4) $$ 16^{x} \cdot 2^{x+1}=\sqrt[5]{4} $$

Question

UpStudy AI · Accepted Answer

To solve the equation:

$$
16^{x} \cdot 2^{x+1} = \sqrt[5]{4}
$$

**Step 1: Express All Terms with the Same Base**

First, rewrite all terms using base 2:

- $$16 = 2^4$$, so $$16^x = (2^4)^x = 2^{4x}$$.
- $$4 = 2^2$$, so $$\sqrt[5]{4} = (2^2)^{1/5} = 2^{2/5}$$.

The equation becomes:

$$
2^{4x} \cdot 2^{x+1} = 2^{2/5}
$$

**Step 2: Combine Exponents on the Left Side**

When multiplying like bases, add the exponents:

$$
2^{4x} \cdot 2^{x+1} = 2^{4x + x + 1} = 2^{5x + 1}
$$

**Step 3: Set the Exponents Equal**

Since the bases are the same, set the exponents equal to each other:

$$
5x + 1 = \frac{2}{5}
$$

**Step 4: Solve for $$x$$**

Subtract 1 from both sides:

$$
5x = \frac{2}{5} - 1 = \frac{2}{5} - \frac{5}{5} = -\frac{3}{5}
$$

Divide both sides by 5:

$$
x = \frac{-\frac{3}{5}}{5} = -\frac{3}{25}
$$

**Answer:**

$$
x = -\frac{3}{25}
$$
$$ x = -\frac{3}{25} $$

4) \( 16^{x} \cdot 2^{x+1}=\sqrt[5]{4} \)