Problem 73 - Algebra and Trigonometry

Question

The given equation involves a power of the variable. Find all real solutions of the equation. \(2 x^{5 / 3}+64=0\)

Step-by-Step Solution

Answer

The real solution is \(x = -8\).

Step 1: Isolate the Variable Term

Start by isolating the term involving the variable. In the equation \(2x^{5/3} + 64 = 0\), subtract 64 from both sides to get:\[2x^{5/3} = -64\]

Step 2: Eliminate the Coefficient

Divide both sides of the equation by 2 to isolate the term with the variable:\[x^{5/3} = -32\]

Step 3: Solve for the Power of x

To solve for \(x\), raise both sides of the equation to the power of \(3/5\) to cancel out the exponent on the left side:\((x^{5/3})^{3/5} = (-32)^{3/5}\)\(x = (-32)^{3/5}\)

Step 4: Simplify the Expression

Calculate \((-32)^{3/5}\). First, find the fifth root of -32, which is -2 (since \(-2^5 = -32\)), then cube the result:\(-2^3 = -8\). So, \(x = -8\).

Key Concepts

Isolating the Variable

When solving equations with exponents, the first crucial step is to isolate the variable. Isolating the variable means getting it by itself on one side of the equation. This involves moving other terms away from the variable by performing operations like addition, subtraction, multiplication, or division.
In the given exercise, we start with the equation \(2x^{5/3} + 64 = 0\). The goal is to make the term \(x^{5/3}\) easier to work with. To do this:

• Subtract 64 from both sides to remove the constant: \(2x^{5/3} = -64\).
• Then, divide every term by 2 to get \(x^{5/3}\) alone: \(x^{5/3} = -32\).

With \(x^{5/3}\) isolated, we are better positioned to handle the power of the variable more easily in the next steps.

Power of a Variable

Understanding the power of a variable is key when solving equations like \(x^{5/3} = -32\). The power, expressed as a fraction, indicates what roots and exponents to apply to both sides of the equation to simplify it.
Here, the exponent \(\frac{5}{3}\) acts as follows:

• The numerator (5) indicates raising the variable to the fifth power.
• The denominator (3) means taking the cube root.

To solve for \(x\), we perform the inverse operation using the reciprocal of the original power, which is \(\frac{3}{5}\). Raising \((x^{5/3})\) to \(\frac{3}{5}\) simplifies it to \(x\). We must do this on both sides of the equation:

• Raise \(-32\) to the power \(\frac{3}{5}\).
• This calculation involves first finding the fifth root of \(-32\) (which is \(-2\)), then cubing the result.

This process effectively cancels out the original exponent, leaving us with the real solution.

Real Solutions

Real solutions in equations involving exponents are crucial as they represent actual values of the variable that satisfy the equation. Not every solution is real; however, in many problems, especially with powers, you are interested in those solutions that can be represented on the number line.
In our exercise:

• We start with \(x^{5/3} = -32\).
• Finding a real solution means resolving \(-32\) raised to \(\frac{3}{5}\), achieving a practical number.

First, figure out the fifth root of \(-32\), which results in \(-2\) since \(-2\) multiplied by itself five times gives \(-32\). Next, cube this \(-2\): \((-2)^3 = -8\). This leads us to the real solution \(x = -8\).
The solution is real because it is a number that fits back into the original equation perfectly, proving that it is viable and valid.