Question: User: 9(9^x) = 27^{y+2} and 4^x + 8^y = 1
1. **State the problem:**
Solve the system of equations:
$$9(9^x) = 27^{y+2}$$
and
$$4^x + 8^y = 1$$
2. **Rewrite the bases as powers of prime numbers:**
Note that:
$$9 = 3^2, \quad 27 = 3^3, \quad 4 = 2^2, \quad 8 = 2^3$$
3. **Rewrite each term:**
$$9(9^x) = 3^2 \times (3^2)^x = 3^2 \times 3^{2x} = 3^{2 + 2x}$$
$$27^{y+2} = (3^3)^{y+2} = 3^{3(y+2)} = 3^{3y + 6}$$
4. **Set the exponents equal since bases are the same:**
$$2 + 2x = 3y + 6$$
Simplify:
$$2x - 3y = 4$$
5. **Rewrite the second equation:**
$$4^x + 8^y = (2^2)^x + (2^3)^y = 2^{2x} + 2^{3y} = 1$$
6. **Introduce variables:**
Let:
$$a = 2x, \quad b = 3y$$
Then the system becomes:
$$2x - 3y = 4 \implies a - b = 4$$
$$2^a + 2^b = 1$$
7. **Express $a$ in terms of $b$:**
$$a = b + 4$$
8. **Substitute into the second equation:**
$$2^{b+4} + 2^b = 1$$
Rewrite:
$$2^b \times 2^4 + 2^b = 1$$
$$2^b (16 + 1) = 1$$
$$17 \times 2^b = 1$$
9. **Solve for $2^b$:**
$$2^b = \frac{1}{17}$$
10. **Take logarithm base 2:**
$$b = \log_2 \left( \frac{1}{17} \right) = -\log_2 17$$
11. **Find $a$:**
$$a = b + 4 = -\log_2 17 + 4$$
12. **Recall $a = 2x$ and $b = 3y$:**
$$2x = 4 - \log_2 17 \implies x = \frac{4 - \log_2 17}{2}$$
$$3y = -\log_2 17 \implies y = -\frac{\log_2 17}{3}$$
**Final answer:**
$$x = \frac{4 - \log_2 17}{2}, \quad y = -\frac{\log_2 17}{3}$$