- Use the rule \(\frac{1}{a^n} = a^{-n}\) to rewrite the first equation, leading to \(g^x = g^{-5}\), which gives \(x = -5\).
- Simplify the second equation by dividing by \(b\) and multiplying by \(c\), applying the exponent rules \(\frac{a^m}{a^n} = a^{m-n}\) and \(a^m \cdot a^n = a^{m+n}\).
- This results in \(y = b^3 c^{-6}\) or \(y = \frac{b^3}{c^6}\).
- Final answers: \(x = \boxed{-5}\), \(y = \boxed{\frac{b^3}{c^6}}\).
Explanation
- Analyzing the Equations
We are given two equations involving variables \(x\) and \(y\). We are asked to find the values (or expressions) for these variables.
- The first equation is \(g^x + h^{-3} = \frac{1}{g^5} + \frac{1}{h^3}\).
- The second equation is \(b^4 c^{-7} = \frac{by}{c}\).
We assume all variables (\(g, h, b, c\)) are non-zero to ensure the expressions are well-defined. We will use exponent rules to simplify these expressions. Recall that \(\frac{1}{a^n} = a^{-n}\).
- Solving for x
Start with the first equation:
\[g^x + h^{-3} = \frac{1}{g^5} + \frac{1}{h^3}\]
Using the rule \(\frac{1}{a^n} = a^{-n}\), we rewrite the right side:
\[g^x + h^{-3} = g^{-5} + h^{-3}\]
By subtracting \(h^{-3}\) from both sides, we isolate the term with \(g\):
\[g^x = g^{-5}\]
Since the bases are the same (assuming \(g
eq 0, 1, -1\)), we can equate the exponents:
\[x = -5\]
Thus, the value of \(x\) is \(-5\).
- Solving for y
Now move to the second equation:
\[b^4 c^{-7} = \frac{by}{c}\]
We want to isolate \(y\). First, note that \(\frac{by}{c} = b \cdot y \cdot c^{-1}\). So the equation is:
\[b^4 c^{-7} = b \cdot y \cdot c^{-1}\]
Divide both sides by \(b\):
\[\frac{b^4}{b} c^{-7} = y \cdot c^{-1}\]
Using the rule \(\frac{a^m}{a^n} = a^{m-n}\), we have \(\frac{b^4}{b^1} = b^{4-1} = b^3\).
So, the equation becomes:
\[b^3 c^{-7} = y \cdot c^{-1}\]
Now, multiply both sides by \(c^1\) to isolate \(y\):
\[y = b^3 c^{-7} \cdot c^1\]
Using the rule \(a^m \cdot a^n = a^{m+n}\), we have \(c^{-7} \cdot c^1 = c^{-7+1} = c^{-6}\).
Therefore:
\[y = b^3 c^{-6} = \frac{b^3}{c^6}\]
Examples
This problem demonstrates how to manipulate algebraic expressions involving exponents, which is a fundamental skill in fields like engineering and physics. For example, when calculating the decay of a substance or the power output of an electrical circuit, one might encounter formulas like \(P = I^2 R\) or exponential decay models. Solving for specific variables within these complex expressions allows for determining parameters such as time, resistance, or initial intensity. Mastery of exponent rules ensures accuracy in modeling these real-world physical phenomena.