Problem 44 - A Graphical Approach to Precalculus with Limits

Question

Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. $$\log x+\log x^{2}=3$$

Step-by-Step Solution

Answer

The exact solution is \( x = 10 \).

Step 1: Combine the Logarithms

We begin by using the property of logarithms that allows us to combine the logarithmic expressions: \( \log a + \log b = \log(ab) \). So, \( \log x + \log x^2 \) becomes \( \log(x \cdot x^2) = \log(x^3) \).

Step 2: Set the Equation

After combining the logarithms, the equation becomes \( \log(x^3) = 3 \). This states that the logarithm with base 10 of \( x^3 \) is 3.

Step 3: Solve for \(x^3\)

To solve for \( x^3 \), rewrite the equation in exponential form. Remember that \( \log_b a = c \) is equivalent to \( b^c = a \). So, \( x^3 = 10^3 \).

Step 4: Solve for \(x\)

Now, solve the equation \( x^3 = 1000 \). Take the cube root of both sides to find \( x \): \( x = \sqrt[3]{1000} \).

Step 5: Calculate the Exact Solution

Calculate the cube root of 1000 to solve for \( x \). Since \( \sqrt[3]{1000} = 10 \), the solution is \( x = 10 \).

Step 6: Verify Using Calculator

Use a calculator to verify the solution. Calculate \( \log 10 + \log 10^2 \) which simplifies to \( \log 1000 \). Since \( \log 1000 = 3 \), this confirms the solution is correct.

Key Concepts

Properties of Logarithms

When you work with logarithmic equations, understanding the properties of logarithms is essential. These properties are mathematical tools that simplify complex log expressions. In this equation, we used the property

• Product Property: The rule states that \( \log a + \log b = \log(ab) \). This means that when you add logarithms with the same base, you can multiply their arguments. For example, the problem \( \log x + \log x^2 \) simplifies to \( \log(x \cdot x^2) = \log(x^3) \).

This property is incredibly useful because it allows us to condense multiple logarithms into a single expression. Simplifying logs in this way often makes it easier to work with the equations, especially when trying to solve for unknown variables. Furthermore, this property is foundational when dealing with more complex logarithmic and exponential operations in math.
Once you get the hang of these properties, using logarithms for solving equations becomes much more manageable. Make sure to practice applying these properties to get comfortable with solving logarithmic equations.

Solving Equations

Solving equations, particularly when logarithms are involved, can seem tricky at first. But don't worry. With a systematic approach, they become easy to manage. After simplifying using the properties of logarithms, you need to focus on isolating the variable.

• In our example, after using the product property, the equation becomes \( \log(x^3) = 3 \).
• Your goal here is to "extract" the variable \( x \) from the logarithm.
• Establish the exponential form to rewrite the equation (we discuss this more in the next section).

To isolate \( x \), you take advantage of the inverse relationship between logarithms and exponents. Turning a logarithm into an exponent allows you to move the variable out. Remember, the main goal in solving equations is to have the unknown variable on one side by itself, so keep operations simple and logical. With practice, these steps become part of an intuitive process you can apply in more complex situations.

Exponential Form

Understanding the exponential form is crucial for solving equations that include logarithms. The key principle is transforming log equations into exponential equations to reveal their solutions.

• Conceptually, if you know that \( \log_b a = c \), then you can convert this to \( b^c = a \).
• This is known as changing a logarithmic form to an exponential form.

In our example, \( \log(x^3) = 3 \) translates to its exponential equivalent, \( x^3 = 10^3 \). This removes the logarithm, leaving you with an equation that is straightforward to solve with basic algebra. By taking the cube root, you quickly arrive at \( x = 10 \).
Converting to exponential form is particularly useful in logarithmic equations as it simplifies the process of solving for the variable. Remember, the exponential form strips away the layer introduced by the logarithm, bringing you closer to finding the variable value. Having a strong understanding of how and when to apply this transformation enhances your capability to solve a wider variety of equations effectively.