Rational Exponents - Algebra 3

1. For $x>0$, what are all real values of $k$ for which $\left(\frac{k}{\sqrt{x}}\right)^{2/3}=\frac{4}{x^{1/3}}$ is true for every $x>0$?
   Answer: $k=8$ or $k=-8$ Rewrite the left side as $k^{2/3}(x^{-1/2})^{2/3}=k^{2/3}x^{-1/3}$. For equality with $4x^{-1/3}$, we need $k^{2/3}=4$. Thus $(k^{1/3})^2=4$, so $k^{1/3}=\pm2$, giving $k=8$ or $k=-8$. The options with only one value omit the other possible real root.
2. If $x^{2/3}=5$ and $x>0$, what is the value of $x^{4/3}+x^{8/3}$?
   Answer: $650$ Using power-to-power, $x^{4/3}=(x^{2/3})^2=25$ and $x^{8/3}=(x^{2/3})^4=625$. Their sum is $650$. The choice $30$ uses $25+5$ instead of $25+625$, $150$ uses $25+125$ by taking the second term as $5^3$, and $625$ omits the first term $25$.
3. For $x>0$, which expression is equivalent to $\sqrt{x\sqrt{x}}$?
   Answer: $x^{3/4}$ The inner radical is $x^{1/2}$, so the radicand is $x\cdot x^{1/2}=x^{3/2}$. The outer square root gives $(x^{3/2})^{1/2}=x^{3/4}$. The choice $x^{1/4}$ ignores the factor $x$ before the inner radical, $x^{1/2}$ treats the whole expression as a single square root, and $x^{3/2}$ forgets to apply the outer radical.
4. Simplify: $(x^{-2/3}y^{3/4})^{-12}, \quad x>0,\ y>0$
   Answer: $\frac{x^8}{y^9}$ When you see a power raised to another exponent, especially a negative one, your first instinct should be: multiply every inner exponent by the outer exponent. The negative outer exponent does not mean "make everything positive"; it means every exponent gets multiplied by $-12$, which changes each sign. Start by distributing the exponent: $$(x^{-2/3}y^{3/4})^{-12} = x^{(-2/3)(-12)}y^{(3/4)(-12)} = x^{8}y^{-9} = \frac{x^8}{y^9}.$$ So the correct simplified form is $\frac{x^8}{y^9}$. Now look at the traps. The choice $x^8y^9$ keeps the $y$ exponent positive, which happens if you forget that multiplying $3/4$ by $-12$ gives $-9$, not $+9$. The choice $\frac{y^9}{x^8}$ comes from rewriting the base as $\frac{y^{3/4}}{x^{2/3}}$ before raising to the 12th power; but the true reciprocal of $x^{-2/3}y^{3/4}$ is $x^{2/3}y^{-3/4}$, not that. The choice $x^8y^{-3}$ gets the $x$ part right but miscalculates the $y$ exponent: $(3/4)(-12)=-9$, not $-3$. A reliable strategy is to distribute the outer exponent to every factor first, then simplify any negative exponents by moving them to the denominator. Check each variable independently — that prevents sign errors and fraction mistakes.
5. Solve for real $x$: $x^{3/4}=8$
   Answer: $16$ Whenever you solve an equation like $x^{3/4}=8$, think in terms of undoing the exponent. The notation $x^{3/4}$ means $(\sqrt[4]{x})^3$, so you can raise both sides to the reciprocal power $\frac{4}{3}$. A clean way is to write 8 as a power of 2: $8=2^3$. If $x=2^k$, then $x^{3/4}=(2^k)^{3/4}=2^{3k/4}$. For this to equal $2^3$, the exponents must match: $\frac{3k}{4}=3$, so $k=4$. Therefore $x=2^4=16$. Quick check: $16^{3/4}=(\sqrt[4]{16})^3=2^3=8$. The other choices are all powers of 2 with the wrong exponent: $4=2^2$ gives $2^{3/2}\approx 2.83$; $32=2^5$ gives $2^{15/4}\approx 13.45$; and $64=2^6$ gives $2^{9/2}\approx 22.6$. None equals 8. These wrong answers typically come from forgetting to invert the fractional exponent or from mishandling the exponents on powers of 2. The fastest routine: for an equation of the form $x^{m/n}=a$, raise both sides to $n/m$ and simplify from the inside out: $x=8^{4/3}=(\sqrt[3]{8})^4=2^4=16$. Remember: undo a fractional exponent with its reciprocal, not by multiplying or guessing powers.
