Problem 68 - College Algebra and Calculus: An Applied Approach

Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\log _{10} 3 z=2\)

Step-by-Step Solution

Answer

The solution to the equation \( \log _{10} 3z=2 \) approximately equals to 33.333 when rounded to three decimal places.

Step 1: Simplifying the equation

The first step is to simplify the original logarithmic equation: \( \log _{10} 3z=2 \)

Step 2: Converting the Logarithmic equation into Exponential equation

Use the rule for changing from logarithmic to exponential form. In exponential form the equation becomes \( 10^2 = 3z \)

Step 3: Solve for the variable z

Solve the exponent equation for z by dividing both sides of the equation with 3. Hence, \( z = 10^2/3 \)

Step 4: Approximating the Result

Approximate the value to three decimal places, the answer will be \( z \approx 33.333 \)

Key Concepts

Transforming Logarithmic to Exponential Form

Understanding how to shift between logarithmic and exponential forms is a crucial skill in solving logarithmic equations. Given the equation from our exercise, \( \text{log}_{10}3z=2 \) we can convert it using the fact that \( \text{log}_b a = c \) is equivalent to \( b^c = a \). In this case, the base \( b \) is 10, and the exponent \( c \) is 2.

Thus, the exponential form of our equation is \( 10^2 = 3z \). It's important to remember that this step is reversible; going from exponential back to logarithmic form can be just as necessary in other problems, ensuring a clear understanding of the relationship between the two forms is essential for algebraic problem-solving.

Algebraic Methods in Calculus

Applying algebraic methods within calculus subjects can streamline solving various types of equations, including logarithmic ones. Algebraic manipulation involves understanding the properties of equalities and operations, such as division and exponentiation. In our exercise, once we have the exponential form \( 10^2 = 3z \), we can isolate \( z \) by dividing both sides by 3:

\( z = \frac{10^2}{3} \).

This process uses fundamental algebraic methods to simplify and solve for the unknown variable. Such algebraic tactics are prevalent in calculus too, for manipulating functions and isolating variables to find derivatives or solving integrals.

Approximating Solutions

Often, exact solutions to equations involving exponential and logarithmic terms are not attainable or practical. In these cases, approximating the solution to a certain degree of accuracy is necessary. After an algebraic solution to an equation is found, like \( z = \frac{10^2}{3} \) in our case, we then approximate \( z \) to the desired precision — three decimal places, resulting in \( z \approx 33.333 \).

It's important to know when an approximation is sufficient for the problem at hand, and understanding the degree of precision required is critical. In many real-world applications, rounded values are used over exact due to their practicality and simplicity for calculations. Tools such as calculators or computational software can assist in these approximations, but the fundamental understanding of how to round to a specified number of decimal places remains an essential skill.