Problem 14 - Intermediate Algebra

Question

For Problems \(1-14\), solve each exponential equation and express solutions to the nearest hundredth. $$ 4 e^{x}-2=26 $$

Step-by-Step Solution

Answer

The solution is approximately 1.95.

Step 1: Isolate the Exponential Term

Start by isolating the exponential term on one side of the equation. The given equation is \(4e^x - 2 = 26\). To isolate \(4e^x\), add 2 to both sides: \(4e^x = 28\).

Step 2: Solve for \(e^x\)

Divide both sides by 4 to solve for \(e^x\): \(e^x = \frac{28}{4} = 7\).

Step 3: Apply the Natural Logarithm

Take the natural logarithm of both sides to solve for \(x\). Natural logarithms are used to cancel out the exponential. So, \(\ln(e^x) = \ln(7)\).

Step 4: Simplify Using Logarithm Properties

Use the property \(\ln(e^x) = x\) to simplify: \(x = \ln(7)\).

Step 5: Evaluate and Express to Nearest Hundredth

Use a calculator to evaluate \(\ln(7)\). The value of \(x\) is approximately 1.95 when rounded to the nearest hundredth.

Key Concepts

Understanding the Natural Logarithm

Natural logarithms are a fundamental concept when dealing with exponential equations. In mathematics, the natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is approximately equal to 2.71828. This number, \( e \), is known as Euler's number and is one of the most important constants in mathematics.

One key property of natural logarithms is that they effectively "undo" the exponential function. This means that the natural logarithm of \( e \) raised to any power \( x \) is simply \( x \) itself. Formally, this can be expressed as:

• If \( a = e^x \), then \( \ln(a) = x \).

Natural logarithms are particularly useful in solving exponential equations because they allow us to isolate the variable in the exponent.

For example, in the equation \( e^x = 7 \), taking the natural logarithm of both sides yields \( \ln(e^x) = \ln(7) \). According to the property of natural logarithms, this simplifies to \( x = \ln(7) \).

Solving Exponential Equations

Solving exponential equations involves several crucial steps that simplify and solve the equation for the unknown variable.

The process begins by isolating the exponential expression. Consider the equation from the original exercise \( 4e^x - 2 = 26 \). Here, we first isolate \( 4e^x \) which involves adding 2 to both sides, resulting in \( 4e^x = 28 \).

The next step is to solve for \( e^x \). This is achieved by dividing both sides by 4, giving \( e^x = 7 \). Once the exponential term is isolated, we apply natural logarithms to both sides of the equation.

• Take the natural logarithm: \( \ln(e^x) = \ln(7) \).

Finally, simplify using logarithm properties to find the variable \( x \), specifically performed by recognizing that \( \ln(e^x) = x \), hence \( x = \ln(7) \). Using a calculator, you find \( x \approx 1.95 \) when rounded to the nearest hundredth. This approach is effective in solving a wide range of exponential equations.

Properties of Logarithms

Logarithms possess several useful properties that simplify calculations, especially when working with exponential equations. These properties are key to transitioning from exponential forms to linear forms that are easier to solve.

One fundamental property is how logarithms convert multiplication into addition. Logarithmatically, it is expressed as:

• \( \log_b(mn) = \log_b(m) + \log_b(n) \)

This property helps simplify the process of handling products of numbers in exponential equations.

Additionally, logarithms turn an exponentiation operation into multiplication, which can be expressed as:

• \( \log_b(m^n) = n \cdot \log_b(m) \)

These properties demonstrate the power of logarithms in reducing complex exponential equations to simple algebraic expressions.

Moreover, natural logarithms have specific characteristics when the base is \( e \) (Euler's number). This is particularly useful because it directly simplifies \( \ln(e^x) \) as \( x \), which simplifies equations significantly. Understanding and applying these properties correctly aids immensely in solving exponential equations efficiently.