This question tests composing exponents from root equations. The relevant rule is that ($\sqrt{x}$ = $x^{1/2}$ = rac{1}{3}), so (x = left(rac{1}{3} $ight)^2$ = rac{1}{9}). Then, $(x^{3/2}$ = $(x^{1/2}$$)^3$ = left(rac{1}{3} $ight)^3$ = rac{1}{27}). Alternatively, (left(rac{1}{9} $ight)^{3/2}$ = rac{1}{9} cdot $\sqrt{rac{1}{9}$} = rac{1}{9} cdot rac{1}{3} = rac{1}{27}). The result is justified by consistent application of exponents. A distractor like choice A, (rac{1}{9}), fails by stopping at (x) instead of applying the 3/2 exponent.

This question tests evaluating fractional exponents on fractions. The relevant rule is that (left(rac{1}{16} $ight)^{3/4}$ = $(16^{-1}$$)^{3/4}$ = $16^{-3/4}$). Since (16 = $2^4$), $(16^{-3/4}$ = $(2^4$$)^{-3/4}$ = $2^{-3}$ = rac{1}{8}). Alternatively, (left(rac{1}{16} $ight)^{1/4}$ = rac{1}{2}), then raised to the third power is (left(rac{1}{2} $ight)^3$ = rac{1}{8}). The result is justified by consistent exponent simplification. A distractor like choice B, (rac{1}{4}), fails by using an incorrect exponent like 1/2 instead of 3/4.

This question tests expressing numbers with the same base to find ratios of exponents. The relevant rule is to rewrite 8 as $(2^3$), so $(2^m$ = $(2^3$$)^n$ = $2^{3n}$). Equating exponents, (m = 3n), so (rac{m}{n} = 3). This holds for positive integers (m) and (n). The result is justified as it satisfies the equation, like (m=3, n=1): $(2^3$ = $8^1$). A distractor like choice A, (rac{1}{3}), fails by inverting the ratio, perhaps from miswriting (8 = $2^{1/3}$).

This question tests simplifying ratios of square roots. The relevant rule is that (rac{$\sqrt{18}$}{$\sqrt{2}$} = $\sqrt{rac{18}{2}$} = $\sqrt{9}$). Simplifying further, ($\sqrt{9}$ = 3). Alternatively, factor as (rac{3$\sqrt{2}$}{$\sqrt{2}$} = 3). The result is justified as it eliminates the roots correctly. A distractor like choice D, 9, fails by incorrectly squaring the entire expression or misapplying exponent rules.

This question tests simplifying root expressions with absolute values for real numbers. The relevant rule is that $($\sqrt{x^6$}$ = $(x^6$$)^{1/2}$ = $|x^3$| = $|x|^3$). Dividing by $(x^2$) gives $(|x|^3$ / $x^2$ = $|x|^3$ / $|x|^2$ = |x|), since $(x^2$ = $|x|^2$). This holds for all real (x eq 0). The result is justified by checks like (x = -2): ($\sqrt{64}$/4 = 8/4 = 2 = |-2|). A distractor like choice A, (x), fails for negative (x), where it would give a negative instead of positive.

This question tests subtracting exponents with the same base. The relevant rule is that ($rac{2^{5/2}$$}{2^{1/2}$} = $2^{5/2 - 1/2}$ = $2^{4/2}$ = $2^2$ = 4). This applies the quotient rule for exponents. The result equals 4, matching $(2^2$). The justification is the exponent arithmetic simplifying correctly. A distractor like choice D, $(2^{2/2}$ = $2^1$ = 2), fails by misapplying the fraction in the exponent.

This question tests the power rule for exponents and understanding of even roots. The rule states that $(a^m)^n = a^{mn}$. Applying this: $(16x^4)^{1/2} = 16^{1/2} \cdot(x^4)^{1/2} = 4 \cdot x^{4 \cdot 1/2} = 4x^2$. Note that $x^2$ is always non-negative for real $x$, so the expression $4x^2$ is well-defined. We can verify: when $x = 2$, $(16 \cdot 2^4)^{1/2} = (16 \cdot 16)^{1/2} = 256^{1/2} = 16$, and $4 \cdot 2^2 = 4 \cdot 4 = 16$ ✓. A common mistake would be to think $(x^4)^{1/2} = |x^2|$ or $2|x|$, not recognizing that $x^2$ is already non-negative.

This question tests understanding of how fractional exponents behave with different bases. We need to check each option for all real $t \neq 0$. Option A: $(t^2)^{1/2} = |t|$, not $t$, since $\sqrt{t^2} = |t|$. Option B: $(t^3)^{1/3} = t$, not $|t|$, since cube roots preserve sign. Option C: $(t^4)^{1/2} = t^2$ is correct because $(t^4)^{1/2} = t^{4 \cdot 1/2} = t^2$, and this works for all real $t$. Option D: $(t^2)^{3/2} = |t|^3$, not $t^3$, when $t < 0$. Option E: $t^{1/2}$ is undefined for $t < 0$. Therefore, only option C is true for all real $t \neq 0$.

This question tests the multiplication rule for radicals with the same index. The rule states that $\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{ab}$ for real numbers where the radicals are defined. Applying this rule: $\sqrt[3]{a^2} \cdot \sqrt[3]{a} = \sqrt[3]{a^2 \cdot a} = \sqrt[3]{a^3}$. Since $\sqrt[3]{a^3} = a$ for all real $a$ (the cube root function is defined for all real numbers and preserves sign), the answer is $a$. A common mistake would be to add the exponents incorrectly, thinking $\sqrt[3]{a^2} \cdot \sqrt[3]{a} = \sqrt[3]{a^{2+1/3}}$, which misunderstands how radical multiplication works.

This question tests your understanding of the relationship between square roots and absolute values. The key principle is that $\sqrt{x^2} = |x|$ for all real numbers $x$. This is because squaring any real number (positive or negative) gives a non-negative result, and the square root function returns the non-negative value. For example, $\sqrt{(-3)^2} = \sqrt{9} = 3 = |-3|$. The expression equals the absolute value of $x$, not just $x$ itself, because when $x$ is negative, $\sqrt{x^2}$ gives the positive value. Choice A ($x$) would be incorrect for negative values of $x$, as it doesn't account for the sign change.