Show that: $$ \begin{array}{ll} ext { (a) } 5 \cdot 2^{x}-2^{x+2}=2^{x} & ext { (b) } 9^{x}+3^{2 x+1}=43^{x x} \ ext { (c) } 2^{2 x-1}+4^{x+1}=2^{x x}\left(\frac{9}{2} ight) & ext { (d) } 2^{x+3}-2^{x+2}=2^{x+2} \ ext { (e) } 2.10^{x}-5^{x+1} \cdot 2^{x}=-3.10^{x} & ext { (f) } 4.3^{1-x}+3^{2-x}=\frac{21}{3^{x}} \ ext { Simplify the following: } & \end{array} $$

Question

Show that: $$ \begin{array}{ll}	ext { (a) } 5 \cdot 2^{x}-2^{x+2}=2^{x} & 	ext { (b) } 9^{x}+3^{2 x+1}=43^{x x} \ 	ext { (c) } 2^{2 x-1}+4^{x+1}=2^{x x}\left(\frac{9}{2}ight) & 	ext { (d) } 2^{x+3}-2^{x+2}=2^{x+2} \ 	ext { (e) } 2.10^{x}-5^{x+1} \cdot 2^{x}=-3.10^{x} & 	ext { (f) } 4.3^{1-x}+3^{2-x}=\frac{21}{3^{x}} \ 	ext { Simplify the following: } & \end{array} $$

UpStudy AI · Accepted Answer

Certainly! Let's simplify each of the given expressions step by step.

---

### **Problem (a)**
**Simplify:**
$$ 5 \cdot 2^{x} - 2^{x+2} = 2^{x} $$

**Solution:**

1. **Express $$2^{x+2}$$ in terms of $$2^{x}$$:**
   $$
   2^{x+2} = 2^{x} \cdot 2^{2} = 4 \cdot 2^{x}
   $$
   
2. **Substitute back into the original equation:**
   $$
   5 \cdot 2^{x} - 4 \cdot 2^{x} = 2^{x}
   $$
   
3. **Combine like terms:**
   $$
   (5 - 4) \cdot 2^{x} = 2^{x} \
   1 \cdot 2^{x} = 2^{x}
   $$
   
4. **Conclusion:**
   $$
   2^{x} = 2^{x}
   $$
   
   This holds true for all real numbers $$x$$. Hence, the equation is an identity.

---

### **Problem (b)**
**Simplify:**
$$ 9^{x} + 3^{2x+1} = 4 \cdot 3^{2x} $$

**Solution:**

*Assuming the right-hand side (RHS) is $$4 \cdot 3^{2x}$$ based on context.*

1. **Express $$9^{x}$$ in terms of $$3^{2x}$$:**
   $$
   9^{x} = (3^{2})^{x} = 3^{2x}
   $$
   
2. **Express $$3^{2x+1}$$ using exponent rules:**
   $$
   3^{2x+1} = 3^{2x} \cdot 3^{1} = 3 \cdot 3^{2x}
   $$
   
3. **Substitute back into the original equation:**
   $$
   3^{2x} + 3 \cdot 3^{2x} = 4 \cdot 3^{2x}
   $$
   
4. **Factor $$3^{2x}$$ from the left-hand side (LHS):**
   $$
   (1 + 3) \cdot 3^{2x} = 4 \cdot 3^{2x}
   $$
   
5. **Simplify:**
   $$
   4 \cdot 3^{2x} = 4 \cdot 3^{2x}
   $$
   
   This confirms that both sides are equal for all real numbers $$x$$. Hence, the equation is an identity.

---

### **Problem (c)**
**Simplify:**
$$ 2^{2x-1} + 4^{x+1} = 2^{x} \left(\frac{9}{2}\right) $$

**Solution:**

1. **Express all terms with base 2:**
   $$
   4^{x+1} = (2^{2})^{x+1} = 2^{2(x+1)} = 2^{2x + 2}
   $$
   
2. **Rewrite the equation:**
   $$
   2^{2x - 1} + 2^{2x + 2} = 2^{x} \cdot \frac{9}{2}
   $$
   
3. **Factor $$2^{2x - 1}$$ from the LHS:**
   $$
   2^{2x - 1} \left(1 + 2^{3}\right) = 2^{x} \cdot \frac{9}{2} \
   2^{2x - 1} \cdot 9 = 2^{x} \cdot \frac{9}{2}
   $$
   
4. **Divide both sides by 9:**
   $$
   2^{2x - 1} = 2^{x} \cdot \frac{1}{2}
   $$
   
5. **Simplify the right-hand side:**
   $$
   2^{2x - 1} = 2^{x - 1}
   $$
   
6. **Equate exponents (since the bases are the same and non-zero):**
   $$
   2x - 1 = x - 1 \
   2x - x = -1 + 1 \
   x = 0
   $$
   
7. **Conclusion:**
   The equation holds true when $$x = 0$$.

