To solve the equation \(2^{3^b} \cdot 7^c = 1176\), we begin by factoring \(1176\) into its prime components.
Prime Factorization of 1176
- First, we divide \(1176\) by \(2\):
- \(1176 \div 2 = 588\)
- \(588 \div 2 = 294\)
- \(294 \div 2 = 147\)
- Next, divide \(147\) by \(3\):
- \(147 \div 3 = 49\)
- Finally, divide \(49\) by \(7\):
- \(49 \div 7 = 7\)
- \(7 \div 7 = 1\)
Thus, the complete prime factorization is:
Factorization Result
- \(1176 = 2^3 \cdot 3^1 \cdot 7^2\)
Equating Exponents
From the equation \(2^{3^b} \cdot 7^c = 2^3 \cdot 3^1 \cdot 7^2\), we can equate the exponents of corresponding bases.
- For base \(2\):
- \(3^b = 3\) ⟹ \(b = 1\)
- For base \(7\):
- \(c = 2\)
Final Evaluation of \(10^a \cdot 7^{-c}\)
Now, we need to find \(10^a \cdot 7^{-c}\). Assuming \(a = 0\) (since \(10^0 = 1\)):
- Substituting the value of \(c\):
- \(10^0 \cdot 7^{-2} = 1 \cdot \frac{1}{7^2} = \frac{1}{49}\)
Conclusion
Thus, the value of \(10^a \cdot 7^{-c} = \frac{1}{49}\).