Problem 15 - Technical Mathematics with Calculus

Question

Solve for \(x\) to three significant digits. $$2^{3 x+1}=3^{2 x+1}$$

Step-by-Step Solution

Answer

Solve for x to get approximately 0.630.

Step 1: Write down the original equation

Start by writing down the original equation presented in the exercise: \(2^{3x+1} = 3^{2x+1}\).

Step 2: Apply logarithms to both sides

To solve for \(x\), take the natural logarithm (denoted as \(\ln\)) of both sides. This allows us to use the properties of logarithms to solve for the exponent. The equation becomes \(\ln(2^{3x+1}) = \ln(3^{2x+1})\).

Step 3: Use the Power Rule for Logarithms

Apply the power rule of logarithms, which states that \(\ln(a^b) = b \cdot \ln(a)\), to bring down the exponents. The equation now is \((3x+1) \cdot \ln(2) = (2x+1) \cdot \ln(3)\).

Step 4: Distribute and gather like terms

Distribute the terms and then collect all the \(x\) terms on one side of the equation: \(3x \cdot \ln(2) + \ln(2) = 2x \cdot \ln(3) + \ln(3)\).

Step 5: Move terms involving \(x\) to one side

Subtract terms involving \(x\) from both sides to isolate the \(x\) terms on one side of the equation: \(3x \cdot \ln(2) - 2x \cdot \ln(3) = \ln(3) - \ln(2)\).

Step 6: Factor out \(x\)

Since \(x\) is a common factor on the left side of the equation, factor it out: \(x \cdot (3 \cdot \ln(2) - 2 \cdot \ln(3)) = \ln(3) - \ln(2)\).

Step 7: Solve for \(x\)

Divide both sides of the equation by \((3 \cdot \ln(2) - 2 \cdot \ln(3))\) to solve for \(x\): \(x = \frac{\ln(3) - \ln(2)}{3 \cdot \ln(2) - 2 \cdot \ln(3)}\).

Step 8: Calculate the value of \(x\) using a calculator

Use a calculator to find the numerical value of \(x\), ensuring that it is in three significant digits: \(x \approx 0.630\).

Key Concepts

Logarithms

Logarithms are instrumental in solving exponential equations where variables reside in the exponent. They translate multiplicative relationships into additive ones, making complex equations more accessible. For instance, the logarithm of a number is the exponent to which a base, typically 10 or the natural number e, must be raised to produce that number.

In our exercise, applying logarithms to both sides of the equation allows us to work with the exponents directly. The beauty of logarithms is that they can unravel the exponent and bring it down to our level, where we can handle it just like any other number. Without logarithms, solving exponential equations would be a much more formidable task.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a logarithm to the base e, where e is an irrational and transcendental number approximately equal to 2.71828.

In the context of our exercise, we use the natural logarithm to simplify the equation. The choice of the natural logarithm is often due to its appearance in many natural phenomena and its properties which facilitate simplification when dealing with exponential functions. \(\ln\) provides a direct path to isolate the variable x in the exponential equation we aim to solve.

Properties of Logarithms

Logarithms come with a toolbox of properties that make manipulating them a breeze. Important properties include the Power Rule, the Product Rule, and the Quotient Rule. The Power Rule states that \(\ln(a^b) = b \cdot \ln(a)\), which we apply in our exercise to move the exponent in front of the logarithm.

Understanding these properties is crucial as they are the bridge between the exponent form and a form where we can solve for x algebraically. By distributing, factoring, and re-organizing terms, we can make full use of logarithmic properties to reach the solution.

Exponents

Exponents signify repeated multiplication and present a shorthand way of expressing large numbers or small decimal values. The equation from our exercise features exponents, which can be untangled using logarithms.

Exponential equations are omnipresent in various fields of study including finance, science, and engineering. Understanding how to manipulate exponents, often through logarithmic functions, is essential. The exponent rules must be thoroughly grasped to proficiently handle and solve equations where variables are exponents, as in our exemplary problem.