---

### **Problem (d)**
**Simplify:**
$$ 2^{x+3} - 2^{x+2} = 2^{x+2} $$

**Solution:**

1. **Express both terms with a common factor:**
   $$
   2^{x+3} = 2^{x} \cdot 2^{3} = 8 \cdot 2^{x} \
   2^{x+2} = 2^{x} \cdot 2^{2} = 4 \cdot 2^{x}
   $$
   
2. **Substitute back into the equation:**
   $$
   8 \cdot 2^{x} - 4 \cdot 2^{x} = 2^{x+2}
   $$
   
3. **Combine like terms:**
   $$
   (8 - 4) \cdot 2^{x} = 4 \cdot 2^{x}
   $$
   
4. **Recognize that $$4 \cdot 2^{x} = 2^{x+2}$$:**
   $$
   4 \cdot 2^{x} = 2^{x} \cdot 2^{2} = 2^{x+2}
   $$
   
5. **Conclusion:**
   $$
   2^{x+2} = 2^{x+2}
   $$
   
   The equation holds true for all real numbers $$x$$. Hence, it is an identity.

---

### **Problem (e)**
**Simplify:**
$$ 2 \cdot 10^{x} - 5^{x+1} \cdot 2^{x} = -3 \cdot 10^{x} $$

**Solution:**

1. **Express $$5^{x+1} \cdot 2^{x}$$ in terms of $$10^{x}$$:**
   $$
   5^{x+1} \cdot 2^{x} = 5 \cdot 5^{x} \cdot 2^{x} = 5 \cdot (5 \cdot 2)^{x} = 5 \cdot 10^{x}
   $$
   
2. **Substitute back into the equation:**
   $$
   2 \cdot 10^{x} - 5 \cdot 10^{x} = -3 \cdot 10^{x}
   $$
   
3. **Combine like terms:**
   $$
   (2 - 5) \cdot 10^{x} = -3 \cdot 10^{x} \
   -3 \cdot 10^{x} = -3 \cdot 10^{x}
   $$
   
4. **Conclusion:**
   The equation holds true for all real numbers $$x$$. Hence, it is an identity.

---

### **Problem (f)**
**Simplify:**
$$ 4 \cdot 3^{1-x} + 3^{2-x} = \frac{21}{3^{x}} $$

**Solution:**

1. **Express all terms with negative exponents:**
   $$
   3^{1-x} = 3^{-(x-1)} = \frac{1}{3^{x-1}} = \frac{3}{3^{x}} \
   3^{2-x} = 3^{-(x-2)} = \frac{1}{3^{x-2}} = \frac{9}{3^{x}}
   $$
   
2. **Rewrite the equation using these expressions:**
   $$
   4 \cdot \frac{3}{3^{x}} + \frac{9}{3^{x}} = \frac{21}{3^{x}}
   $$
   
3. **Combine like terms:**
   $$
   \left(4 \cdot 3 + 9\right) \cdot \frac{1}{3^{x}} = \frac{21}{3^{x}} \
   (12 + 9) \cdot \frac{1}{3^{x}} = \frac{21}{3^{x}} \
   21 \cdot \frac{1}{3^{x}} = \frac{21}{3^{x}}
   $$
   
4. **Conclusion:**
   $$
   \frac{21}{3^{x}} = \frac{21}{3^{x}}
   $$
   
   The equation holds true for all real numbers $$x$$. Hence, it is an identity.

---

### **Summary**

All the given equations simplify to identities that hold true for all real numbers $$x$$, except for Problem (c), which holds when $$x = 0$$.
All the given equations simplify to identities that hold true for all real numbers $$x$$, except for Problem (c), which holds when $$x = 0$$.

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