Problem 77 - College Algebra with Modeling and Visualization

Question

Solve each equation. Approximate answers to four decimal places when appropriate. (a) \(\log x=2\) (b) \(\log x=-3\) (c) \(\log x=1.2\)

Step-by-Step Solution

Answer

(a) 100, (b) 0.001, (c) 15.8489

Step 1: Understanding the Equation

The equation given is in the form of \( \log_{10} x = y \), where we will convert this logarithmic form to its equivalent exponential form. To solve equations involving logarithms, we convert \( \log_{10} x \) to \( x = 10^{y} \).

Step 2: Solve Equation (a)

For equation (a), given \( \log x = 2 \), the exponential form can be written as \( x = 10^2 \). Calculating this, we find \( x = 100 \).

Step 3: Solve Equation (b)

For equation (b), given \( \log x = -3 \), convert to exponential form: \( x = 10^{-3} \). Calculate to find \( x = 0.001 \).

Step 4: Solve Equation (c)

For equation (c), given \( \log x = 1.2 \), convert to exponential form: \( x = 10^{1.2} \). Calculate using a calculator to approximate: \( x \approx 15.8489 \).

Key Concepts

Logarithmic Form

The logarithmic form is a way to express algebraic equations where logarithms are involved. A logarithm answers the question: "To what exponent must the base be raised, in order to produce a given number?" For example, in

• the equation \( \log_{10} x = y \),
• \( 10 \) is the base, \( x \) is the input value, and \( y \) is the exponent.

When solving problems like these, our primary goal is to convert them into a more "workable" format, usually into an exponential form. This conversion helps simplify the equation for solving. Essentially, as you will see, logarithmic equations can be tricky at a glance, but with the proper conversion, they become much easier to solve.

Exponential Form

Converting a logarithmic equation into exponential form is one of the first steps to solving it. Here’s how it works:

• When you see \( \log_{10} x = y \), imagine it as telling you to find \( x \), the power of \( 10 \) that equals the result \( y \).
• So, it becomes \( x = 10^y \).

In our exercise, we made these conversions:

• For \( \log x = 2 \), the exponential form is \( x = 10^2 \), which equals \( 100 \).
• For \( \log x = -3 \), we have \( x = 10^{-3} \), leading to \( x = 0.001 \).
• Lastly, \( \log x = 1.2 \) turns into \( x = 10^{1.2} \), which is approximately \( 15.8489 \).

This method shows how powerful exponential form is for understanding and solving logarithmic equations. It turns a theoretical concept into something more tangible.

Approximating Solutions

Approximating solutions is key when exact values aren't easily calculated, especially when dealing with logarithms resulting in non-integer exponents. While it's straightforward to find powers like \( 10^2 \) or \( 10^{-3} \), something like \( 10^{1.2} \) requires more precision.
Here’s how you can do it:

• Using a scientific calculator or software, you can calculate \( 10^{1.2} \) directly.
• When approximating, it’s common to round results to four decimal places for better accuracy, as shown with \( 10^{1.2} \approx 15.8489 \).

These approximations are important in science and engineering, where exact answers aren't always necessary, but a high degree of accuracy is crucial. This approach ensures you have the most precise numbers possible when working through logarithmic problems.