Problem 12 - Algebra and Trigonometry

Question

\(7-18\) Evaluate the expression. $$ \log _{12} 9+\log _{12} 16 $$

Step-by-Step Solution

Answer

The expression evaluates to 2.

Step 1: Identify the Logarithm Properties

Recognize that the expression \( \log_{12} 9 + \log_{12} 16 \) can be combined into a single logarithm using the property of logarithms that states: \( \log_b x + \log_b y = \log_b (xy) \).

Step 2: Combine the Logarithms

Using the logarithm property identified in the previous step, combine the two logarithms:\[\log_{12} 9 + \log_{12} 16 = \log_{12} (9 \times 16)\]

Step 3: Perform the Multiplication

Calculate the product inside the logarithm:\[9 \times 16 = 144\]

Step 4: Simplify the Logarithm

Substitute the product back into the logarithm to obtain:\[\log_{12} 144\]

Step 5: Evaluate the Logarithm

Determine if 144 can be expressed as a power of 12. We notice that \( 144 = 12^2 \). Thus,\[\log_{12} 144 = \log_{12} (12^2)\] Using the property of logarithms \( \log_b (a^c) = c \cdot \log_b a \), we have:\[\log_{12} (12^2) = 2 \cdot \log_{12} 12 \] Since \( \log_{12} 12 = 1 \), it follows that:\[2 \cdot \log_{12} 12 = 2\]

Key Concepts

Logarithm Properties

Logarithms have some helpful properties that simplify complex expressions into more manageable forms. The properties are essential to unlocking the potential of logarithms in solving mathematical problems efficiently.

• The **addition property** of logarithms tells us that the sum of two logarithms with the same base can be combined: \( \log_b x + \log_b y = \log_b (xy) \). This property is particularly handy when handling multiplication inside logarithms efficiently.
• The **exponentiation property** states that if a number is raised to a power inside a logarithm, it can be taken out as a multiplier: \( \log_b (a^c) = c \times \log_b a \). This property simplifies expressions where numbers can be represented as powers.
• The **identity property** tells us that the logarithm of a number with the base equal to that number is 1: \( \log_b b = 1 \). This is because any number raised to the power of 1 is itself.

Recognizing these properties is crucial in combining or simplifying logarithmic expressions as seen in the original exercise. By using these, the process of finding answers becomes more straightforward.

Logarithmic Equations

Solving logarithmic equations involves manipulating the equations using the properties of logarithms. Logarithmic equations can often involve adding, subtracting, or comparing logarithmic expressions.

• The key to solving many logarithmic equations is combining multiple logarithmic terms into a single logarithm using the properties mentioned before.
• Once the logarithms are combined, as in \( \log_b (xy) \) from the exercise, the task usually involves expressing the inside terms in a recognizable form.
• In the original solution, the equation \( \log_{12} 9 + \log_{12} 16 \) was combined to form \( \log_{12} (9 \times 16) = \log_{12} 144 \).

Recognizing patterns like these in logarithmic equations allows you to transition from what might initially seem like complex expressions to much simpler and solvable forms, making these equations approachable.

Exponents in Logarithms

Exponents often play a pivotal role in simplifying logarithmic expressions, as they allow us to see relationships in a new light. When we link logarithms and exponents, it often enables a straightforward solution path.

• In the exercise, by identifying that \( 144 = 12^2 \), we were able to simplify the logarithmic expression to take advantage of the exponentiation property.
• Once exponents are recognized, they can be factored out according to the property \( \log_b (a^c) = c \cdot \log_b a \). Thus, in this instance, \( \log_{12} (12^2) = 2 \cdot \log_{12} 12 \).
• Simplifying further with the identity property, knowing \( \log_{12} 12 = 1 \), we find \( 2 \cdot \log_{12} 12 = 2 \), thus solving the problem.

By understanding how exponents are intertwined with logarithms, you can often transform a seemingly difficult problem into an easier one, leveraging the straightforward nature of base and exponent relationships.