Problem 21 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 11-24: Find the exact solution, using common logarithms, and a two- decimal-place approximation of each solution, when appropriate. $$ 5^{x}+125\left(5^{-x}\right)=30 $$

Step-by-Step Solution

Answer

The solutions are \(x = 2\) and \(x = 1\).

Step 1: Substitute and simplify

Given the equation \(5^x + 125(5^{-x}) = 30\), let \(y = 5^x\). Then \(5^{-x} = \frac{1}{y}\). Substitute these expressions into the equation to get \(y + 125\cdot\frac{1}{y} = 30\).

Step 2: Create a quadratic equation

Multiply through by \(y\) to eliminate the fraction: \(y^2 + 125 = 30y\). Rearrange this to form a standard quadratic equation: \(y^2 - 30y + 125 = 0\).

Step 3: Use the quadratic formula

The quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is applicable here. Plug in \(a = 1\), \(b = -30\), and \(c = 125\) to find \(y\). Thus, \(y = \frac{30 \pm \sqrt{900 - 500}}{2}\). Simplify: \(y = \frac{30 \pm \sqrt{400}}{2} = \frac{30 \pm 20}{2}\).

Step 4: Solve for y

Calculate the roots: \(y = \frac{50}{2} = 25\) and \(y = \frac{10}{2} = 5\). Since \(y = 5^x\), set \(5^x = 25\) and \(5^x = 5\).

Step 5: Solve for x using logarithms

For \(5^x = 25\), take the common logarithm: \(x\log(5) = \log(25)\). Thus, \(x = \frac{\log(25)}{\log(5)} = 2\). For \(5^x = 5\), \(x\log(5) = \log(5)\), giving \(x = 1\).

Step 6: Verify solutions

Substituting \(x = 2\) back into the original equation gives \(5^2 + 125\cdot5^{-2} = 25 + 5 = 30\), which is correct. Similarly, \(x = 1\) gives \(5^1 + 125\cdot5^{-1} = 5 + 25 = 30\), also verifying the solution.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool to solve equations in the form \(ax^2 + bx + c = 0\). It allows us to find the roots of a quadratic equation by substituting the coefficients into

• \(a\) representing the coefficient of \(x^2\)
• \(b\) the coefficient of \(x\)
• \(c\) the constant term

The formula is given by:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In our problem, after applying some substitutions, the quadratic equation \(y^2 - 30y + 125 = 0\) emerges. Here, \(a = 1\), \(b = -30\), and \(c = 125\). Substituting these into the formula and simplifying helps us find the value of \(y\), which initially comes from \(y = 5^x\). Breaking down and calculating, we find \(y = 25\) and \(y = 5\). These are the critical steps leading us to determine \(x\) in the original problem.

Common Logarithms

Common logarithms can be very useful for solving exponential equations. They are logarithms with base 10. We often write the common logarithm of a number \(n\) as \(\log(n)\). This notation allows us to express powers in a format that's easier to manipulate algebraically.

In this exercise, common logarithms are used to solve for \(x\) once we have a simpler equation. If the exponential form is \(5^x = n\), taking the common logarithm of both sides gives:

• \(x \log(5) = \log(n)\)
• Solving for \(x\) results in \(x = \frac{\log(n)}{\log(5)}\)

For our particular solutions \(5^x = 25\) and \(5^x = 5\), substituting \(n = 25\) and \(n = 5\) allows us to calculate \(x = 2\) and \(x = 1\), respectively. Understanding this process is key to solving similar logarithmic equations.

Substitution Method

In many mathematical problems, including this one, substitution is a valuable method for simplifying and solving equations. By letting \(y = 5^x\), and recognizing that \(5^{-x} = \frac{1}{y}\), the original exponential equation transforms into a more manageable quadratic: \(y + 125\cdot \frac{1}{y} = 30\).

This substitution streamlines the problem, making it easier to transform and ultimately solve via the quadratic formula.

• Eliminates complexity by reducing equation forms
• Reveals underlying relationships such as \(5^x\)

By using substitution, we reduce complications and thus form a clear path to solving the more challenging components of the equation. This step is crucial for effectively utilizing the quadratic formula later in the solution process.

Verifying Solutions

Verifying solutions is an important step to confirm the accuracy of our findings. After solving an equation and finding potential solutions, we must check that these values satisfy the original equation. This step ensures that no errors have occurred during substitution or computation.

For the equation \(5^x + 125(5^{-x}) = 30\), found solutions \(x = 2\) and \(x = 1\) need verification:

• For \(x = 2\): \(5^2 + 125 \cdot 5^{-2} = 25 + 5 = 30\)
• For \(x = 1\): \(5 + 125 \cdot \frac{1}{5} = 5 + 25 = 30\)

In both cases, the calculated \(x\) values confirm the equation holds, thus the solutions are verified as correct. This breakdown not only ensures correctness but also reinforces the problem-solving techniques used.