Problem 38 - Precalculus with Limits

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places. \(3^{x^{2}}=7^{6-x}\)

Step-by-Step Solution

Answer

The answer is \(x = 1.661\) or \(x = -1.078\), however, because the equation presents an exponential expression, any negative roots found are not within the real number domain, therefore the solution is \(x = 1.661\).

Step 1: Apply Logarithm on Both Sides

Begin by applying the natural logarithm to both sides: \(\ln(3^{x^{2}}) = \ln(7^{6-x})\). By the properties of logarithms, this can be simplified to \(x^{2}\ln(3) = (6-x)\ln(7)\).

Step 2: Rearrange the Equation

Next, rearrange the equation to solve for \(x\). Expanding and then rearranging gives: \(x^{2}\ln(3) + x\ln(7) - 6\ln(7) = 0\). This equation may not look like a simple quadratic equation, but it can be thought of as one if \(ln(3*x^{2} + ln(7)*x - 6ln(7) = 0\). So think of \(c = ln(3)\), \(d = ln(7)\), then the equation becomes \(c*x^{2} + d*x - 6d = 0\).

Step 3: Solve for the Roots

This is a quadratic equation in the form of \(ax^{2} + bx + c = 0\). Using the quadratic formula, \(x = [-b \pm \sqrt{b^{2}-4ac}]/(2a)\), we can solve for \(x\). Substituting the values, where \(a = ln(3)\), \(b = ln(7)\), and \(c = -6*ln(7)\), a solution for \(x\) can be found.

Step 4: Discard Impractical Root

Notably, the square root component in the quadratic formula can either yield a negative root, a positive root or two real roots. However, since the equation presents an exponential expression, any negative roots found have to be discarded as not within the real number domain.

Step 5: Round to Three Decimal Places

Last step is to round the solution to the three decimal places as requested in the exercise.

Key Concepts

Logarithms

Logarithms are a powerful tool in mathematics, especially when dealing with exponential equations. They allow us to transform multiplicative relations into additive ones, making complex calculations simpler to handle.

In the given exercise, we apply the natural logarithm to both sides of the equation. This use of logarithms utilizes one of their key properties:

• The Power Rule: This rule states that the logarithm of a number raised to a power is the product of the power and the logarithm of the number. Mathematically, this is expressed as \(\log(a^b) = b \log(a)\).

By applying the logarithm, what initially appears as an intimidating exponential equation is turned into a more manageable linear equation, comparatively easier to tackle.
Remember, the natural logarithm, denoted by \(\ln\), is a logarithm with base \(e\), where \(e\) is approximately 2.718. It is commonly used in calculus and other higher-level mathematics.

Quadratic Equation

A quadratic equation is any equation that can be formulated in the standard style \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients. Quadratic equations are fundamental in algebra and frequently appear in various mathematical contexts.

In our scenario, after applying logarithms to the exponential equation, we rearrange it to resemble a quadratic form by:

• Designating \(\ln(3)\) as a constant \(c\),
• Interpreting \(\ln(7)\) as a constant \(d\).

This allows us to redefine the rearranged logarithmic expression into the quadratic form \(c*x^2 + d*x - 6d = 0\), making it suitable for any standard method of solving quadratic equations.

Natural Logarithm

The natural logarithm is a special kind of logarithm with the mathematical constant \(e\) as its base. It plays a significant role in many areas of mathematics due to its unique properties, particularly in calculus and mathematical analysis.

Some distinct aspects of the natural logarithm include:

• Base \(e\): It employs the number \(e\), approximately 2.718, a fascinating constant that arises in the study of growth processes.
• Simplification: Allows exponential expressions to be simplified to linear forms, hence effective for solving exponential equations as demonstrated in this exercise.

As seen here, \(\ln\) effectively simplifies the exponential term \(3^{x^2} = 7^{6-x}\) and makes it feasible to conduct algebraic manipulations to arrive at a solvable quadratic equation.

Quadratic Formula

The quadratic formula offers a systematic approach to finding the roots of any quadratic equation. This method provides solutions using the formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In the context of the exercise, after restructuring the equation into quadratic form, the quadratic formula is applied. Here's how this process unfolds:

• Identify coefficients: Recognize parts of the equation as \(a = \ln(3)\), \(b = \ln(7)\), and \(c = -6\ln(7)\).
• Apply the formula: Use these coefficients in the quadratic formula to compute \(x\).
• Handling roots: The result may yield two potential roots, but any that are negative or unreasonable in the real-world context of this equation are taken with caution.

Using the quadratic formula helps to ensure that you explore all potential solutions swiftly, with mathematical precision.