The problem has two parts. Part (a) requires us to solve the equation $(\frac{2}{3})^{x+2} = (\frac{3}{2})^{2-3x}$ for $x$. Part (b) gives us a diagram with $|PQ| = 4$ cm, $|QR| = 2$ cm, $|QT| = 2$ cm, and $|PT| = |TS| = 5$ cm, and asks us to find $|RS|$.

1. Problem Description

The problem has two parts. Part (a) requires us to solve the equation $(\frac{2}{3})^{x+2} = (\frac{3}{2})^{2-3x}$ for $x$. Part (b) gives us a diagram with $|PQ| = 4$ cm, $|QR| = 2$ cm, $|QT| = 2$ cm, and $|PT| = |TS| = 5$ cm, and asks us to find $|RS|$.

2. Solution Steps

(a) To solve the equation $(\frac{2}{3})^{x+2} = (\frac{3}{2})^{2-3x}$, we can rewrite the right side of the equation using the property that $(\frac{a}{b})^{-1} = \frac{b}{a}$.

$(\frac{3}{2})^{2-3x} = (\frac{2}{3})^{-(2-3x)} = (\frac{2}{3})^{-2+3x}$.

Now we have $(\frac{2}{3})^{x+2} = (\frac{2}{3})^{-2+3x}$.

Since the bases are equal, we can set the exponents equal to each other:

$x+2 = -2+3x$.

Subtract $x$ from both sides: $2 = -2 + 2x$.

Add 2 to both sides: $4 = 2x$.

Divide by 2: $x = 2$.

(b) Since $|PQ|=4$ and $|QR|=2$, the ratio $\frac{|PQ|}{|QR|} = \frac{4}{2} = 2$. Since $|PT|=5$ and $|QT|=2$, the ratio $\frac{|PT|}{|QT|} = \frac{5}{2} = 2.5$. The sides are not proportional.

We can say triangles $PQT$ and $QRS$ are similar if and only if $\frac{PQ}{QR} = \frac{QT}{RS} = \frac{PT}{QS}$ and $\angle PQT = \angle QRS$, $\angle QTP = \angle RSS$ and $\angle TPQ = \angle SQR$.

The length of RS can be found by observing that triangle PQT and triangle QRS may be similar.

The given information is $|PQ| = 4$, $|QR| = 2$, $|QT| = 2$, $|PT| = 5$, $|TS| = 5$. We need to find $|RS|$.

Since QT is perpendicular to PS, triangles PQT and QTS are right triangles.

However, triangles PQT and QRS are not similar in general.

Let's consider the two right triangles PQT and a hypothetical right triangle similar to PQT with hypotenuse QR of length

2. In triangle PQT, we have $|PQ|=4$, $|QT|=2$, $|PT|=5$.

If triangle QRS is similar to triangle PQT, then $\frac{QR}{PQ} = \frac{RS}{PT} = \frac{QS}{QT}$ and $\frac{2}{4} = \frac{1}{2}$.

$\frac{|QR|}{|PQ|} = \frac{|QT|}{|?|}$ means $\frac{2}{4} = \frac{2}{|?|}$. So $|?| = 4$.

$\frac{|QR|}{|PQ|} = \frac{|RS|}{|PT|}$ means $\frac{2}{4} = \frac{|RS|}{5}$. So $|RS| = \frac{2}{4} * 5 = \frac{5}{2} = 2.5$.

We are not given that QRS is a right triangle. So the right triangle properties can not be used here.

$\angle PTQ = 90$. So $QT \perp PS$. In order to have triangle PQT similar to triangle QRS, we must have $\angle QTS$ a right angle and also $|QR|/|PQ|=|RS|/|PT|$, which gives us

$|RS|=(|QR|/|PQ|)*|PT|=(2/4)*5=2.5$.

Final Answer:

3. Final Answer

(a) $x=2$

(b) $|RS| = 2.5$ cm