Problem 39 - Beginning and Intermediate Algebra

Question

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log _{2}(10+3 x)=5$$

Step-by-Step Solution

Answer

We followed these steps to solve the equation \(\log_{2}(10+3x) = 5\): 1. Converted the logarithmic equation into the exponential equation: \(2^5 = 10 + 3x\). 2. Solved the exponential equation algebraically and obtained the solution: \(x = \frac{22}{3}\). 3. Checked the solution using a graphing calculator and verified that it is correct.

Step 1: Convert the logarithmic equation into an exponential equation

We start by converting the logarithmic equation \(\log_{2}(10+3x) = 5\) into an exponential equation. We do this using the following logarithmic property: \[ \log_b{a} = c \Rightarrow b^c = a \] So, in our case, we have: \[ 2^5 = 10 + 3x \]

Step 2: Solve the exponential equation

Now we need to solve the exponential equation \(2^5 = 10 + 3x\). First, evaluate \(2^5\): \[ 2^5 = 32 \] Now we have the equation: \[ 32 = 10 + 3x \] Next, we'll isolate the term with x on one side of the equation: \[ 32 - 10 = 3x \] \[ 22 = 3x \] Now, divide both sides by 3: \[ x = \frac{22}{3} \]

Step 3: Check the solution using a graphing calculator

For the final step, use a graphing calculator to check the solution. Enter the function \(y = \log_{2}(10+3x)\) and find the point where \(y = 5\). The x-coordinate of the intersection point should be equal to the solution we found, which is \(\frac{22}{3}\). You will find that the intersection point's x-coordinate matches the value of x that we found algebraically. So, the correct solution is \(x = \frac{22}{3}\).

Key Concepts

Exponential Equations

An exponential equation is an equation where the variables appear in the form of an exponent. For example, in the equation \(2^5 = 10 + 3x\), the base is 2, and 5 is the exponent. This equation states that 2 raised to the power of 5 equals the expression on the right-hand side, \(10 + 3x\).

To solve this, we rely on the basic understanding of exponentials:

• Identify the exponential part and evaluate it. Here \(2^5 = 32\).
• Set the evaluated exponential term equal to the expression on the right side.
• Solve for the variable by isolating \(x\) on one side.
• Simplify to find the value of \(x\).

In this way, we transform a logarithmic equation into an exponential one to find a solution. Remember, the ability to convert between these forms is a critical skill as it helps in solving equations that initially seem complex.

Solving Logarithms

Solving logarithmic equations typically involves several steps, starting with understanding the relationship between logarithms and exponentials. A logarithm such as \( \log_{2}(10 + 3x) = 5\) can be converted to its exponential form to make solving straightforward. This leverages the property that \( \log_b{a} = c \Rightarrow b^c = a \).

Here’s how it works:

• Identify the logarithmic statement, \( \log_{b}(a) \), and convert it using the equivalence: if \( \log_{2}(10 + 3x) = 5 \), then it can be expressed as \( 2^5 = 10 + 3x \).
• Use the conversion to form an exponential equation.
• Once in exponential form, follow the standard procedure to solve for the unknown variable.

By transforming logarithmic equations into exponential form, it simplifies the process by narrowing down operations to basic algebra. It's also helpful to check your work by substituting your solution back into the original equation.

Graphing Calculator

Graphing calculators are excellent tools for checking solutions to algebraic equations, including logarithmic and exponential ones. After solving an equation like \( \log_{2}(10 + 3x) = 5\) algebraically, you can use a graphing calculator to visually verify the solution.

Here’s the process to use a graphing calculator:

• Enter the function into the calculator. For our example, it's \( y = \log_{2}(10 + 3x) \).
• Graph the function and look for the y-value that corresponds to 5.
• Locate the x-coordinate of the intersection point where the graph meets \( y = 5 \). This point should match the algebraically determined value of \( x = \frac{22}{3} \).

This method of verification provides a visual confirmation of your solution, enhancing understanding and confidence. It’s useful for cross-checking work, especially in complex equations, and developing an intuition for function behavior on graphs.