6. Rewrite $x^{3/4}y^{-1/2}$ using only positive exponents, for $x>0,\ y>0$.
   Answer: $\frac{\sqrt[4]{x^3}}{\sqrt{y}}$ Whenever you see fractional and negative exponents together, separate the two rules: a fractional exponent like $\frac{3}{4}$ means a power and a root, while a negative exponent means "take the reciprocal." For $x^{3/4}$, the numerator 3 is the power and the denominator 4 is the root, so it becomes $\sqrt[4]{x^3}$. For $y^{-1/2}$, the negative exponent flips it to the denominator, and the $\frac12$ exponent becomes a square root: $y^{-1/2} = \frac{1}{\sqrt{y}}$. Combining gives $\frac{\sqrt[4]{x^3}}{\sqrt{y}}$. Now check the traps. The choice with $\frac{\sqrt[4]{x^3}}{y^2}$ treats the exponent $-\frac12$ as if it meant squaring in the denominator — that would be $y^{-2}$, not $y^{-1/2}$. The choice with $\frac{\sqrt[3]{x^4}}{\sqrt{y}}$ swaps the numerator and denominator of the fractional exponent, giving $x^{4/3}$ instead of $x^{3/4}$. The choice with $\frac{\sqrt[4]{x^3}}{y}$ drops the root from the denominator: $y^{-1/2}$ is not $y^{-1}$, so it must become $\sqrt{y}$, not just $y$. Your takeaway: rewrite each piece separately — "root by the bottom, power by the top, negative means flip." That order prevents most exponent-law errors on this exam.
7. Simplify: $\left(\frac{x^{1/2}}{y^{1/3}}\right)^{-6}, \quad x>0,\ y>0$
   Answer: $\frac{y^2}{x^3}$ When you see a fraction raised to a negative exponent, remember that the negative sign tells you to take the reciprocal, and the exponent applies to both the numerator and denominator. Here, distributing $-6$ gives $(x^{1/2})^{-6}/(y^{1/3})^{-6} = x^{-3}/y^{-2}$. Since $x^{-3}=1/x^3$ and $y^{-2}=1/y^2$, the expression simplifies to $(1/x^3)/(1/y^2) = y^2/x^3$. That is the correct answer. Now, why are the other choices wrong? The option $\frac{x^3}{y^2}$ comes from ignoring the negative exponent entirely and using $+6$ instead of $-6$. The option $x^3y^2$ arises if you correctly flip the fraction first—because $(a/b)^{-n} = (b/a)^n$—but then mistakenly multiply the results instead of dividing: $(y^{1/3}/x^{1/2})^6$ should give $y^2/x^3$, not $y^2 \cdot x^3$. The option $\frac{1}{x^3y^2}$ results from applying the exponent to the numerator and denominator separately but then multiplying those outcomes, treating the original division as if it were multiplication. A useful habit: after simplifying, check whether your result has the correct reciprocal relationship—for a negative exponent, the variables should swap positions across the fraction bar. And when raising a fraction to a power, always divide the numerator's result by the denominator's result, never multiply.
8. Which is the completely factored form of $2x^{-3/2}-x^{-5/2}+4x^{-7/2}$, with the smallest power of $x$ factored out?
   Answer: $x^{-7/2}(2x^2-x+4)$ The smallest exponent is $-7/2$. Subtracting it from each exponent gives $-3/2-(-7/2)=2$, $-5/2-(-7/2)=1$, and $-7/2-(-7/2)=0$, so the factored form is $x^{-7/2}(2x^2-x+4)$. The options that factor out $-3/2$ or $-1/2$ do not use the smallest power, and the option leaving $x^{-1}$ and $x^{-2}$ inside results from sign errors when subtracting exponents.
