Question

Solve: \(2^{\frac{2}{3} x+1}-3 \cdot 2^{\frac{1}{3} x}-20=0\)

Short answer

The solution to the equation is \( x = 6 \).

Step-by-step solution

- Simplify the Exponent

Let \( y = 2^{\frac{1}{3}x} \). Thus, the given equation transforms to \( 2^{\frac{2}{3}x + 1} = 2 \times (2^{\frac{2}{3}x}) \).

- Substitute and Rewrite

Rewrite \( 2^{\frac{2}{3}x+1} \) in terms of \( y \). Since \( 2^{\frac{2}{3}x+1} = 2 \times (y^2) \), the equation becomes: \( 2y^2 - 3y - 20 = 0 \).

- Solve the Quadratic Equation

This is a quadratic equation in terms of \( y \): \( 2y^2 - 3y - 20 = 0 \). To solve for \( y \), use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Identifying \( a = 2 \), \( b = -3 \), and \( c = -20 \), it becomes: \[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-20)}}{2 \cdot 2} \]

- Simplify the Discriminant

Calculate the discriminant: \( (-3)^2 - 4 \cdot 2 \cdot (-20) = 9 + 160 = 169 \). Substitute back into the formula: \[ y = \frac{3 \pm \sqrt{169}}{4} \]

- Solve for y

Simplify further: \( y = \frac{3 \pm 13}{4} \). This yields two solutions: \[ y = \frac{3 + 13}{4} = 4 \] and \[ y = \frac{3 - 13}{4} = -2.5 \]. Since \( y = 2^{\frac{1}{3}x} \) must be positive, \( y = -2.5 \) is not valid. Thus, \( y = 4 \).

- Back-substitute for x

Recall that \( y = 2^{\frac{1}{3}x} \). So, \( 2^{\frac{1}{3}x} = 4 \). Rewriting 4 as a power of 2, \( 2^{\frac{1}{3}x} = 2^2 \). Therefore, \( \frac{1}{3}x = 2 \), and solving for \( x \): \( x = 6 \).

Key concepts

Quadratic Equations

A quadratic equation is a second-order polynomial equation in a single variable, often written in the standard form: \[ ax^2 + bx + c = 0 \]
where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable.
The solutions to a quadratic equation can be found using various methods, including:

• Factoring
• Completing the square
• Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

A key feature of quadratic equations is their graph, which is a parabola that can open upwards or downwards depending on the sign of \(a\).

Substitution Method

The substitution method is a strategy used to solve equations, including exponential and quadratic types.
It involves replacing a complex part of the equation with a simpler variable to make the equation easier to solve.
For instance, in the given problem:
\(2^{\frac{2}{3} x+1}-3 \cdot 2^{\frac{1}{3} x}-20=0\), we let \( y = 2^{\frac{1}{3}x} \) to simplify the equation.
By substituting, the original equation transforms to a more traditional form, making it feasible to apply known solving techniques.

Exponent Rules

Understanding exponent rules is vital when working with exponential equations.
Some fundamental rules are:

• Product of powers: \( a^m \times a^n = a^{m+n} \)
• Quotient of powers: \( a^m \/ a^n = a^{m-n} \)
• Power of a power: \( (a^m)^n = a^{mn} \)

In our problem, rewriting
\( 2^{\frac{2}{3}x+1} \) as \( 2 \times (2^{\frac{2}{3}x})\) required using these exponent rules.
These rules simplify complex expressions and are crucial in solving exponential equations.

Discriminant in Quadratic Equations

The discriminant is a key component in determining the nature of the roots of a quadratic equation.
It's found within the quadratic formula, given by: \[ \text{Discriminant} = b^2 - 4ac \]
The value of the discriminant reveals the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root, meaning the parabola touches the x-axis at one point.
• If the discriminant is negative, there are no real roots but two complex roots.

In our solution, calculating the discriminant for
\(2y^2 - 3y - 20 = 0\), yielded 169, a positive number suggesting two real solutions.