Problem 23 - 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. $$ 4^{x}-3\left(4^{-x}\right)=8 $$

Step-by-Step Solution

Answer

The exact solution is x = \( \frac{\log_{10}(4+\sqrt{19})}{\log_{10}(4)} \), approximately 1.40.

Step 1: Substitute Variables

To solve the equation \( 4^x - 3(4^{-x}) = 8 \), let \( y = 4^x \). Then, \( 4^{-x} = \frac{1}{y} \). This substitution transforms our equation into: \( y - \frac{3}{y} = 8 \).

Step 2: Clear the Denominator

Multiply both sides of the equation \( y - \frac{3}{y} = 8 \) by \( y \) to eliminate the fraction: \[ y^2 - 3 = 8y \]. This simplifies to \[ y^2 - 8y - 3 = 0 \].

Step 3: Solve the Quadratic Equation

Solve the quadratic equation \( y^2 - 8y - 3 = 0 \) using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -8, \text{ and } c = -3 \). Substitute these values into the formula:\[ y = \frac{8 \pm \sqrt{64 + 12}}{2} = \frac{8 \pm \sqrt{76}}{2} \].

Step 4: Simplify the Quadratic Formula Solution

Simplify the quadratic formula solution: \[ y = \frac{8 \pm \sqrt{76}}{2} \].Simplifying further gives:\[ y = 4 \pm \sqrt{19} \].

Step 5: Evaluate Possible Values of y

Since \( y = 4^x \) must be positive, we have two possible values for \( y \): \( y = 4 + \sqrt{19} \) and \( y = 4 - \sqrt{19} \). However, \( 4 - \sqrt{19} \) is negative, not valid for \( 4^x \).Thus, \( y = 4 + \sqrt{19} \) is valid.

Step 6: Solve for x

Since \( y = 4^x = 4 + \sqrt{19} \), we take the logarithm to find \( x \). Using base 10 log, we have \( x = \log_{10}(4 + \sqrt{19}) / \log_{10}(4) \).

Step 7: Calculate Logarithms

Compute the logarithms: \( x = \frac{\log_{10}(4 + \sqrt{19})}{\log_{10}(4)} \). Using a calculator, approximate these values: \( \log_{10}(4 + \sqrt{19}) \approx 0.8450 \) and \( \log_{10}(4) \approx 0.6021 \).

Step 8: Final Calculation

Calculate \( x \) with the approximate log values: \( x = \frac{0.8450}{0.6021} \approx 1.40 \) (with two decimal places).

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \). Each term is significant:

• \( a \) is the coefficient of \( x^2 \),
• \( b \) is the coefficient of \( x \), and
• \( c \) is the constant term.

Quadratic equations are called 'quadratic' because they involve the square of the variable, \( x^2 \).
To solve these equations, you can use several methods including factoring, completing the square, or the quadratic formula.
In our exercise, the equation transformed to a quadratic form: \( y^2 - 8y - 3 = 0 \). Here, we've set the equation to zero, a key step in setting up a quadratic equation properly.

Substitution Method

The substitution method is a common technique in algebra to simplify complex equations. By substituting part of the expression with a new variable, the equation often becomes easier to handle.
The essential steps are:

• Identify a complex part in the equation to substitute with a simpler variable.
• Rewrite the entire equation using this new variable.
• Solve the simpler equation for the new variable.

In our problem, we set \( y = 4^x \) to turn an equation with negative exponents into a quadratic one, making it easier to solve.
This strategic simplification is key, paving the way for finding solutions more efficiently.

Logarithmic Functions

Logarithms are the inverses of exponential functions and are used to solve equations where the variable is an exponent.
They are useful for transforming multiplicative relationships into additive ones, simplifying calculations. The common logarithm, denoted as \( \log_{10} \), is the logarithm with base 10.
Here’s why logarithms are crucial in this context:

• They help in solving equations where the variable is in the exponent.
• Logarithms convert exponentiation into multiplication, which is easier to handle.

In the exercise, solving for \( x \) involved using common logarithms to convert \( y = 4^x \) back into a tractable form, allowing us to isolate \( x \).

Exact Solution

An exact solution means finding a precise answer to an equation without approximating. It involves solving and expressing the solution in its true form with no rounding or estimation.
In mathematics, precision is essential, especially when dealing with foundational problems like quadratic equations or those involving logarithms.
Finding an exact solution:

• Rearrange the equation until the variable of interest is isolated.
• Use mathematical operations such as logarithms or square roots with precision.
• Avoid unnecessary approximations until final calculations.

In the given exercise, we determined \( y = 4 + \sqrt{19} \) is the valid exact solution for \( 4^x \). This clear solution acts as the foundation for further approximation or real-world application, ensuring accuracy first before estimation.