Problem 27 - Algebra 2

Question

Solve each equation. Check your solutions. \(\log _{10} 16-\log _{10}(2 t)=\log _{10} 2\)

Step-by-Step Solution

Answer

The solution is \(t = 4\).

Step 1: Apply the Logarithm Property

The equation is \(\log_{10} 16 - \log_{10} (2t) = \log_{10} 2\). Use the logarithmic property \(\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)\) to combine the left side: \(\log_{10} \left(\frac{16}{2t}\right) = \log_{10} 2\).

Step 2: Equate the Arguments

Since the bases of the logarithms are the same, we set the arguments equal to each other: \(\frac{16}{2t} = 2\).

Step 3: Solve the Equation for t

Multiply both sides by \(2t\) to eliminate the fraction: \(16 = 4t\). Then, divide both sides by 4 to isolate \(t\): \(t = 4\).

Step 4: Check the Solution

Substitute \(t = 4\) back into the original equation: \(\log_{10} 16 - \log_{10} (8) = \log_{10} 2\). Simplify the left side using logarithm properties: \(\log_{10} \left(\frac{16}{8}\right) = \log_{10} 2\), which is true. Therefore, the solution \(t = 4\) is correct.

Key Concepts

Logarithmic Properties

Logarithmic properties are essential for simplifying equations involving logarithms. One of the most common properties is the difference rule:

• \( \log_b A - \log_b B = \log_b \left(\frac{A}{B}\right) \)

This property helps condense or combine terms that have a common logarithmic base. In our exercise, we used this property to simplify the left-hand side of the equation from two separate logarithmic terms into a single term.
Another crucial property is the change of base formula. However, it wasn't directly needed here, it is good to know:

• \( \log_b A = \frac{\log_c A}{\log_c b} \)

These properties let us manipulate and solve logarithmic equations more easily.

Equation Solving

Solving equations often involves getting rid of fractions or combining like terms to simplify the problem. In the exercise, once we simplified the logarithmic expression using logarithmic properties:

• The equation simplified to: \( \log_{10} \left(\frac{16}{2t}\right) = \log_{10} 2 \).
• This allowed us to equate the arguments: i.e., \( \frac{16}{2t} = 2 \).

When dealing with equations, the goal is to isolate the variable. Here, multiplying both sides by \(2t\) helped clear the way to solve for \(t\):

• Multiply both sides: \( 16 = 4t \).
• Then, solve for \( t \) by dividing by 4: \( t = 4 \).

Equation solving skills ensure we can confidently manipulate and unravel the given expression towards finding a solution.

Logarithmic Equations

In logarithmic equations, similar bases and properties play a crucial role in simplification and solution. Logarithms are exponents that indicate the power to which a base number must be raised to produce a certain value. The typical form is:

• \( \log_b x = y \Rightarrow b^y = x \)

Our exercise began by simplifying the equation using the property of logarithms, allowing for an easier comparison of the arguments. Once simplified, it's important to turn the logarithmic equation into a more straightforward format where the unknown can be solved easily.
Checking solutions involves substituting the answer back into the original equation to ensure the left side equals the right side. For us, substituting \( t = 4 \) confirmed that both sides equaled \( \log_{10} 2 \), verifying our solution was correct. This reassures accuracy and understanding of the problem's requirements.