QUESTION 1 Simplify the following completely: $$ 1.1 \frac{10^{2 x+3} \cdot 4^{1-x}}{25^{2+x}} $$ predice to $$ 1.2 \frac{3^{t}-3^{t-2}}{2.3^{t}-3^{t}} $$ QUESTION 2 Solve for $$ x $$ $$ 2.1 \quad 5^{x}=\frac{1}{125} $$ $$ 2.2\left(\frac{1}{2}\right)^{3 x+1}=32 $$ $$ 2.3 \quad 4.2^{x}=(0,5)^{x-2} $$

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**QUESTION 1**

**1.1** Simplify 
$$
\frac{10^{2x+3}\cdot 4^{1-x}}{25^{2+x}}.
$$

1. Express each number in terms of its prime factors:
   - $$10=2\cdot5$$ so 
     $$
     10^{2x+3}=(2\cdot5)^{2x+3}=2^{2x+3}\cdot 5^{2x+3}.
     $$
   - $$4=2^2$$ so 
     $$
     4^{1-x}=(2^{2})^{1-x}=2^{2(1-x)}=2^{2-2x}.
     $$
   - $$25=5^2$$ so 
     $$
     25^{2+x}=(5^2)^{2+x}=5^{2(2+x)}=5^{4+2x}.
     $$
   
2. Substitute these into the expression:
   $$
   \frac{2^{2x+3}\cdot 5^{2x+3}\cdot 2^{2-2x}}{5^{4+2x}}.
   $$

3. Combine the powers of 2 in the numerator:
   $$
   2^{2x+3}\cdot 2^{2-2x} = 2^{(2x+3)+(2-2x)} = 2^{5}.
   $$
   
4. The expression becomes:
   $$
   \frac{2^{5}\cdot 5^{2x+3}}{5^{4+2x}}.
   $$
   
5. Combine the powers of 5 by subtracting exponents:
   $$
   5^{2x+3- (4+2x)}=5^{(2x+3-4-2x)}=5^{-1}.
   $$
   
6. Hence,
   $$
   \frac{10^{2x+3}\cdot 4^{1-x}}{25^{2+x}} = 2^{5}\cdot 5^{-1}=\frac{32}{5}.
   $$

---

**1.2** Simplify 
$$
\frac{3^{t}-3^{t-2}}{2\cdot 3^{t}-3^{t}}.
$$

1. Factor $$3^{t-2}$$ out of the numerator:
   $$
   3^{t}-3^{t-2}=3^{t-2}\left(3^{2}-1\right)=3^{t-2}(9-1)=8\cdot 3^{t-2}.
   $$
   
2. Factor $$3^{t}$$ out of the denominator:
   $$
   2\cdot 3^{t}-3^{t}=3^{t}(2-1)=3^{t}.
   $$
   
3. Therefore,
   $$
   \frac{8\cdot 3^{t-2}}{3^{t}}=8\cdot 3^{t-2-t}=8\cdot 3^{-2}=\frac{8}{3^2}=\frac{8}{9}.
   $$

---

**QUESTION 2**

**2.1** Solve for $$x$$ in
$$
5^{x}=\frac{1}{125}.
$$

1. Notice that $$125=5^3$$ so $$\frac{1}{125}=5^{-3}$$.

2. Equate the exponents:
   $$
   5^{x}=5^{-3}\quad\Rightarrow\quad x=-3.
   $$

---

**2.2** Solve for $$x$$ in
$$
\left(\frac{1}{2}\right)^{3x+1}=32.
$$

1. Write $$\frac{1}{2}$$ as $$2^{-1}$$ and $$32$$ as $$2^{5}$$:
   $$
   \left(2^{-1}\right)^{3x+1}=2^{-3x-1} \quad \text{and} \quad 32=2^5.
   $$
   
2. Equate the exponents:
   $$
   -3x-1=5.
   $$
   
3. Solve:
   $$
   -3x=6\quad\Rightarrow\quad x=-2.
   $$

---

**2.3** Solve for $$x$$ in
$$
4\cdot 2^{x}=(0,5)^{x-2}.
$$

1. Interpret $$(0,5)$$ as the decimal $$0.5$$ (common in European notation). Write $$0.5$$ as $$\frac{1}{2}$$, hence
   $$
   (0,5)^{x-2}=\left(\frac{1}{2}\right)^{x-2}=2^{-(x-2)}=2^{-x+2}.
   $$
   
2. Rewrite $$4$$ as $$2^2$$. The equation becomes:
   $$
   2^2\cdot 2^x=2^{-x+2}.
   $$
   
3. Combine the powers on the left:
   $$
   2^{x+2}=2^{-x+2}.
   $$
   
4. Equate exponents:
   $$
   x+2=-x+2.
   $$
   
5. Solve:
   $$
   2x=0\quad\Rightarrow\quad x=0.
   $$
**Solutions:** **QUESTION 1** 1.1. Simplify $$ \frac{10^{2x+3} \cdot 4^{1-x}}{25^{2+x}} $$: $$ \frac{10^{2x+3} \cdot 4^{1-x}}{25^{2+x}} = \frac{32}{5} $$ 1.2. Simplify $$ \frac{3^{t} - 3^{t-2}}{2 \cdot 3^{t} - 3^{t}} $$: $$ \frac{3^{t} - 3^{t-2}}{2 \cdot 3^{t} - 3^{t}} = \frac{8}{9} $$ **QUESTION 2** 2.1. Solve $$ 5^{x} = \frac{1}{125} $$: $$ x = -3 $$ 2.2. Solve $$ \left(\frac{1}{2}\right)^{3x+1} = 32 $$: $$ x = -2 $$ 2.3. Solve $$ 4 \cdot 2^{x} = (0.5)^{x-2} $$: $$ x = 0 $$

QUESTION 1 Simplify the following completely: \( 1.1 \frac{10^{2 x+3} \cdot 4^{1-x}}{25^{2+x}} \) predice to \( 1.2 \frac{3^{t}-3^{t-2}}{2.3^{t}-3^{t}} \) QUESTION 2 Solve for \( x \) \( 2.1 \quad 5^{x}=\frac{1}{125} \) \( 2.2\left(\frac{1}{2}\right)^{3 x+1}=32 \) \( 2.3 \quad 4.2^{x}=(0,5)^{x-2} \)

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