The product $$2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \ldots$$ to $$\infty$$ is equal to:

Let the given product be $$P$$:

$$P = 2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \dots \text{to } \infty$$

Step 1: Express all terms with a common base of 2.

• $$4^{\frac{1}{16}} = (2^2)^{\frac{1}{16}} = 2^{\frac{2}{16}} = 2^{\frac{1}{8}}$$
• $$8^{\frac{1}{48}} = (2^3)^{\frac{1}{48}} = 2^{\frac{3}{48}} = 2^{\frac{1}{16}}$$
• $$16^{\frac{1}{128}} = (2^4)^{\frac{1}{128}} = 2^{\frac{4}{128}} = 2^{\frac{1}{32}}$$

Substitute these back into the original expression:

$$P = 2^{\frac{1}{4}} \cdot 2^{\frac{1}{8}} \cdot 2^{\frac{1}{16}} \cdot 2^{\frac{1}{32}} \dots \text{to } \infty$$

Step 2: Add the exponents.

Since the bases are all the same, we can use the exponent rule $$x^a \cdot x^b = x^{a+b}$$:

$$P = 2^{\left(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots \text{to } \infty \right)}$$

Step 3: Evaluate the infinite series in the exponent.

The exponent forms an infinite Geometric Progression (G.P.):

$$\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots$$

• First term, $$a = \frac{1}{4}$$
• Common ratio, $$r = \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{1}{2}$$

Using the formula for the sum of an infinite G.P. ($$S_{\infty} = \frac{a}{1 - r}$$):

$$S_{\infty} = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$$

Step 4: Find the final product.

Substitute the sum back into the exponent:

$$P = 2^{\frac{1}{2}} = \sqrt{2}$$