If y = x^{1/3} cdot x^{1/9} cdot x^{1/27} cdo | vedclass

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(C) Given $y = x^{1/3} \cdot x^{1/9} \cdot x^{1/27} \cdot \dots \infty$. Using the property of exponents $a^m \cdot a^n = a^{m+n}$,we get: $y = x^{(1/

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$x^{1/2}$

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(C) Given $y = x^{1/3} \cdot x^{1/9} \cdot x^{1/27} \cdot \dots \infty$. Using the property of exponents $a^m \cdot a^n = a^{m+n}$,we get: $y = x^{(1/3 + 1/9 + 1/27 + \dots \infty)}$. The exponent is an infinite geometric series with first term $a = 1/3$ and common ratio $r = 1/3$. The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$. $S = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = 1/2$. Therefore,$y = x^{1/2}$.

If $y = x^{1/3} \cdot x^{1/9} \cdot x^{1/27} \cdot \dots \infty$,then $y = \dots$

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