### Instruction: Determine the value of the infinite product $(2^{1/3})(4^{1/9})(8^{1/27})(16^{1/81}) \dotsm.$  Enter your answer in the form "\sqrt[a]{b}", which stands for $\sqrt[a]{b}.$.### Response: To solve this problem, I need to find a pattern in the product and then use a formula for infinite products.
I notice that the bases of the exponents are all powers of 2, and the exponents are decreasing powers of 3.
So I can rewrite the product as $(2^{1/3})(2^{2/9})(2^{3/27})(2^{4/81}) \dotsm.$
Now I can factor out a 2 from each term and get $2^{(1/3 + 2/9 + 3/27 + 4/81 + \dotsm)}.$
The exponent is an infinite series of the form $\sum_{n=1}^{\infty} \frac{n}{3^n}.$
I recall that there is a formula for such series: $\sum_{n=0}^{\infty} nx^n = \frac{x}{(1-x)^2}$ for any $|x| < 1.$
To apply this formula, I need to shift the index of the series by one and subtract the first term, which is zero. So I get $\sum_{n=1}^{\infty} \frac{n}{3^n} = \sum_{n=0}^{\infty} \frac{n}{3^n} - 0 = \frac{1/3}{(1-1/3)^2}.$
Simplifying the fraction, I get $\frac{1/3}{(2/3)^2} = \frac{1/3}{4/9} = \frac{3}{4}.$
Therefore, the exponent of 2 is 3/4, and the product is $2^{3/4}.$