9. Simplify: $2\sqrt[3]{x}\cdot 3x^{-1/2}, \quad x>0$
   Answer: $6x^{-1/6}$ When you see a radical multiplied by a power of $x$, your first move should be to rewrite the radical as a fractional exponent: $\sqrt[3]{x}=x^{1/3}$. Once everything is in exponent form, you're just multiplying same bases, so multiply coefficients and add exponents. Multiply the coefficients: $2\cdot 3=6$. Then add the exponents: $$x^{1/3}\cdot x^{-1/2}=x^{\frac13+\left(-\frac12\right)} =x^{\frac{2}{6}-\frac{3}{6}} =x^{-1/6}.$$ So the simplified expression is $$6x^{-1/6}.$$ Now let's expose the traps in the other choices. - $6x^{1/6}$ comes from subtracting the exponents as $\frac12-\frac13$, but multiplication of powers requires addition, not subtraction. - $6x^{5/6}$ comes from adding $\frac13+\frac12$ while ignoring the negative sign on $-1/2$. The exponent is negative, so that sign must be included. - $6x^{-5/6}$ comes from treating $\sqrt[3]{x}$ as $x^{-1/3}$ rather than $x^{1/3}$. A cube root is a positive fractional exponent, not a negative one. A quick way to guard against these mistakes: rewrite every term using fractional exponents, then add the exponents with their signs. You can even approximate: $1/3-1/2\approx 0.333-0.5=-0.166$, which is $-1/6$. That check alone would eliminate three of the four answer choices.
10. Simplify: $\sqrt[3]{x\sqrt{x}}, \quad x>0$
    Answer: $\sqrt{x}$ Whenever you see nested radicals with a variable base, the cleanest move is to rewrite every radical using fractional exponents. Here, $\sqrt{x}=x^{1/2}$, so \sqrt[3]{x\sqrt{x}}=\sqrt[3]{x\cdot x^{1/2}}}=\sqrt[3]{x^{1+1/2}}} =\sqrt[3]{x^{3/2}}}=(x^{3/2})^{1/3}=x^{1/2}=\sqrt{x}. Thus the simplified form is $\sqrt{x}$. The other choices each lose one layer of the expression: $x^{1/6}$ is what you get if you cube-root only the $\sqrt{x}$ and drop the leading $x$; $\sqrt[3]{x}$ drops the $\sqrt{x}$ factor entirely; and $x^{3/2}$isthe product $x\sqrt{x}$ before applying the outer cube root at all. Study tip: rewrite every radical as a fractional exponent, then add exponentsinside and multiply by theouter root's exponent. Because the cube root applies to the whole product $x\sqrt{x}$, not to just one piece, make sure the final exponent reflects both factors and the outer root.
11. Simplify $\left(\frac{8a^6}{27b^{-3}}\right)^{-2/3}$. Assume $a,b>0$.
    Answer: $\frac{9}{4a^4b^2}$ Rewrite the base as $\frac{8a^6b^3}{27}$. Then $8^{-2/3}=\frac14$, $(a^6)^{-2/3}=a^{-4}$, $(b^3)^{-2/3}=b^{-2}$, and $27^{-2/3}=\frac19$. Dividing by $\frac19$ gives $\frac{9}{4a^4b^2}$. The option with $\frac{4a^4b^2}{9}$ uses $+2/3$ instead of $-2/3$, $\frac{9b^2}{4a^4}$ misreads $b^{-3}$ as $b^3$ in the denominator, and $\frac{4}{9a^4b^2}$ treats $8^{-2/3}$ as $4$ rather than $\frac14$.
