Solve for $$ x $$ : (a) $$ 3^{x}=1 $$ (d) $$ \left(\frac{1}{9}\right)^{x}=27 $$ (g) $$ 4\left(\frac{1}{4}\right)^{x-1}=8 $$ (j) $$ \frac{3^{2 x-1}}{3^{x}}=3 $$ (m) $$ 8^{-x}-2.4^{x-1}=0 $$

Question

UpStudy AI · Accepted Answer

Let's solve each equation step by step.

---

### **Problem (a):** $$ 3^{x} = 1 $$

**Solution:**

We know that any non-zero number raised to the power of 0 is 1. Therefore:

$$
3^{x} = 1 \implies x = 0
$$

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

---

### **Problem (d):** $$ \left(\frac{1}{9}\right)^{x} = 27 $$

**Solution:**

First, express both sides with the same base. Notice that $$ \frac{1}{9} = 9^{-1} = 3^{-2} $$ and $$ 27 = 3^{3} $$.

$$
\left(3^{-2}\right)^{x} = 3^{3} \implies 3^{-2x} = 3^{3}
$$

Since the bases are the same, set the exponents equal:

$$
-2x = 3 \implies x = -\frac{3}{2}
$$

**Answer:** $$ x = -\frac{3}{2} $$

---

### **Problem (g):** $$ 4\left(\frac{1}{4}\right)^{x-1} = 8 $$

**Solution:**

First, express $$ \frac{1}{4} $$ and 8 with base 2:

$$
\frac{1}{4} = 2^{-2}, \quad 8 = 2^{3}
$$

Substitute:

$$
4 \cdot \left(2^{-2}\right)^{x-1} = 2^{3} \implies 4 \cdot 2^{-2(x-1)} = 2^{3}
$$

Express 4 as $$ 2^2 $$:

$$
2^{2} \cdot 2^{-2x + 2} = 2^{3} \implies 2^{2 - 2x + 2} = 2^{3} \implies 2^{4 - 2x} = 2^{3}
$$

Set exponents equal:

$$
4 - 2x = 3 \implies -2x = -1 \implies x = \frac{1}{2}
$$

**Answer:** $$ x = \frac{1}{2} $$

---

### **Problem (j):** $$ \frac{3^{2x-1}}{3^{x}} = 3 $$

**Solution:**

Simplify the left side using the laws of exponents:

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

Set equal to the right side:

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

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

---

### **Problem (m):** $$ 8^{-x} - 2 \cdot 4^{x-1} = 0 $$

**Solution:**

First, express 8 and 4 with base 2:

$$
8 = 2^{3}, \quad 4 = 2^{2}
$$

Substitute into the equation:

$$
(2^{3})^{-x} - 2 \cdot (2^{2})^{x-1} = 0 \implies 2^{-3x} - 2 \cdot 2^{2x - 2} = 0
$$

Simplify:

$$
2^{-3x} - 2^{2x - 1} = 0
$$

Let $$ y = 2^{x} $$. Then:

$$
y^{-3} - \frac{y^{2}}{2} = 0 \implies \frac{1}{y^{3}} = \frac{y^{2}}{2}
$$

Multiply both sides by $$ 2y^{3} $$:

$$
2 = y^{5} \implies y = 2^{1/5}
$$

Since $$ y = 2^{x} $$:

$$
2^{x} = 2^{1/5} \implies x = \frac{1}{5}
$$

**Answer:** $$ x = \frac{1}{5} $$

---
Solutions: - (a) $$ x = 0 $$ - (d) $$ x = -\frac{3}{2} $$ - (g) $$ x = \frac{1}{2} $$ - (j) $$ x = 2 $$ - (m) $$ x = \frac{1}{5} $$

Solve for \( x \) : (a) \( 3^{x}=1 \) (d) \( \left(\frac{1}{9}\right)^{x}=27 \) (g) \( 4\left(\frac{1}{4}\right)^{x-1}=8 \) (j) \( \frac{3^{2 x-1}}{3^{x}}=3 \) (m) \( 8^{-x}-2.4^{x-1}=0 \)

Real Tutor Solution

Answer

Solution

Extra Insights

Related Questions

Latest Algebra Questions