i $$ 3^{x}-3^{x-2}=24 $$ if $$ 3^{x}+3^{x+2}=10 $$ iii $$ 2^{x}+2^{x}+2^{x}=96 $$ iv $$ 12^{x}+12^{x}=288 $$ $$ 2^{x}+4^{x+1}=80 $$

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Sure, let's solve each of the given exponential equations step by step to find the value of $$ x $$.

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### **i. $$ 3^{x} - 3^{x-2} = 24 $$**

**Step 1:** Factor out $$ 3^{x-2} $$

$$
3^{x} - 3^{x-2} = 3^{x-2} \left( 3^{2} - 1 \right) = 3^{x-2} \times (9 - 1) = 3^{x-2} \times 8
$$

So,

$$
3^{x-2} \times 8 = 24
$$

**Step 2:** Solve for $$ 3^{x-2} $$

$$
3^{x-2} = \frac{24}{8} = 3
$$

**Step 3:** Express 3 as a power of 3

$$
3 = 3^{1}
$$

So,

$$
3^{x-2} = 3^{1} \implies x - 2 = 1 \implies x = 3
$$

**Answer:** $$ x = 3 $$

---

### **ii. $$ 3^{x} + 3^{x+2} = 10 $$**

**Step 1:** Factor out $$ 3^{x} $$

$$
3^{x} + 3^{x+2} = 3^{x} \left(1 + 3^{2}\right) = 3^{x} \times (1 + 9) = 3^{x} \times 10
$$

So,

$$
3^{x} \times 10 = 10
$$

**Step 2:** Solve for $$ 3^{x} $$

$$
3^{x} = \frac{10}{10} = 1
$$

**Step 3:** Express 1 as a power of 3

$$
1 = 3^{0}
$$

So,

$$
3^{x} = 3^{0} \implies x = 0
$$

**Answer:** $$ x = 0 $$

---

### **iii. $$ 2^{x} + 2^{x} + 2^{x} = 96 $$**

**Step 1:** Combine like terms

$$
3 \times 2^{x} = 96
$$

**Step 2:** Solve for $$ 2^{x} $$

$$
2^{x} = \frac{96}{3} = 32
$$

**Step 3:** Express 32 as a power of 2

$$
32 = 2^{5}
$$

So,

$$
2^{x} = 2^{5} \implies x = 5
$$

**Answer:** $$ x = 5 $$

---

### **iv. $$ 12^{x} + 12^{x} = 288 $$**

**Step 1:** Combine like terms

$$
2 \times 12^{x} = 288
$$

**Step 2:** Solve for $$ 12^{x} $$

$$
12^{x} = \frac{288}{2} = 144
$$

**Step 3:** Express 144 as a power of 12

Note that $$ 12^{2} = 144 $$.

So,

$$
12^{x} = 12^{2} \implies x = 2
$$

**Answer:** $$ x = 2 $$

---

### **v. $$ 2^{x} + 4^{x+1} = 80 $$**

**Step 1:** Express $$ 4^{x+1} $$ in terms of base 2.

Since $$ 4 = 2^{2} $$,

$$
4^{x+1} = (2^{2})^{x+1} = 2^{2(x+1)} = 2^{2x + 2}
$$

**Step 2:** Rewrite the equation

$$
2^{x} + 2^{2x + 2} = 80
$$

**Step 3:** Let’s set $$ y = 2^{x} $$. Then $$ 2^{2x} = y^{2} $$.

So,

$$
y + y^{2} \times 2^{2} = 80 \implies y + 4y^{2} = 80
$$

**Step 4:** Rearrange the equation

$$
4y^{2} + y - 80 = 0
$$

**Step 5:** Solve the quadratic equation

Using the quadratic formula $$ y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} $$, where $$ a = 4 $$, $$ b = 1 $$, and $$ c = -80 $$,

$$
y = \frac{-1 \pm \sqrt{1 + 1280}}{8} = \frac{-1 \pm \sqrt{1281}}{8}
$$

Calculate $$ \sqrt{1281} \approx 35.8 $$,

$$
y \approx \frac{-1 + 35.8}{8} \approx \frac{34.8}{8} \approx 4.35
$$

We discard the negative solution since $$ y = 2^{x} > 0 $$.

**Step 6:** Solve for $$ x $$

$$
2^{x} \approx 4.35
$$

Take the natural logarithm on both sides:

$$
x \ln 2 \approx \ln 4.35 \implies x \approx \frac{\ln 4.35}{\ln 2} \approx \frac{1.470}{0.693} \approx 2.12
$$

**Answer:** $$ x \approx 2.12 $$

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**Summary of Solutions:**

1. **i.** $$ x = 3 $$
2. **ii.** $$ x = 0 $$
3. **iii.** $$ x = 5 $$
4. **iv.** $$ x = 2 $$
5. **v.** $$ x \approx 2.12 $$
Solutions for the equations are: 1. $$ x = 3 $$ 2. $$ x = 0 $$ 3. $$ x = 5 $$ 4. $$ x = 2 $$ 5. $$ x \approx 2.12 $$

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