Show that: $$ \begin{array}{ll} ext { (a) } 5 \cdot 2^{x}-2^{x+2}=2^{x} & ext { (b) } 9^{x}+3^{2 x+1}=4.3^{2 x} \ ext { (c) } 2^{2 x-1}+4^{x+1}=2^{2 x}\left(\frac{9}{2} ight) & ext { (d) } 2^{x+3}-2^{x+2}=2^{x+2} \ ext { (e) } 2.10^{x}-5^{x+1} \cdot 2^{x}=-3 \cdot 10^{x} & ext { (f) } 4.3^{1-x}+3^{2-x}=\frac{21}{3^{x}} \ ext { Simplify the following: } & \ 10^{x} \cdot 25^{x+1} & ext { (b) } 6^{n+2} imes 10^{n-2}\end{array} $$ (c) $$ \frac{6^{x} \cdot 9^{x+1} \cdot 2}{} $$

Question

Show that: $$ \begin{array}{ll}	ext { (a) } 5 \cdot 2^{x}-2^{x+2}=2^{x} & 	ext { (b) } 9^{x}+3^{2 x+1}=4.3^{2 x} \ 	ext { (c) } 2^{2 x-1}+4^{x+1}=2^{2 x}\left(\frac{9}{2}ight) & 	ext { (d) } 2^{x+3}-2^{x+2}=2^{x+2} \ 	ext { (e) } 2.10^{x}-5^{x+1} \cdot 2^{x}=-3 \cdot 10^{x} & 	ext { (f) } 4.3^{1-x}+3^{2-x}=\frac{21}{3^{x}} \ 	ext { Simplify the following: } & \ 10^{x} \cdot 25^{x+1} & 	ext { (b) } 6^{n+2} 	imes 10^{n-2}\end{array} $$ (c) $$ \frac{6^{x} \cdot 9^{x+1} \cdot 2}{} $$

UpStudy AI · Accepted Answer

Let's solve each part step by step.

### Part (a)
We need to show that:
$$
5 \cdot 2^{x} - 2^{x+2} = 2^{x}
$$

**Step 1:** Rewrite $$2^{x+2}$$:
$$
2^{x+2} = 4 \cdot 2^{x}
$$

**Step 2:** Substitute into the equation:
$$
5 \cdot 2^{x} - 4 \cdot 2^{x} = 2^{x}
$$

**Step 3:** Simplify:
$$
(5 - 4) \cdot 2^{x} = 2^{x}
$$
$$
1 \cdot 2^{x} = 2^{x}
$$

Thus, part (a) is verified.

### Part (b)
We need to show that:
$$
9^{x} + 3^{2x+1} = 4 \cdot 3^{2x}
$$

**Step 1:** Rewrite $$9^{x}$$ and $$3^{2x+1}$$:
$$
9^{x} = (3^{2})^{x} = 3^{2x}
$$
$$
3^{2x+1} = 3^{2x} \cdot 3
$$

**Step 2:** Substitute into the equation:
$$
3^{2x} + 3 \cdot 3^{2x} = 4 \cdot 3^{2x}
$$

**Step 3:** Factor out $$3^{2x}$$:
$$
(1 + 3) \cdot 3^{2x} = 4 \cdot 3^{2x}
$$
$$
4 \cdot 3^{2x} = 4 \cdot 3^{2x}
$$

Thus, part (b) is verified.

### Part (c)
We need to show that:
$$
2^{2x-1} + 4^{x+1} = 2^{2x} \left(\frac{9}{2}\right)
$$

**Step 1:** Rewrite $$4^{x+1}$$:
$$
4^{x+1} = 2^{2(x+1)} = 2^{2x+2}
$$

**Step 2:** Substitute into the equation:
$$
2^{2x-1} + 2^{2x+2} = 2^{2x} \left(\frac{9}{2}\right)
$$

**Step 3:** Factor out $$2^{2x-1}$$:
$$
2^{2x-1} (1 + 2^{3}) = 2^{2x} \left(\frac{9}{2}\right)
$$
$$
2^{2x-1} (1 + 8) = 2^{2x} \left(\frac{9}{2}\right)
$$
$$
2^{2x-1} \cdot 9 = 2^{2x} \left(\frac{9}{2}\right)
$$

Thus, part (c) is verified.

### Part (d)
We need to show that:
$$
2^{x+3} - 2^{x+2} = 2^{x+2}
$$

**Step 1:** Factor out $$2^{x+2}$$:
$$
2^{x+2} (2 - 1) = 2^{x+2}
$$
$$
2^{x+2} = 2^{x+2}
$$

Thus, part (d) is verified.

