Problem 100 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the logarithmic equation algebraically. Round the result to three decimal places. Verify your answer(s) using a graphing utility. $$3+2 \ln x=10$$

Step-by-Step Solution

Answer

The logarithmic equation yields the solution \(x \approx 33.115\).

Step 1: Arrange The Equation

First isolate the term with logarithm. Subtract 3 from both sides: \[2 \ln x = 10 - 3\] which simplifies to \[2 \ln x = 7\].

Step 2: Remove the Coefficient of ln

Next, divide both sides of the equation by 2 to isolate \(\ln x\): \[\ln x = 7/2 = 3.5\].

Step 3: Un-Log the Equation

Getting rid of the natural logarithm on the left side would leave just x. To undo the natural logarithm, use the property that \(e^{\ln a} = a\). Thus, you get \[x = e^{3.5}\].

Step 4: Calculate X

Using a calculator, \(x = \approx 33.115\), rounding to three decimal places.

Step 5: Check The Solution Graphically

A graph of the equation \(y = 3 + 2 \ln x\) and \(y = 10\) would intersect at \(x = 33.115\) verifying that this is the correct solution.

Key Concepts

Logarithms

Logarithms are an integral part of algebra and provide a different way to represent exponents. The logarithm of a number is the exponent to which another fixed value, the base, must be raised to produce that number. For example, in the equation \( a = b^c \), the logarithm of \( a \) to base \( b \) is \( c \) or \( \log_b(a) = c \).

Understanding logarithms is key to solving equations that involve exponential relationships. When you encounter a logarithmic equation, such as \( 3+2 \ln x=10 \), it's crucial to recognize that the logarithmic term, \( \ln x \), represents the power to which the number \( e \) must be raised to produce \( x \). This understanding allows you to isolate and eliminate the logarithmic part of the equation, as seen in the step-by-step algebraic solution.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \) where \( e \) is an irrational constant approximately equal to 2.71828. The equation \( \ln x = y \), therefore, means that \( e \), raised to the power \( y \), equals \( x \) or \( e^y = x \).

Using the natural logarithm simplifies the handling of equations involving \( e \) since the relationship between \( e \) and \( \ln \) is direct. This property is incredibly useful when solving natural logarithmic equations, as it allows for the application of exponential rules to 'un-log' an expression—rendering a complex logarithmic equation into a simpler exponential one.

Exponential and Logarithmic Properties

To efficiently solve logarithmic equations, it is vital to understand key exponential and logarithmic properties. These properties make it easier to manipulate and ultimately solve equations that contain logarithms or exponents.

Some of these fundamental properties include:

• The product rule: \( \ln(a \cdot b) = \ln(a) + \ln(b) \)
• The quotient rule: \( \ln(\frac{a}{b}) = \ln(a) - \ln(b) \)
• The power rule: \( \ln(a^c) = c \ln(a) \)
• The inverse property: \( e^{\ln a} = a \)

These properties assist in rewriting equations into a more solvable form, as we do when removing the coefficient from the \( \ln \) term or when converting a natural logarithm to its base 'e' exponent.

Graphing Calculator Usage

A graphing calculator can be an invaluable tool for verifying solutions to logarithmic equations. After solving an equation algebraically, a graphing calculator can graph the functions involved to visually confirm where they intersect.

For example, to verify the solution of \( 3+2 \ln x=10 \) graphically, you can enter the original equation into one graphing line and \( y=10 \) into another. The intersection point of \( 3 + 2 \ln x \) and \( y=10 \) on the graph corresponds to the \( x \) value that satisfies both equations—in this case, \( x \approx 33.115 \). Seeing the graphical representation can provide confidence in your algebraic solution and enhance your comprehension of logarithmic behavior.