If x = y a, y = z band z = x c, then the value of abc is
1
Solving Exponential Equations with Multiple Variables
The problem provides us with a system of three equations involving variables x, y, and z, and exponents a, b, and c. We are asked to find the value of the product abc.
The given equations are:
- \(\text{x} = \text{y}^\text{a}\)
- \(\text{y} = \text{z}^\text{b}\)
- \(\text{z} = \text{x}^\text{c}\)
We can solve this system by using substitution. Our goal is to express one variable in terms of itself, but with a product of the exponents (a, b, and c) involved.
Let's start with the first equation and substitute the expression for y from the second equation into it:
From (1), we have \(\text{x} = \text{y}^\text{a}\).
Substitute \(\text{y} = \text{z}^\text{b}\) (from equation 2) into this equation:
\(\text{x} = (\text{z}^\text{b})^\text{a}\)
Using the rule of exponents \((m^n)^p = m^{np}\), we simplify this to:
\(\text{x} = \text{z}^{\text{ab}}\)
Now we have an equation relating x and z with the product of exponents ab. Next, we can substitute the expression for z from the third equation into this new equation.
Substitute \(\text{z} = \text{x}^\text{c}\) (from equation 3) into \(\text{x} = \text{z}^{\text{ab}}\):
\(\text{x} = (\text{x}^\text{c})^{\text{ab}}\)
Again, using the rule of exponents \((m^n)^p = m^{np}\), we simplify:
\(\text{x} = \text{x}^{\text{c} \times \text{ab}}\)
\(\text{x} = \text{x}^{\text{abc}}\)
We are left with the equation \(\text{x} = \text{x}^{\text{abc}}\). For this equation to hold true for variables x, y, and z that are typically assumed to be positive and not equal to 1 in such problems (to avoid indeterminate forms or trivial cases), the exponents must be equal. Since x on the left side has an implied exponent of 1 (\(\text{x} = \text{x}^1\)), we can compare the exponents:
\(1 = \text{abc}\)
Thus, the value of abc is 1.
This assumes that \(\text{x} \neq 0\) and \(\text{x} \neq 1\). If \(\text{x}=1\), then \(1 = 1^\text{abc}\), which is true for any value of abc. If \(\text{x}=0\), the original equations might involve \(0^a\) or \(0^c\), which are undefined for certain values of a or c (like \(a \le 0\) or \(c \le 0\)). Standard problems of this type usually imply positive bases not equal to 1.
Step-by-Step Solution for abc
Let's summarize the steps:
- Start with the equation: \(\text{x} = \text{y}^\text{a}\)
- Substitute y using \(\text{y} = \text{z}^\text{b}\): \(\text{x} = (\text{z}^\text{b})^\text{a} = \text{z}^{\text{ab}}\)
- Substitute z using \(\text{z} = \text{x}^\text{c}\): \(\text{x} = (\text{x}^\text{c})^{\text{ab}} = \text{x}^{\text{c} \times \text{ab}} = \text{x}^{\text{abc}}\)
- Compare exponents in \(\text{x}^1 = \text{x}^{\text{abc}}\) (assuming \(\text{x} > 0, \text{x} \neq 1\)): \(1 = \text{abc}\)
The final value for the product abc is 1.
Revision Table: Exponents and Variables
| Concept | Explanation | Application in Problem |
|---|---|---|
| Exponent Rule: \((\text{m}^\text{n})^\text{p} = \text{m}^{\text{np}}\) | When raising a power to another power, multiply the exponents. | Used to simplify \((\text{z}^\text{b})^\text{a}\) to \(\text{z}^{\text{ab}}\) and \((\text{x}^\text{c})^{\text{ab}}\) to \(\text{x}^{\text{abc}}\). |
| Substitution Method | Replacing a variable with its equivalent expression from another equation. | Used to combine the three initial equations into a single equation involving x and abc. |
| Equating Exponents | If \(\text{b}^\text{m} = \text{b}^\text{n}\) and \(\text{b} > 0, \text{b} \neq 1\), then \(\text{m} = \text{n}\). | Used to conclude that \(1 = \text{abc}\) from \(\text{x}^1 = \text{x}^{\text{abc}}\). |
Additional Information: Conditions for Exponent Equality
The step where we equate the exponents from \(\text{x}^1 = \text{x}^{\text{abc}}\) requires certain conditions on the base, x.
- If \(\text{x} = 0\), the original equations might be undefined. For example, \(0^{-1}\) is undefined. If defined (e.g., a, b, c are positive integers), \(0 = 0^{\text{abc}}\) is true if abc > 0, but \(0^0\) is indeterminate.
- If \(\text{x} = 1\), then \(1 = 1^\text{a}\), \(1 = 1^\text{b}\), \(1 = 1^\text{c}\). Substituting gives \(1 = 1^{\text{abc}}\), which is true for any value of abc. In this trivial case where x=y=z=1, abc is not uniquely determined.
- If \(\text{x} = -1\), the situation is more complex as powers of -1 alternate between -1 and 1. \((-1)^n\) is 1 if n is even, and -1 if n is odd. \((-1)^1 = (-1)^{\text{abc}}\) implies that 1 and abc must have the same parity if base is -1 and exponents are integers. This does not uniquely determine abc=1.
Given that 1 is provided as a definite answer among the options, the problem implicitly assumes the case where the base is a positive number not equal to 1, which allows us to confidently equate the exponents and find \(\text{abc} = 1\). This is a standard assumption in such algebraic problems unless otherwise specified.
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