12. Simplify: $(\sqrt[3]{x^2})^6 \div \sqrt{x^3}, \quad x>0$
    Answer: $x^{5/2}$ Whenever you see radical expressions raised to powers and divided, rewrite each radical as a fractional exponent first. For this expression, $\sqrt[3]{x^2}=x^{2/3}$ and $\sqrt{x^3}=x^{3/2}$. The original becomes $(x^{2/3})^6 \div x^{3/2}$. Use the power rule: $(x^{2/3})^6 = x^{(2/3)\cdot 6}=x^4$. Then for division, subtract exponents: $x^4 \div x^{3/2}=x^{4-3/2}=x^{5/2}$. That is the correct simplification. The trap choices all come from misapplying exponent rules. $x^{8/3}$ results from dividing the exponents ($4 \div 3/2 = 8/3$) instead of subtracting them. $x^{11/2}$ comes from adding the exponents ($4+3/2=11/2$) when the operation is division. $x^{7/2}$ usually comes from misreading $\sqrt{x^3}$ as $\sqrt{x}$, which is $x^{1/2}$; dropping the exponent 3 under the radical gives $4 - 1/2 = 7/2$. Remember that $\sqrt{x^3}=x^{3/2}$, not $x^{1/2}$. Strategy: On any exponent-simplification problem, convert radicals to fractional powers, apply the power rule by multiplying, and apply the quotient rule by subtracting. Always carry the full exponent from inside the radical when converting.
13. Which equation has no real solution?
    Answer: $x^{2/3}=-4$ For real $x$, $x^{2/3}=(x^{1/3})^2$, which is never negative, so it cannot equal $-4$. The first equation has solution $x=-8$. The second equation has a real solution: $x=(-32)^{5/3}=-2^{25/3}$. The third has solution $x=16$.
14. Simplify: $\sqrt[4]{\frac{x^3}{y^2}}\cdot\sqrt[3]{y^2}, \quad x>0,\ y>0$
    Answer: $x^{3/4}y^{1/6}$ Whenever you see radicals like $\sqrt[4]{...}$ and $\sqrt[3]{...}$, rewrite them as fractional exponents: $\sqrt[n]{a^m} = a^{m/n}$. Also, a variable in the denominator gets a negative exponent. Here, $\sqrt[4]{\frac{x^3}{y^2}} = (x^3 y^{-2})^{1/4} = x^{3/4} y^{-1/2}$. Then multiply by $\sqrt[3]{y^2} = y^{2/3}$. Adding the exponents for $y$: $-1/2 + 2/3 = (-3+4)/6 = 1/6$. So the result is $x^{3/4} y^{1/6}$. The other options each come from a common slip. The one with $y^{7/6}$ arises if you forget the negative sign on the denominator and compute $1/2 + 2/3 = 7/6$. The one with $y^{-1/6}$ results from mistakenly writing $\sqrt[3]{y^2}$ as $y^{1/3}$ (instead of $y^{2/3}$) and then adding $-1/2 + 1/3 = -1/6$. The option with $y^{5/6}$ combines two errors: ignoring the negative exponent on the $y$ in the denominator and using $y^{1/3}$ for the cube root, giving $1/2 + 1/3 = 5/6$. Study tip: Always convert radicals to fractional exponents first, and be careful with signs when a variable is in the denominator. Write each exponent explicitly, then combine like bases using addition. Double-check your $y$-exponent by plugging in a simple value like $y=4$ if you're unsure.
15. For $x>0$, which expression is NOT equivalent to $\sqrt[3]{x^2}$?
    Answer: $\sqrt[6]{x^3}$ When you see radical expressions like $\sqrt[3]{x^2}$, your first move should be to rewrite them using fractional exponents: the index becomes the denominator, and the exponent inside becomes the numerator. So $\sqrt[3]{x^2} = x^{2/3}$. That is your target expression. Now check each option by converting to fractional exponents. The expression $\sqrt[6]{x^3}$ becomes $x^{3/6} = x^{1/2}$, which is not the same as $x^{2/3}$ — that is the one that is NOT equivalent. Every other option simplifies to your target: $x^{2/3}$ is exactly the target; $(\sqrt[3]{x})^2 = (x^{1/3})^2 = x^{2/3}$; and $\sqrt[6]{x^4} = x^{4/6} = x^{2/3}$. The trap here is confusing the order of operations: when you take a root and then raise to a power, the exponents multiply, but when you have a power inside a root, the exponents add in the fraction. Always reduce the fractional exponent fully (like simplifying $4/6$ to $2/3$) before comparing. A quick strategy: for any "which is NOT equivalent" question, rewrite every choice as a power of $x$ and simplify the fraction. Then compare to the original exponent. This works because $x>0$ guarantees the rules of exponents apply without worrying about negative bases.
