Question

If \(x=y^2, y=z^2\) and \((x y z)^4 / 2=64, x\) is: (A) 2 (B) 4 (C) 8 (D) 1 (E) 16

Short answer

The correct answer is (A) 2.

Step-by-step solution

- Understand the given equations

You are given three relationships between variables: 1. \(x = y^2\)2. \(y = z^2\)3. \((xyz)^4 / 2 = 64\)

- Substitute values

First, express all variables in terms of one variable. Using the second equation, substitute \(y = z^2\) into the first equation to get \(x = (z^2)^2 = z^4\).

- Combine into a single equation

Next, substitute the expressions for \(x\) and \(y\) into the third equation: \[ \left( z^4 \times z^2 \times z \right)^4 / 2 = 64\].

- Simplify the equation

Simplify the combined term inside the parentheses: \(z^4 \times z^2 \times z = z^{4+2+1} = z^7\). The equation becomes: \[ (z^7)^4 / 2 = 64\].

- Solve for z

Apply the exponent: \(z^{28} / 2 = 64\). Multiply both sides by 2 to isolate the exponent term: \(z^{28} = 128\). Take the 28th root of both sides: \(z = 128^{1/28}\). Since 128 = 2^7, this simplifies to: \(z = 2^{7/28} = 2^{1/4},\) so \(z = \frac{1}{2}\).

- Find x

Recall that \(x = z^4\). Substitute \(z = \frac{1}{2}\) into this equation: \(x = \left(\frac{1}{2}\right)^4 = \frac{1}{16}\).

Key concepts

Variable Substitution

In this exercise, variable substitution is key. Given the equations: \ \(x = y^2\), \ \(y = z^2\), \ and \ \((xyz)^4 / 2 = 64\), we start by expressing everything in terms of one variable. This makes solving easier and more systematic. For example, using \(y = z^2\), you can substitute \(z\) into the equations for \(x\) and \(y\). This gives you: \ \(x = (z^2)^2\) which simplifies to \(x = z^4\). Now, both \(x\) and \(y\) are expressed in terms of \(z\), making subsequent calculations straightforward.

Algebraic Equations

Solving the given algebraic equations involves combining and simplifying them. After substitution, the third equation becomes: \ \((z^4 \times z^2 \times z)^4 / 2 = 64\). \ Simplify inside the parentheses by combining the exponents: \ \((z^4 \times z^2 \times z) = z^{4+2+1} = z^7\). \ This reduces the equation to: \ \((z^7)^4 / 2 = 64\). By raising \(z^7\) to the power of 4, you get \(z^{28}\), so the equation now is: \ \(z^{28} / 2 = 64\). \ Multiplying both sides by 2, you solve for \(z^{28}\), which gives: \ \(z^{28} = 128\). \ Taking the 28th root of both sides: \ \(z = 128^{1/28}\). Finally, knowing \(128 = 2^7\), this simplifies to \(z = 2^{1/4}\).

Problem-Solving Steps

Let's break down the solution steps: \ 1. **Understand the given equations**: Identify relationships \(x = y^2\), \(y = z^2\), and \((xyz)^4 / 2 = 64\). \ 2. **Substitute values**: Use \(y = z^2\) to express \(x\) as \(x = z^4\). Now all variables are in terms of \(z\). \ 3. **Combine into a single equation**: Substitute back into \((xyz)^4 / 2 = 64\). \ 4. **Simplify the equation**: Simplify the variables inside the parentheses and solve for \(z\). \ 5. **Solve for \(z\)**: Isolate \(z\) by taking roots. \ 6. **Find \(x\)**: Substitute \(z\) back into \(x = z^4\) and simplify to find \(x\). In this case, we solved \(128^{1/28}\) to get \(z = 2^{1/4}\) and then found \(x = (2^{-2})^4 = 2^{-8}\). \ Each of these steps builds upon the previous, making complex problems more manageable with organized thinking and methodical problem-solving.