Problem 60 - Intermediate Algebra

Question

Solve each equation. $$ \log _{2} 5 x-\log _{2} 3=4 $$

Step-by-Step Solution

Answer

The solution is \(x = 9.6\).

Step 1: Use the Logarithm Rule

First, use the property of logarithms that states \(\log_b a - \log_b c = \log_b \left(\frac{a}{c}\right) \). This allows us to combine the logarithms:\[\log_{2} \left( \frac{5x}{3} \right) = 4\]

Step 2: Convert Logarithm to Exponent

Next, we convert the logarithmic equation into an exponential equation. Recall that if \( \log_b A = C \), then \( A = b^C \). Applying this gives us:\[\frac{5x}{3} = 2^4\]

Step 3: Evaluate Exponent

Calculate the value of \( 2^4 \). Since \( 2^4 = 16 \), we substitute that into the equation:\[\frac{5x}{3} = 16\]

Step 4: Solve for x

To isolate \(x\), multiply both sides of the equation by 3 to clear the fraction:\[5x = 48\]Now, divide both sides by 5 to solve for \(x\):\[x = \frac{48}{5} = 9.6\]

Step 5: Solution Review

The final solution is \(x = 9.6\). It's often helpful to quickly verify the solution by substituting \(x\) back into the original logarithmic equation.

Key Concepts

Logarithm Properties

Logarithms have several useful properties that simplify complex equations. Understanding these properties is crucial when dealing with logarithmic equations. One important property is the difference of logarithms:

• If you have an equation like \( \log_b a - \log_b c = \log_b \left(\frac{a}{c}\right) \), you can combine the two logarithms into one.
• This property stems from the idea that logarithms are exponents. So, subtracting exponents is the same as dividing their bases.

This property is widely used to simplify equations and make them easier to solve. In our exercise, it allows us to go from two separate logarithms to one. Breaking down the problem using this property is usually a great first step, as it reduces the number of terms you have to work with. This sets the stage for solving the equation efficiently.

Exponential Equations

Exponential equations appear once you transform a logarithmic equation by using the inverse property of logarithms. Understanding how to convert between these forms is key. Remember, an equation \( \log_b A = C \) can be rewritten as \( A = b^C \).

• This inverse relationship between logarithms and exponents helps to solve the equation, making exponential problems more accessible.
• In the example, we converted \( \log_2 \left( \frac{5x}{3} \right) = 4 \) to \( \frac{5x}{3} = 2^4 \).
• Solving the exponential equation entails calculating the power—simple arithmetical operations then follow.

By rewiring your thinking from logarithmic to exponential, you can leverage the straightforwardness of exponentiation to find a solution. This forms the core of solving any logarithmic equation.

Problem-Solving Steps

Solving complex equations becomes simpler with a structured approach. The original step-by-step method outlined in the exercise lays a great foundation for tackling these problems.

• Start by fully simplifying the logarithmic components using logarithm properties. This reduces the equation to a more manageable form.
• Transform the logarithm into an exponential equation, which is typically easier to solve.
• Solve the resulting equation through arithmetic operations, focusing on isolating the variable of interest.
• Perform operations like multiplication or division to find the solution for the variable.

Adhering to this methodical sequence not only solves the equation at hand but also reinforces the mathematical principles involved, providing a solid framework applicable to a variety of similar problems.

Verifying Solutions

Always consider verifying your solution in any problem-solving process. Making sure that your derived solution holds true in the original equation is essential.

• To verify, plug the calculated value back into the original equation.
• Check if both sides of the original equation balance out post-substitution.
• If they do, the solution is accurate; if not, revisit your steps to identify potential errors.

For example, if we substitute \( x = 9.6 \) back into the original logarithmic equation, both sides of the equation equal out, thus confirming the validity of the solution. Verification serves as a critical checkpoint ensuring your final solution is both accurate and reliable.