Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(2 \ln x=7\)

Short answer

The solution to the equation is x = 33.115

Step-by-step solution

Isolate the Logarithm

The logarithm in this case is \(2 \ln x\), and it is already isolated on the left side of the equation. The equation can be rewritten as \(2 \ln x = 7\).

Remove the Coefficient

The coefficient of the natural logarithm is 2. Divide both sides by 2 to get \(\ln x = 7/2\) or \(\ln x = 3.5\).

Convert to Exponential Form

Knowing that the natural logarithm is the inverse of the exponential function, you can remove the natural logarithm by expressing the equation in exponential form. This gives \(e^{3.5} = x\).

Calculate the Value of x

Using a calculator, \(e^{3.5}\) can be calculated as approximately 33.115. Therefore, \(x = 33.115\).

Key concepts

Understanding Natural Logarithms

The natural logarithm is a special type of logarithm with a base of Euler's number, denoted as \( e \). Euler's number \( e \) is an irrational number approximately equal to 2.71828. Natural logarithms are used frequently in mathematical models involving growth and decay, such as in finance or sciences.
The natural logarithm of a number \( x \) is written as \( \ln(x) \). It answers the question: "To what power must \( e \) be raised to obtain \( x \)?". For example, if \( \ln(x) = 2 \), then \( e^2 = x \).

• The natural logarithm is the inverse operation of exponentiation with base \( e \).
• It simplifies solving exponential equations because you can switch back and forth between exponential and logarithmic form.
• Natural logarithms can transform multiplicative processes into additive ones.

In the exercise, we manipulated the natural logarithm \( 2 \ln x = 7 \) by dividing both sides by 2 to isolate \( \ln x \). By doing this, it is easier to convert into an exponential form in the next steps.

Exploring Exponential Functions

Exponential functions are mathematical expressions in which variables appear as exponents. An exponential function is often expressed as \( f(x) = a \cdot b^x \), where \( b \) is the base and \( x \) is the exponent.
In context with natural logarithms, the base \( b \) is Euler's number \( e \), that makes the function \( f(x) = e^x \).
Exponential functions grow rapidly. For instance, for base \( e \), this growth reflects naturally occurring phenomena with constant relative rates, such as compound interest or population growth.

• Exponential functions are the inverse of logarithmic functions, meaning if \( f(x) = e^x \), then \( \ln(f(x)) = x \).
• They are vital in calculus, particularly in derivatives and integrals, where the natural exponential function \( e^x \) has unique properties.

In the solution, converting \( \ln x = 3.5 \) into exponential form \( e^{3.5} = x \) simplifies solving for \( x \). It makes use of the inverse relationship between natural logarithms and exponential functions.

Solving Equations Algebraically

Algebraic solutions involve using algebraic methods to solve equations, such as rearranging terms and performing inverse operations. These methods make it easier to isolate variables and find exact or approximate solutions.
In the problem, the equation \( 2 \ln x = 7 \) required using algebraic techniques to solve for \( x \) step-by-step.

• Begin by isolating the logarithmic term, ensuring all components involving the variable are on one side of the equation.
• Divide to remove coefficients, simplifying the equation to its simplest logarithmic form.
• Convert the logarithmic equation into an exponential equation, which allows for easier calculation of numeric solutions.

Using a calculator was vital for the final step of solving \( e^{3.5} \), approximating \( x \) to be about 33.115. This approach demonstrates the power of algebraic manipulation when working with both simple and complex equations.