16. If $x>0$ and $x^{1/3}+x^{-1/3}=3$, what is $x^{2/3}+x^{-2/3}$?
    Answer: $7$ Square both sides: $(x^{1/3}+x^{-1/3})^2=x^{2/3}+2+x^{-2/3}=9$. Therefore $x^{2/3}+x^{-2/3}=7$. The choice $9$ comes from distributing the square incorrectly, the choice $11$ adds $2$ instead of subtracting it, and the choice $5$ subtracts $2^2$ instead of $2$.
17. Solve $x^{1/2}-3x^{1/4}+2=0$ for $x>0$.
    Answer: $x=1$ or $x=16$ Let $u=x^{1/4}$. Then $x^{1/2}=u^2$, so $u^2-3u+2=0$. Thus $u=1$ or $u=2$. Since $x=u^4$, the solutions are $x=1$ and $x=16$. The choices that keep only one root discard a valid root, and $x=8$ comes from using $x=u^3$ instead of $x=u^4$.
18. Solve $(2x+1)^{2/3}=25$ for all real $x$.
    Answer: $x=62$ or $x=-63$ Let $u=2x+1$. Since $u^{2/3}=(u^{1/3})^2=25$, we need $u^{1/3}=\pm5$, so $u=\pm125$. Thus $2x+1=125$ gives $x=62$, and $2x+1=-125$ gives $x=-63$. The choice $x=62$ only ignores the negative root; $x=-63$ only keeps the negative root; and $x=-62$ comes from an arithmetic sign error when solving $2x+1=-125$.
19. Which equation is true for all real $x$?
    Answer: $(\sqrt[3]{x})^3=x$ For all real $x$, the cube root of $x$, cubed, returns $x$. The first option should be $|x|$, not $x$, the second is defined only for $x\ge0$, and the third should be $|x|^{1/2}$, not $x^{1/2}$, for negative $x$.
20. Simplify: $\sqrt[3]{\frac{8x^6}{y^{-9}}}, \quad y\ne0$
    Answer: $2x^2y^3$ When you see a cube root of a rational expression, your first move is to separate the parts: cube root the numerator, cube root the denominator, and simplify each piece. Remember that a negative exponent means a reciprocal, so $y^{-9}$ in the denominator actually moves to the numerator as $y^9$. Start with the cube root of the whole expression: $\sqrt[3]{\frac{8x^6}{y^{-9}}}$. Because dividing by $y^{-9}$ is the same as multiplying by $y^9$, rewrite the radicand as $8x^6 \cdot y^9$. Now take the cube root of each factor: $\sqrt[3]{8}=2$, $\sqrt[3]{x^6}=x^2$ since $6 \div 3 = 2$, and $\sqrt[3]{y^9}=y^3$. Multiply these together to get $2x^2y^3$. Why the others miss the mark: $\frac{2x^2}{y^3}$ comes from treating $y^{-9}$ as if it stayed in the denominator after cubing — but the negative exponent means it belongs upstairs. $8x^2y^3$ is close, but it forgets to take the cube root of 8; the cube root of 8 is 2, not 8. $2x^3y^3$ makes the common exponent error of dividing 6 by 2 instead of 3 when taking a cube root; $x^6$ cubed gives $x^2$, not $x^3$. Your takeaway: for a cube root, divide every exponent by 3, and always flip negative exponents before simplifying. That combination will keep you safe on radical–exponent questions.

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