Problem 19 - Algebra A Combined Function

Question

Solve each equation. $$ \log _{2} 6-\log _{2} x=3 $$

Step-by-Step Solution

Answer

x = \(\frac{3}{4}\)

Step 1: Use the Logarithm Property

The equation given is \( \log_{2} 6 - \log_{2} x = 3 \). According to the property of logarithms, \( \log_{b} A - \log_{b} B = \log_{b} \left( \frac{A}{B} \right) \). Apply this property: \( \log_{2} \left( \frac{6}{x} \right) = 3 \).

Step 2: Rewrite the Logarithmic Equation as Exponential

The equation \( \log_{2} \left( \frac{6}{x} \right) = 3 \) implies that \( \frac{6}{x} = 2^{3} \), because a logarithmic equation \( \log_{b} A = C \) is equivalent to \( A = b^C \).

Step 3: Solve for x

Now \( \frac{6}{x} = 2^{3} \) simplifies to \( \frac{6}{x} = 8 \). Solve this equation by cross-multiplying to get \( 6 = 8x \).

Step 4: Isolate x

To find \( x \), divide both sides of the equation \( 6 = 8x \) by 8, resulting in \( x = \frac{6}{8} \). Simplify this fraction to get \( x = \frac{3}{4} \).

Key Concepts

Properties of Logarithms

The properties of logarithms are essential tools for simplifying and solving logarithmic equations. One important property is the "quotient rule." This tells us that the logarithm of a quotient is the difference of the logarithms of the numerator and denominator.
For example, the expression \( \log_{b} A - \log_{b} B \) becomes \( \log_{b} \left( \frac{A}{B} \right) \). This property helps simplify the original equation \( \log _{2} 6 - \log _{2} x = 3 \) to \( \log_{2} \left( \frac{6}{x} \right) = 3 \).
Once you understand this property, you can see how it helps in rewriting and solving equations involving logarithms. Without this property, the equation would be much more complex to solve. Remember that these properties apply only to logarithms with the same base.

• Quotient Rule: \( \log_{b} \left( \frac{A}{B} \right) = \log_{b} A - \log_{b} B \)
• Make sure the bases are the same when applying log properties.

Exponential Equations

An exponential equation is one where variables appear as exponents. A core skill in solving logarithmic equations is converting them to exponential form. This is because both forms are different ways to express the same relationship.
In essence, if \( \log_{b} A = C \), then it can be rewritten as \( A = b^C \). This conversion is crucial for solving the original problem, where \( \log_{2} \left( \frac{6}{x} \right) = 3 \) was rewritten as \( \frac{6}{x} = 2^3 \).
Understanding this relationship allows you to switch between logarithmic and exponential forms with ease. Once in exponential form, it’s often simpler to solve for the unknown variable using basic algebra. Keep practicing rewriting equations—this skill simplifies many complex problems.

• Convert log form to exponential: \( \log_{b} A = C \) is \( A = b^C \).
• Solving in exponential form often uses basic algebra and can be simpler.

Solving Logarithmic Equations

Solving logarithmic equations involves several steps, primarily using properties of logarithms and transforming to exponential equations. In step-by-step solutions, it helps to methodically break down each part of the process.
After applying the properties of logarithms to simplify, converting the equation into an exponential form is often the next step. This transition helps make the equation easier to handle with standard algebraic techniques.
For our example \( \log _{2} 6- \log _{2} x=3 \), using these techniques leads to the equation \( \frac{6}{x} = 8 \). From here, solve the equation by cross-multiplying and isolating \( x \) to find that \( x = \frac{3}{4} \).

• Use quotient rule to simplify logs: \( \log_{b} A - \log_{b} B = \log_{b} \left( \frac{A}{B} \right) \)
• Transform logs to exponentials to simplify the solve.
• Isolate the unknown variable using algebraic techniques.

By applying these strategies, solving logarithmic equations becomes more systematic and less intimidating. Practice will build confidence in working through these types of problems effectively.