### Part (e)
We need to show that:
$$
2 \cdot 10^{x} - 5^{x+1} \cdot 2^{x} = -3 \cdot 10^{x}
$$

**Step 1:** Rewrite $$10^{x}$$:
$$
10^{x} = 2^{x} \cdot 5^{x}
$$

**Step 2:** Substitute into the equation:
$$
2 \cdot (2^{x} \cdot 5^{x}) - 5^{x+1} \cdot 2^{x} = -3 \cdot (2^{x} \cdot 5^{x})
$$

**Step 3:** Factor out $$2^{x} \cdot 5^{x}$$:
$$
2 \cdot 2^{x} \cdot 5^{x} - 5 \cdot 2^{x} \cdot 5^{x} = -3 \cdot 2^{x} \cdot 5^{x}
$$
$$
(2 - 5) \cdot 2^{x} \cdot 5^{x} = -3 \cdot 2^{x} \cdot 5^{x}
$$
$$
-3 \cdot 2^{x} \cdot 5^{x} = -3 \cdot 2^{x} \cdot 5^{x}
$$

Thus, part (e) is verified.

### Part (f)
We need to show that:
$$
4 \cdot 3^{1-x} + 3^{2-x} = \frac{21}{3^{x}}
$$

**Step 1:** Rewrite $$3^{1-x}$$ and $$3^{2-x}$$:
$$
4 \cdot \frac{3}{3^{x}} + \frac{9}{3^{x}} = \frac{21}{3^{x}}
$$

**Step 2:** Combine the fractions:
$$
\frac{4 \cdot 3 + 9}{3^{x}} = \frac{21}{3^{x}}
$$
$$
\frac{12 + 9}{3^{x}} = \frac{21}{3^{x}}
$$
$$
\frac{21}{3^{x}} = \frac{21}{3^{x}}
$$

Thus, part (f) is verified.

### Simplification
Now, let's simplify the following expressions:

#### (a) $$10^{x} \cdot 25^{x+1}$$
**Step 1:** Rewrite $$25^{x+1}$$:
$$
25^{x+1} = 25^{x} \cdot 25 = (5^{2})^{x} \cdot 25 = 5^{2x} \cdot 5^{2} = 5^{2x+2}
$$

**Step 2:** Substitute:
$$
10^{x} \cdot 25^{x+1} = 10^{x} \cdot 5^{2x+2}
$$

**Step 3:** Rewrite $$10^{x}$$:
$$
10^{x} = 2^{x} \cdot 5^{x}
$$
$$
= 2^{x}
$$10^{x} \cdot 25^{x+1} = 2^{x} \cdot 5^{2x+2}$$ ### Simplification of $$6^{n+2} \times 10^{n-2}$$ **Step 1:** Rewrite $$6^{n+2}$$ and $$10^{n-2}$$: \[ 6^{n+2} = 6^{n} \cdot 6^{2} = 6^{n} \cdot 36 $$ $$ 10^{n-2} = 10^{n} \cdot 10^{-2} = 10^{n} \cdot \frac{1}{100} $$ **Step 2:** Multiply the two expressions: $$ 6^{n} \cdot 36 \times 10^{n} \cdot \frac{1}{100} = 36 \cdot \frac{6^{n} \cdot 10^{n}}{100} $$ **Step 3:** Simplify: $$ 36 \cdot \frac{6^{n} \cdot 10^{n}}{100} = \frac{36 \cdot 6^{n} \cdot 10^{n}}{100} = \frac{36 \cdot 6^{n} \cdot 10^{n}}{100} $$ Thus, the simplified form is: $$ \frac{36 \cdot 6^{n} \cdot 10^{n}}{100} $$ ### Simplification of $$ \frac{6^{x} \cdot 9^{x+1} \cdot 2}{} $$ **Step 1:** Rewrite $$9^{x+1}$$: $$ 9^{x+1} = 9^{x} \cdot 9 = (3^{2})^{x} \cdot 9 = 3^{2x} \cdot 9 $$ **Step 2:** Substitute: $$ 6^{x} \cdot 9^{x+1} \cdot 2 = 6^{x} \cdot 3^{2x} \cdot 9 \cdot 2 $$ **Step 3:** Simplify: $$ 6^{x} \cdot 3^{2x} \cdot 18 = 6^{x} \cdot 3^{2x} \cdot 18 $$ Thus, the simplified form is: $$ 6^{x} \cdot 3^{2x} \cdot 18 $$

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