Problem 89 - Elementary and Intermediate Algebra

Question

a. \(\log 5 x=1.7\) b. \(\ln 5 x=1.7\)

Step-by-Step Solution

Answer

a. \( x \approx 10.024 \); b. \( x \approx 1.0948 \).

Step 1: Understand the Problem Statement for Part A

You are given the equation \( \log 5x = 1.7 \). This means that the logarithm (base 10) of \( 5x \) is 1.7. The goal is to solve for \( x \).

Step 2: Isolate the Logarithmic Expression (Part A)

Rewrite the equation \( \log 5x = 1.7 \) in its exponential form. The equation becomes \( 5x = 10^{1.7} \), because the base of a common logarithm (log) is 10.

Step 3: Calculate Exponential and Solve for x (Part A)

Calculate \( 10^{1.7} \) using a calculator to get approximately 50.119. Then, to find \( x \), divide both sides by 5: \( x = \frac{50.119}{5} \). Simplifying, we find \( x \approx 10.024 \).

Step 4: Understand the Problem Statement for Part B

You now have the equation \( \ln 5x = 1.7 \). This means that the natural logarithm (base \( e \)) of \( 5x \) is 1.7, and you need to solve for \( x \).

Step 5: Isolate the Logarithmic Expression (Part B)

Rewrite the equation \( \ln 5x = 1.7 \) in its exponential form as \( 5x = e^{1.7} \), because the base of the natural logarithm is \( e \).

Step 6: Calculate Exponential and Solve for x (Part B)

Calculate \( e^{1.7} \) using a calculator to get approximately 5.4739. Then, to find \( x \), divide both sides by 5: \( x = \frac{5.4739}{5} \). Simplifying, we find \( x \approx 1.0948 \).

Key Concepts

Understanding Logarithms

Logarithms might seem complex at first, but they're really just a way to answer this question: **What power must we raise a number (the base) to get another number?**
When you see an expression like \( \log 5x = 1.7 \), it means the base 10 logarithm of \( 5x \) equals 1.7. This base of 10 is the key to understanding the equation, as it tells us that \( 10^{1.7} \) will give us the value of \( 5x \).
To convert a logarithmic equation into its exponential form, you reverse this process; it allows us to easily solve for \( x \). For example:

• The expression \( \log 5x = 1.7 \) in exponential form becomes \( 5x = 10^{1.7} \).
• From here, we can solve for \( x \) by performing simple arithmetic operations like division.

Thus, understanding logarithms involves a shift from thinking about multiplication to thinking about powers—it's a direct but fundamental change in approach.

Natural Logarithms Explained

Natural logarithms are just like regular logarithms, but they use a special base: Euler's number, \( e \). Euler's number \( e \) is approximately 2.71828 and is a fundamental constant in mathematics.
When you come across the term or notation \( \ln \), it signifies a natural logarithm. So, when we see \( \ln 5x = 1.7 \), it implies that \( 5x \) is equal to \( e^{1.7} \). This is similar to the common logarithm case, but with the mystical \( e \) taking the place of 10.
The steps to solve natural logarithmic equations mimic those of regular log equations. For instance:

• Establish the exponential form: \( 5x = e^{1.7} \).
• Use the exponential value to solve for \( x \) by dividing both sides as necessary.

This conversion helps make sense of equations that initially seem complex, giving us a straightforward way to find unknowns.

Diving into Exponential Functions

Exponential functions define situations where growth or decay occurs exponentially; that is, every increase or decrease is by a proportion. They're written as \( a^x \), where \( a \) is a positive constant.
These functions are crucial for understanding not just logarithms but also real-life phenomena like population growth, radioactive decay, and even finance, such as compound interest.
In solving equations like \( \log 5x = 1.7 \) or \( \ln 5x = 1.7 \), converting them into exponential forms is key:

• For \( \log 5x = 1.7 \), we get \( 5x = 10^{1.7} \).
• For \( \ln 5x = 1.7 \), it becomes \( 5x = e^{1.7} \).

These conversions allow us to solve for unknowns with clarity. So, exponential functions bridge the gap between logarithmic expressions and more intuitive, solvable forms for practical use.