$$ \begin{array}{ll} ext { (4) } \frac{5^{x+4}-5^{x+3}}{100.5^{x+1}} & ext { (5) } \frac{4^{x}+3 \cdot 2^{2 x+1}}{7 \cdot 2^{2 x+1}} \ \begin{array}{lll} ext { (7) } \frac{8^{x} \cdot 2^{x}+2.16^{x+1}}{11.2^{x+1}} & ext { (8) } \frac{12^{x}+4^{x} \cdot 3^{x+1}}{2^{2 x+4} \cdot 3^{x}} & ext { (6) } \frac{\left(3^{x} ight)^{2}-9^{x-1}}{9^{x-1}} \ 5.2^{x+1}-7 \cdot 3^{x-1}\end{array} \ 37\end{array} $$

Question

$$ \begin{array}{ll}	ext { (4) } \frac{5^{x+4}-5^{x+3}}{100.5^{x+1}} & 	ext { (5) } \frac{4^{x}+3 \cdot 2^{2 x+1}}{7 \cdot 2^{2 x+1}} \ \begin{array}{lll}	ext { (7) } \frac{8^{x} \cdot 2^{x}+2.16^{x+1}}{11.2^{x+1}} & 	ext { (8) } \frac{12^{x}+4^{x} \cdot 3^{x+1}}{2^{2 x+4} \cdot 3^{x}} & 	ext { (6) } \frac{\left(3^{x}ight)^{2}-9^{x-1}}{9^{x-1}} \ 5.2^{x+1}-7 \cdot 3^{x-1}\end{array} \ 37\end{array} $$

UpStudy AI · Accepted Answer

To solve the given expressions, we will simplify each one step by step. Let's denote the expressions as follows:

1. $$ E_4 = \frac{5^{x+4}-5^{x+3}}{100 \cdot 5^{x+1}} $$
2. $$ E_5 = \frac{4^{x}+3 \cdot 2^{2x+1}}{7 \cdot 2^{2x+1}} $$
3. $$ E_6 = \frac{(3^{x})^{2}-9^{x-1}}{9^{x-1}} $$
4. $$ E_7 = \frac{8^{x} \cdot 2^{x}+2 \cdot 16^{x+1}}{11 \cdot 2^{x+1}} $$
5. $$ E_8 = \frac{12^{x}+4^{x} \cdot 3^{x+1}}{2^{2x+4} \cdot 3^{x}} $$

Let's simplify each expression one by one.

### Simplifying $$ E_4 $$

$$
E_4 = \frac{5^{x+4} - 5^{x+3}}{100 \cdot 5^{x+1}} 
$$

Factoring out $$ 5^{x+3} $$ from the numerator:

$$
= \frac{5^{x+3}(5 - 1)}{100 \cdot 5^{x+1}} = \frac{4 \cdot 5^{x+3}}{100 \cdot 5^{x+1}} 
$$

Now simplifying:

$$
= \frac{4 \cdot 5^{2}}{100} = \frac{100}{100} = 1
$$

### Simplifying $$ E_5 $$

$$
E_5 = \frac{4^{x} + 3 \cdot 2^{2x+1}}{7 \cdot 2^{2x+1}} 
$$

Rewriting $$ 4^{x} $$ as $$ (2^2)^{x} = 2^{2x} $$:

$$
= \frac{2^{2x} + 3 \cdot 2^{2x+1}}{7 \cdot 2^{2x+1}} = \frac{2^{2x} + 6 \cdot 2^{2x}}{7 \cdot 2^{2x+1}} 
$$

Factoring out $$ 2^{2x} $$:

$$
= \frac{7 \cdot 2^{2x}}{7 \cdot 2^{2x+1}} = \frac{1}{2}
$$

### Simplifying $$ E_6 $$

$$
E_6 = \frac{(3^{x})^{2} - 9^{x-1}}{9^{x-1}} 
$$

Rewriting $$ 9^{x-1} $$ as $$ (3^2)^{x-1} = 3^{2x-2} $$:

$$
= \frac{3^{2x} - 3^{2x-2}}{3^{2x-2}} = \frac{3^{2x} - \frac{1}{9} \cdot 3^{2x}}{3^{2x-2}} 
$$

Factoring out $$ 3^{2x} $$:

$$
= \frac{3^{2x}(1 - \frac{1}{9})}{3^{2x-2}} = \frac{3^{2x} \cdot \frac{8}{9}}{3^{2x-2}} = \frac{8 \cdot 3^{2}}{9} = \frac{72}{9} = 8
$$

### Simplifying $$ E_7 $$

$$
E_7 = \frac{8^{x} \cdot 2^{x} + 2 \cdot 16^{x+1}}{11 \cdot 2^{x+1}} 
$$

Rewriting $$ 8^{x} $$ and $$ 16^{x+1} $$:

$$
= \frac{(2^3)^{x} \cdot 2^{x} + 2 \cdot (2^4)^{x+1}}{11 \cdot 2^{x+1}} = \frac{2^{4x} + 2 \cdot 2^{4x + 4}}{11 \cdot 2^{x+1}} 
$$

Factoring out $$ 2^{4x} $$:

$$
= \frac{2^{4x}(1 + 2^{5})}{11 \cdot 2^{x+1}} = \frac{2^{4x} \cdot 33}{11 \cdot 2^{x+1}} = \frac{3 \cdot 2^{4x}}{2^{x+1}} = 3 \cdot 2^{3x-1}
$$

### Simplifying $$ E_8 $$

$$
E_8 = \frac{12^{x} + 4^{x} \cdot 3^{x+1}}{2^{2x+4} \cdot 3^{x}} 
$$

Rewriting $$ 12^{x} $$ and $$ 4^{x} $$:

$$
= \frac{(2^2 \cdot 3)^{x} + (2^2)^{x} \cdot 3^{x+1}}{2^{2x+4} \cdot 3^{x}} = \frac{2^{2x} \cdot 3^{x} + 2^{2x} \cdot 3^{x} \cdot 3}{2^{2x+4} \cdot 3^{x}} 
$$

Factoring out $$ 2^{2x} \cdot 3^{x} $$:

$$
= \frac{2^{2x} \cdot 3^{x}(1 + 3)}{2^{2x+4} \cdot 3^{x}} = \frac{4 \cdot 2^{2x}}{2^{2x+4}} = \frac{4}{16} = \frac{1}{4}
$$

### Summary of Results

1. $$ E_4 = 1 $$
2. $$ E_5 = \frac{1}{2} $$
3. $$ E_6 = 8 $$
4. $$ E_7 = 3 \cdot 2^{3x-1} $$
5. $$ E_8 = \frac{1}{4} $$

Now, we can summarize the results or proceed with any further calculations if needed. If you have any
Here are the simplified results for each expression: 1. $$ E_4 = 1 $$ 2. $$ E_5 = \frac{1}{2} $$ 3. $$ E_6 = 8 $$ 4. $$ E_7 = 3 \cdot 2^{3x-1} $$ 5. $$ E_8 = \frac{1}{4} $$

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