_{16} \frac{36}{9} = \log_{16} 4 \), and since \( \frac{36}{9} = 4 \), the equation holds true.

Key Concepts

Logarithmic Identities

Logarithmic identities are very useful tools in solving logarithmic equations. They help us rewrite expressions in a way that can be easier to work with. These identities are based on the properties of logarithms.

One common identity is the power rule for logarithms. It states:

• \( a \log_b x = \log_b x^a \)

This rule allows us to bring the exponent of a logarithmic term inside the log as a power of its argument. For example, if we have \( 2 \log_{16} 3 \), using the power rule, it becomes \( \log_{16} 3^2 \) or \( \log_{16} 9 \).

This identity was used in the first step of the solution where we simplified \( 2 \log_{16} 3 \) to \( \log_{16} 9 \). It made the equation easier to solve by allowing us to work with simpler logarithmic expressions. Understanding and applying logarithmic identities like this can break complex logarithmic equations into more manageable parts.

Quotient Rule of Logarithms

The quotient rule of logarithms helps us combine or simplify logarithmic expressions that involve subtraction. This rule states:

• \( \log_b x - \log_b y = \log_b \frac{x}{y} \)

This means that if we have a subtraction of two logs with the same base, we can express it as the log of the quotient of their arguments.

For instance, the original equation included \( \log_{16} c - \log_{16} 9 \) during the second step, which using the quotient rule becomes \( \log_{16} \frac{c}{9} \).

By applying the quotient rule, the equation becomes simpler and thus easier to solve. This rule is particularly helpful when solving equations, as it allows us to work with a single logarithmic term instead of multiple terms, making it more straightforward to solve for the variable involved.

Exponential Form of Logarithms

Once we have simplified a logarithmic equation using identities and rules such as the quotient rule, we often convert the problem into an exponential form to solve for the unknown variable. This is because the logarithm can be interpreted as an exponent, making it easier to manipulate the equation.

The definition of a logarithm is:

• \( \log_b y = x \)

This means that in exponential terms it translates to:

• \( b^x = y \)

In the exercise, once we had \( \log_{16} \frac{c}{9} = \log_{16} 4 \), we recognized that since the logs were equal, their arguments must be equal as well: \( \frac{c}{9} = 4 \). Converting this into a simple algebraic equation allows for straightforward solutions, such as multiplying both sides by 9 to solve for \( c \).

Understanding how to switch between logarithmic forms and exponential forms allows us to leverage the power of properties of exponents, thus making solving logarithmic equations far easier.