Problem 13 - Algebra 2

Question

Solve each equation or inequality. Round to the nearest ten-thousandth. \(e^{x}>30\)

Step-by-Step Solution

Answer

The solution is \( x > 3.4012 \).

Step 1: Define the inequality

The given inequality is \( e^{x} > 30 \). We need to solve for \( x \).

Step 2: Apply the natural logarithm

To solve for \( x \), we apply the natural logarithm (\( \ln \)) to both sides of the inequality: \( \ln(e^{x}) > \ln(30) \).

Step 3: Simplify using logarithm properties

Using the property \( \ln(e^{x}) = x \), the inequality becomes: \( x > \ln(30) \).

Step 4: Calculate the natural logarithm of 30

Find \( \ln(30) \) using a calculator. \( \ln(30) \approx 3.4012 \) when rounded to the nearest ten-thousandth.

Step 5: Write the solution

Since \( x > \ln(30) \), the solution to the inequality \( e^{x} > 30 \) is \( x > 3.4012 \).

Key Concepts

natural logarithm

The natural logarithm, denoted as \( \ln \), is a fundamental concept in mathematics which helps us deal with exponential expressions. It is based on the constant \( e \), which is approximately equal to 2.71828. This particular logarithm is useful because it simplifies expressions where \( e \) is the base.
Natural logarithms have several properties that can make calculations simpler:

• The natural logarithm of 1 is always 0, \( \ln(1) = 0 \), because any number raised to the power of 0 is 1.
• The natural logarithm of \( e \) itself is 1, \( \ln(e) = 1 \), because \( e^1 = e \).

In solving the inequality \( e^{x} > 30 \), we use the natural logarithm to transform the inequality into a more manageable linear form. By applying \( \ln \) to both sides as in \( \ln(e^{x}) > \ln(30) \), we leverage its properties to solve for \( x \). Remember, the natural logarithm is particularly useful in calculus and complex equations involving growth processes.

properties of logarithms

Logarithms have several key properties that make solving equations and inequalities more straightforward. Some important properties include:

• Power Rule: \( \ln(a^b) = b \cdot \ln(a) \). This is the rule used when you have an exponent within the argument of a logarithm.
• Product Rule: \( \ln(a \cdot b) = \ln(a) + \ln(b) \).
• Quotient Rule: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).

In the context of solving the inequality \( e^{x} > 30 \), we use the power rule: \( \ln(e^{x}) = x \cdot \ln(e) \). Since \( \ln(e) = 1 \), this simplifies to \( x \), turning the inequality into \( x > \ln(30) \).
These properties are invaluable when simplifying expressions or equations because they transform complex expressions into simpler, linear forms which are easier to handle mathematically.

solving inequalities

Solving inequalities is a crucial skill in algebra, and it involves finding the range of values that satisfy a given expression. The steps to solve an inequality like \( e^x > 30 \) follow a systematic approach:

• Start by rewriting the inequality to isolate the variable. In this example, apply \( \ln \) to both sides to get \( \ln(e^x) > \ln(30) \).
• Simplify the expression using logarithmic properties, resulting in \( x > \ln(30) \).
• Compute the mathematical value needed on the right side, \( \ln(30) \), and round it to the required precision. In this example, it is approximately \( 3.4012 \) when rounded to the nearest ten-thousandth.
• The solution to the inequality is then expressed as \( x > 3.4012 \).

Understanding these steps allows you to tackle a variety of exponential inequalities confidently. Remember, the main challenge is often in manipulating complex expressions into simpler forms via exponent rules and logarithms, allowing us to focus on the variable itself.