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Chapter 7: Problem 118

$$ 4^{\sqrt{3 x^{2}-2 x}+1}+2=9 \cdot 2^{\sqrt{3 x^{2}-2 x}} $$

Short Answer

Expert verified
Rewrite the given equation with bases 4 and 9 as powers of 2, then simplify and apply the exponential property. Rearrange the equation to isolate the radical term. Solve for x using the quadratic formula, and then check the validity of the solutions in the original equation.

Step by step solution

01

Rewrite bases 4 and 9 as powers of 2

Rewrite the given equation with bases 4 and 9 as powers of 2: $$ (2^2)^{\sqrt{3 x^{2}-2 x}+1}+2=9 \cdot 2 \cdot (2^{\sqrt{3 x^{2}-2 x}}) $$
02

Simplify the equation

Simplify the equation by applying exponentiation rules and combining terms: $$ 2^{2(\sqrt{3 x^{2}-2 x}+1)}+2=18 \cdot 2^{\sqrt{3 x^{2}-2 x}} $$
03

Apply the exponential property

The exponential property states that if \(a^{m}=a^{n}\), then \(m=n\). Apply this property to the given equation: $$ 2(\sqrt{3 x^{2}-2 x}+1)=\log_2(18) + \sqrt{3 x^{2}-2 x} $$
04

Rearrange the equation to isolate the radical term

Rearrange the equation by isolating the radical term on one side: $$ 2\sqrt{3 x^{2}-2 x} = \log_2(18) - 2 $$
05

Solve for x

Proceed by dividing both sides by 2 and squaring each side, then rearrange to form a quadratic equation: $$ 3x^2-2x=\left(\frac{\log_2(18)-2}{2}\right)^2 $$ Use the quadratic formula to solve for x: $$ x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} $$
06

Check the validity of the solutions

Finally, check if the solutions found in Step 5 are valid in the original equation. If they satisfy the original equation, then they are valid solutions. Otherwise, discard any invalid solutions. This concludes the step-by-step solution for the given exercise.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Properties of Exponents
Understanding the properties of exponents is crucial when dealing with exponential equations. These properties simplify expressions and make complex equations easier to handle.

Let's start with the basic rules:
  • The Product of Powers: When multiplying two expressions with the same base, you add the exponents. For example, \(a^m \cdot a^n = a^{m+n}\).
  • The Power of a Power: When raising an exponent to another power, you multiply the exponents: \((a^m)^n = a^{m \cdot n}\).
  • The Power of a Product: When raising a product to a power, each factor gets raised to the power: \((ab)^n = a^n \cdot b^n\).
In the given exercise, you observe these principles. For instance, the base 4 is written as \(2^2\), and we raise it to another power. As seen in the solution, \((2^2)^{\sqrt{3x^2 - 2x} + 1}\) becomes \(2^{2(\sqrt{3x^2 - 2x} + 1)}\), illustrating the power of a power property.
Quadratic Equations
Quadratic equations are polynomials of degree two, typically having the form \(ax^2 + bx + c = 0\). Solving them accurately is important for many math problems, including those involving exponents.

Quadratic equations can be solved using several methods:
  • Factoring: Breaking down the quadratic into two binomial expressions.
  • Completing the Square: Adjusting the equation to form a perfect square trinomial.
  • Quadratic Formula: Using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find solutions when other methods are challenging or impossible.
In the exercise, after simplifying the exponential equation, it transforms into a quadratic equation \(3x^2 - 2x = \left(\frac{\log_2(18) - 2}{2}\right)^2\). Using the quadratic formula is often the most straightforward way to solve such equations, especially when they involve complicated terms like logarithms.
Logarithms
Logarithms are the inverse of exponents. They help to solve equations where the unknown variable is in the exponent, which is necessary in many exponential equations.

The core components include:
  • Logarithm Definition: \(\log_b(a)\) is the exponent to which the base \(b\) must be raised to produce \(a\).
  • Properties: These include the product, quotient, and power properties:
    • Product: \(\log_b(MN) = \log_b(M) + \log_b(N)\)
    • Quotient: \(\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)\)
    • Power: \(\log_b(M^p) = p \cdot \log_b(M)\)
In the provided solution, \(\log_2(18)\) is calculated when equating exponents during simplification. This showcases the need to understand logarithms well so you can apply them in exponential equations effectively.

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