Notice that 3²x+4 = (3^(x+2))². Substituting u = 3^(x+2) gives the quadratic equation u² - 2u + 1 = 0, which factors as (u - 1)² = 0. Thus, 3^(x+2) = 1 = 3^0, giving x + 2 = 0, so x = -2.

Express 5²x+1 as 5 · (5^x)². Letting u = 5^x, the equation becomes 5u² - 6u + 1 = 0. Factoring gives (5u - 1)(u - 1) = 0, which yields 5^x = 5^(-1) or 5^x = 5^0. Hence, x = -1 or x = 0.

Rewrite 2^(x+3) as 8 · 2^x and move -2^4 = -16 to the left side to get (2^x)² - 8(2^x) + 16 = 0. Factoring as a perfect square gives (2^x - 4)² = 0, leading to 2^x = 4 = 2², so x = 2.

From 2^(1-(x)/(2)) = (2²)^y, equating powers of 2 gives 1 - (x)/(2) = 2y, or x + 4y = 2. Next, write (7^(1+x)) × (7²)^(-2y) = 7^(1+x-4y) = 7^0 = 1, which gives x - 4y = -1. Adding the two equations yields 2x = 1 ⇒ x = (1)/(2), and subtracting yields 8y = 3 ⇒ y = (3)/(8).

From 2^(1-(x)/(2)) = (2²)^y, equating powers of 2 gives 1 - (x)/(2) = 2y, or x + 4y = 2. Next, write (7^(1+x)) × (7²)^(-2y) = 7^(1+x-4y) = 7^0 = 1, which gives x - 4y = -1. Adding the two equations yields 2x = 1 ⇒ x = (1)/(2), and subtracting yields 8y = 3 ⇒ y = (3)/(8).

From 2^x = 2^4 · 2^y = 2^(y+4), we have x = y + 4, or x - y = 4. From (3³)^x = 3² · 3²y, we have 3³x = 3²y+2, giving 3x - 2y = 2. Substituting x = y + 4 gives 3(y + 4) - 2y = 2, which simplifies to y + 12 = 2 ⇒ y = -10, and x = -10 + 4 = -6.

From 2^x = 2^4 · 2^y = 2^(y+4), we have x = y + 4, or x - y = 4. From (3³)^x = 3² · 3²y, we have 3³x = 3²y+2, giving 3x - 2y = 2. Substituting x = y + 4 gives 3(y + 4) - 2y = 2, which simplifies to y + 12 = 2 ⇒ y = -10, and x = -10 + 4 = -6.

Solve the following equations for $x$ 11. $3^{2x+4} + 1 = 2 \cdot 3^{x+2}$ 12. $5^{2x+1} = 6 \cdot 5^x - 1$ 13. $2^{2x} - 2^{x+3} = -2^4$ Solve for $x$ and $y$ : 14. $9^x = 3^{y-2}$ , $81^y = 3^2 \times (27)^x$ 15. $2^{1-\frac{x}{2}} = 4^y$ , $(7^{1+x}) \times (49)^{-2y} = 1$ 16. $2^x = 16 \times 2^y$ , $(27)^x = 9 \times 3^{2y}$

Answer: 11. $x = -2$ 12. $x = -1$ or $x = 0$ 13. $x = 2$ 14. $x = -\frac{6}{5},\; y = -\frac{2}{5}$ 15. $x = \frac{1}{2},\; y = \frac{3}{8}$ 16. $x = -6,\; y = -10$

Step-by-step solution

Step 1: Solve Equation 11

Notice that $3^{2x+4} = (3^{x+2})^2$ . Substituting $u = 3^{x+2}$ gives the quadratic equation $u^2 - 2u + 1 = 0$ , which factors as $(u - 1)^2 = 0$ . Thus, $3^{x+2} = 1 = 3^0$ , giving $x + 2 = 0$ , so $x = -2$ .

Step 2: Solve Equation 12

Express $5^{2x+1}$ as $5 \cdot (5^x)^2$ . Letting $u = 5^x$ , the equation becomes $5u^2 - 6u + 1 = 0$ . Factoring gives $(5u - 1)(u - 1) = 0$ , which yields $5^x = 5^{-1}$ or $5^x = 5^0$ . Hence, $x = -1$ or $x = 0$ .

Step 3: Solve Equation 13

Rewrite $2^{x+3}$ as $8 \cdot 2^x$ and move $-2^4 = -16$ to the left side to get $(2^x)^2 - 8(2^x) + 16 = 0$ . Factoring as a perfect square gives $(2^x - 4)^2 = 0$ , leading to $2^x = 4 = 2^2$ , so $x = 2$ .

Step 4: Solve System 14 for x and y

Express all terms with base 3: $9^x = 3^{2x} = 3^{y-2}$ , so $2x = y - 2$ . From $81^y = 3^{4y}$ and $3^2 \times (27)^x = 3^{3x+2}$ , we obtain $4y = 3x + 2$ . Substituting $y = 2x + 2$ into the second equation yields $4(2x + 2) = 3x + 2$ , giving $5x = -6$ , wait, $8x + 8 = 3x + 2 \implies 5x = -6$ , so $x = -\frac{6}{5}$ and $y = -\frac{2}{5}$ .

Step 5: Solve System 15 for x and y

From $2^{1-\frac{x}{2}} = (2^2)^y$ , equating powers of 2 gives $1 - \frac{x}{2} = 2y$ , or $x + 4y = 2$ . Next, write $(7^{1+x}) \times (7^2)^{-2y} = 7^{1+x-4y} = 7^0 = 1$ , which gives $x - 4y = -1$ . Adding the two equations yields $2x = 1 \implies x = \frac{1}{2}$ , and subtracting yields $8y = 3 \implies y = \frac{3}{8}$ .

Step 6: Solve System 16 for x and y

From $2^x = 2^4 \cdot 2^y = 2^{y+4}$ , we have $x = y + 4$ , or $x - y = 4$ . From $(3^3)^x = 3^2 \cdot 3^{2y}$ , we have $3^{3x} = 3^{2y+2}$ , giving $3x - 2y = 2$ . Substituting $x = y + 4$ gives $3(y + 4) - 2y = 2$ , which simplifies to $y + 12 = 2 \implies y = -10$ , and $x = -10 + 4 = -6$ .

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