Problem 10 - Basic Engineering Mathematics

Question

Solve the equation \(2^{x+1}=3^{2 x-5}\) correct to 2 decimal places.

Step-by-Step Solution

Answer

The solution to the equation is approximately \(x = 4.11\).

Step 1: Simplify the Equation

We start with the equation \(2^{x+1} = 3^{2x-5}\). First, note that \(2^{x+1} = 2^x \cdot 2\). So, we can rewrite the equation as \(2 \cdot 2^x = 3^{2x-5}\). The goal is to express both sides of the equation in terms of powers making it easier to solve for \(x\).

Step 2: Apply Logarithms

Taking the natural logarithm (ln) on both sides gives us \(\ln(2 \cdot 2^x) = \ln(3^{2x-5})\). Using the logarithm property \(\ln(a \cdot b) = \ln(a) + \ln(b)\), we can expand the left side to \(\ln(2) + \ln(2^x)\). The right side becomes \((2x-5) \ln(3)\) using the property \(\ln(a^b) = b \ln(a)\).

Step 3: Simplify Further

Now our equation is \(\ln(2) + x \ln(2) = (2x-5) \ln(3)\). We can simplify the terms as follows: separate terms with \(x\) on one side leading to \(x \ln(2) - 2x \ln(3) = -5 \ln(3) - \ln(2)\).

Step 4: Factor and Solve for x

Factor \(x\) on the left-hand side to get \(x(\ln(2) - 2 \ln(3)) = -5 \ln(3) - \ln(2)\). Solve for \(x\) to find \(x = \frac{-5 \ln(3) - \ln(2)}{\ln(2) - 2 \ln(3)}\).

Step 5: Compute the Value of x

Insert the natural log values: \(\ln(2) \approx 0.693\) and \(\ln(3) \approx 1.099\). Substitute these into the equation to get \(x = \frac{-5 \times 1.099 - 0.693}{0.693 - 2 \times 1.099}\). Calculate the numerator: \(-5.495 - 0.693 = -6.188\), and the denominator: \(0.693 - 2.198 = -1.505\). Thus, \(x \approx \frac{-6.188}{-1.505} \approx 4.11\).

Step 6: Final Answer

Rounding to two decimal places, the final value of \(x\) is 4.11.

Key Concepts

Logarithms

Logarithms are powerful mathematical tools used to simplify complex calculations, especially when dealing with exponential equations. When we take the logarithm of both sides of an equation, we can transform a multiplicative problem into an additive one, making it easier to handle.

Consider the original equation: \(2^{x+1} = 3^{2x-5}\). To simplify, we use logarithms to break down the exponents:

• Apply the natural logarithm (\(\ln\)) to both sides.
• This yields: \(\ln(2^{x+1}) = \ln(3^{2x-5})\).

By using the logarithmic properties such as \(\ln(a^b) = b\ln(a)\), we can express exponents as coefficients. This transformation is an essential step in solving exponential equations. Rewriting the equation in this form helps isolate the variable for further solving.

Solving Equations

Solving equations involves finding the value of a variable that satisfies the given mathematical equation. In the context of exponential equations, this often requires multiple steps and the use of different mathematical techniques.

Let's walk through the process using our example equation:

• First, rewrite the equation with equal bases, using properties of logarithms.
• After applying logarithms, collect all terms involving the variable on one side.
• In our example: \(\ln(2) + x \ln(2) = (2x-5) \ln(3)\) is rewritten, keeping terms with \(x\) together to simplify further.

This method enables us to eventually isolate \(x\), allowing us to evaluate it numerically. Understanding this step-by-step approach is crucial for solving not only exponential equations but also various algebraic equations you may encounter.

Mathematical Properties

Mathematical properties serve as the foundation for transforming and solving equations. When working with logarithms in exponential equations, specific properties help simplify the terms:

• Product property: \(\ln(a \cdot b) = \ln(a) + \ln(b)\), used to separate or combine logarithmic terms.
• Power property: \(\ln(a^b) = b\ln(a)\), crucial for turning exponents into coefficients.

Incorporating these properties into the solution process helps in manipulating equations to a form that is easier to solve numerically or algebraically.

When you grasp these fundamental properties, solving complex equations becomes more manageable. They allow you to move seamlessly from one form of an equation to another, maintaining equality and consistency throughout the process. Mastering these properties ensures a better understanding of how mathematics operates across different problem-solving scenarios.