The image shows three math problems. Let's solve them one by one: (a) $8^{3x-2} = 16$ (b) $9^{2x-3} > 27^{x-4}$ (c) $2^{x+2} + 4^x = 5$

1. Problem Description

The image shows three math problems. Let's solve them one by one:

(a) $8^{3x-2} = 16$

(b) $9^{2x-3} > 27^{x-4}$

(c) $2^{x+2} + 4^x = 5$

2. Solution Steps

(a) $8^{3x-2} = 16$

We can rewrite both sides of the equation using the base

2. $8 = 2^3$ and $16 = 2^4$.

So the equation becomes:

$(2^3)^{3x-2} = 2^4$

$2^{3(3x-2)} = 2^4$

$2^{9x-6} = 2^4$

Since the bases are equal, the exponents must be equal:

$9x - 6 = 4$

$9x = 10$

$x = \frac{10}{9}$

(b) $9^{2x-3} > 27^{x-4}$

We can rewrite both sides of the inequality using the base

3. $9 = 3^2$ and $27 = 3^3$.

So the inequality becomes:

$(3^2)^{2x-3} > (3^3)^{x-4}$

$3^{2(2x-3)} > 3^{3(x-4)}$

$3^{4x-6} > 3^{3x-12}$

Since the base is greater than 1, we can compare the exponents:

$4x - 6 > 3x - 12$

$4x - 3x > 6 - 12$

$x > -6$

(c) $2^{x+2} + 4^x = 5$

We can rewrite the equation as:

$2^{x+2} + (2^2)^x = 5$

$2^x \cdot 2^2 + 2^{2x} = 5$

$4 \cdot 2^x + (2^x)^2 = 5$

Let $y = 2^x$. Then the equation becomes:

$4y + y^2 = 5$

$y^2 + 4y - 5 = 0$

$(y+5)(y-1) = 0$

So, $y = -5$ or $y = 1$.

Since $y = 2^x$, we know that $y$ must be positive. Thus, $y = -5$ is not a valid solution.

If $y = 1$, then $2^x = 1$.

$2^x = 2^0$

$x = 0$

3. Final Answer

(a) $x = \frac{10}{9}$

(b) $x > -6$

(c) $x = 0$