Problem 76 - Elementary and Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \log x^{3}=3 $$

Step-by-Step Solution

Answer

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

Step 1: Understand the given equation

The given equation is \( \log x^3 = 3 \). This equation expresses that the logarithm of \( x^3 \) is equal to 3. Our goal is to find the value of \( x \) that satisfies this equation.

Step 2: Rewrite the expression using properties of logarithms

Using the property of logarithms, \( \log x^a = a \log x \), we can rewrite the equation as \( 3 \log x = 3 \). This simplifies our equation and isolates the logarithmic part.

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

Divide both sides of the equation by 3 to solve for \( \log x \): \[ \log x = 1 \]

Step 4: Convert the logarithmic equation to an exponential equation

Convert the logarithmic form \( \log x = 1 \) to its equivalent exponential form: \[ x = 10^1 \] This is because the base of the logarithm is 10 when not specified.

Step 5: Calculate the value of \( x \)

Now directly calculate the value of \( x \) from the exponential equation \[ x = 10 \] Therefore, the value of \( x \) that satisfies \( \log x^3 = 3 \) is 10.

Key Concepts

Properties of Logarithms

Understanding the properties of logarithms is crucial to solving logarithmic equations efficiently. Logs have specific properties that make calculations a lot simpler and help transform complex equations into more manageable ones.
One of these properties is the power property of logarithms, which states:

• \( \log_b(x^a) = a \log_b(x) \)

This property allows us to move the exponent of a log inside, making it easier to solve as demonstrated in the exercise. Here, \( x^3 \) becomes \( 3 \log x \). As you can see, applying this property helped rewrite the original equation into a simpler form. This step not only facilitates solving the equation but also helps us to better handle more complex algebraic expressions involving logarithms.Additionally, having a grasp of other logarithmic properties such as the product and quotient rules can be instrumental for tackling more challenging problems in the future.

Exponential Equations

After transforming a logarithmic equation using its properties, the next logical step is often rearranging it into an exponential form. Understanding how to maneuver between logarithmic and exponential equations is fundamental.

• An exponential equation is one where the variable appears in the exponent, for example, \( x = 10^y \).

When we simplified \( \log x = 1 \) to the exponential form \( x = 10^1 \), we essentially solved for \( x \) by finding what number \( x \) must be so that the log of \( x \) equals a specific value.
This conversion is particularly effective because the relationship between logs and exponents is inverse; the base of the logarithm becomes the base of the exponent in an exponential equation, and the result of the logarithm becomes the exponent. Always remember this relationship as it transforms the equation and brings you closer to finding \( x \).

Exact and Approximate Solutions

In mathematics, not every solution comes out as a neat integer or simple fraction. It's often necessary to provide both exact and approximate solutions to fully satisfy the requirements of a problem.
In the exercise, we reached an exact solution: \( x = 10 \). Sometimes, though, equations won't be so straightforward, and you'll need to approximate solutions, especially when dealing with irrational numbers or when calculations require more precision.

• An approximate solution usually involves rounding a number to a specified number of decimal places, often noted in instructions or standards such as "to four decimal places."

By understanding the nature of exact and approximate solutions, you can present your answers more professionally and cater to various requirements, particularly in scientific and engineering contexts where precision can be crucial. While our exercise concluded with an exact integer, this practice of approximation sets a solid foundation for tackling more intricate problems in your mathematical journey.