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Rational Exponents: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Between two integers

Without a calculator or decimals, find the two consecutive positive integers between which $137^{1/4}$ lies, and justify your answer.
1. Hint 1
  The principal fourth root of a positive number is the positive number whose fourth power is that number.
2. Hint 2
  Compare $137$ with the fourth powers of small positive integers.
Answer

$3<137^{1/4}<4$ .

Full solution

The number $137^{1/4}$ is the positive number whose fourth power is $137$ .

Since $3^4=81$ and $4^4=256$ ,
$81<137<256$

For positive numbers, a larger number has a larger positive fourth root, so the order survives taking fourth roots:
$3<137^{1/4}<4$

Answer

$3<137^{1/4}<4$ .

Key idea

On positive numbers, taking a positive root keeps order, so known powers bracket an unknown root.
Problem 1: Mark complete
Problem 2 An exponent slot

Find the rational number $r$ for which $(3^r)^6=3^5$ .
1. Hint 1
  For a positive base, a power raised to another power multiplies the exponents.
2. Hint 2
  With base $3$ , equal powers have equal exponents.
Answer

$r=\frac56$ .

Full solution

The base is positive, so the power law applies.
$6r=5$

Divide by $6$ .
$r=\frac56$

Checking multiplies $5/6$ by $6$ and recovers the exponent $5$ .

Answer

$r=\frac56$ .

Key idea

The power law determines which fractional exponent undoes a given integer power.
Problem 2: Mark complete
Problem 3 The input interval

Find the real domain of $f(t)=\sqrt[4]{12-3t}$ .
1. Hint 1
  An even root requires a zero or positive radicand.
2. Hint 2
  Solve the inequality $12-3t\ge0$ .
Answer

$t\le4$ .

Full solution

The fourth root is real exactly when its radicand is zero or positive.
$12-3t\ge0$

Subtract $12$ and divide by the negative number $-3$ , reversing the inequality.
$t\le4$

At $t=4$ the root is $0$ , so the endpoint is included.

Answer

$t\le4$ .

Key idea

An even root restricts its input through the sign of its radicand.
Problem 3: Mark complete
Problem 4 Two stored values

A calculator stores $A=(36/121)^{1/2}$ and $B=(36/121)^{-1/2}$ . Find $A/B$ as a reduced fraction.
1. Hint 1
  The negative exponent makes one stored value the reciprocal of the other.
2. Hint 2
  Find the positive square root of $36/121$ , then divide the two values.
Answer

$\frac{36}{121}$ .

Full solution

The positive square root gives
$A=\frac6{11}$

The reciprocal gives
$B=\frac{11}6$

The divisor is nonzero.
Divide by multiplying by its reciprocal.
$\frac AB=\frac{36}{121}$

Answer

$\frac{36}{121}$ .

Key idea

A negative exponent changes a positive value to its reciprocal.
Problem 4: Mark complete
Problem 5 A cube model

The surface area of a cube with volume $V$ cubic centimeters is $6V^{2/3}$ square centimeters. Find the surface area when $V=64/125$ . Give an exact value.
1. Hint 1
  The exponent first finds the edge length and then squares it.
2. Hint 2
  Take the cube root of the numerator and denominator before squaring.
Answer

$\frac{96}{25}$ square centimeters.

Full solution

The cube root gives the edge length in centimeters.
$V^{1/3}=\frac45$

Square this value and multiply by the six faces.
$6V^{2/3}=\frac{96}{25}$

An edge of $4/5$ cm gives volume $64/125$ cubic centimeters, matching the input.

Answer

$\frac{96}{25}$ square centimeters.

Key idea

Rooting first keeps fractional-power calculations small.
Problem 5: Mark complete
Problem 6 A shared building block

Positive numbers $a$ and $b$ satisfy $a^{1/4}=b^{1/3}$ . Write $b$ as a power of $a$ , then find the exact value of $\frac{a^{1/2}}{b^{2/3}}$ .
1. Hint 1
  Raise both sides to a power that leaves $b$ on its own.
2. Hint 2
  Once $b$ is a power of $a$ , use the power law on $b^{2/3}$ .
Answer

$b=a^{3/4}$ ; the quotient is $1$ .

Full solution

Both bases are positive, so the power law applies.
Cubing both sides gives
$b=a^{3/4}$

Then
$\begin{aligned}b^{2/3}&=(a^{3/4})^{2/3}\\&=a^{1/2}\end{aligned}$

So the quotient is $\frac{a^{1/2}}{a^{1/2}}=1$ , which is defined because $a^{1/2}>0$ .

Answer

$b=a^{3/4}$ ; the quotient is $1$ .

Key idea

When $a^r=b^s$ for positive $a$ and $b$ and $s\ne0$ , the base $b$ equals $a^{r/s}$ , so expressions in both bases compare directly.
Problem 6: Mark complete
Problem 7 An adjustable gain

A device uses a positive setting $x$ and computes
$G(x)=\frac{x^{1/3}+x^{-1/3}}{x^{1/3}}$
Write the gain as a sum of two terms with no negative exponents, and explain why the restriction $x>0$ permits the exponent-law steps and the division.
1. Hint 1
  Divide each term in the numerator by the common denominator.
2. Hint 2
  Subtract exponents for each quotient and rewrite any negative power as a reciprocal.
Answer

$G(x)=1+\frac1{x^{2/3}}$ ; with $x>0$ the exponent laws apply and the divisor is not zero.

Full solution

Since $x>0$ , the denominator is positive and the power laws apply.
The first quotient is $1$ , and the second has exponent $-1/3-1/3$ .
$G(x)=1+x^{-2/3}$

Write the negative power as a reciprocal.
$G(x)=1+\frac1{x^{2/3}}$

At $x=1$ , both forms give $2$ .

Answer

$G(x)=1+\frac1{x^{2/3}}$ ; with $x>0$ the exponent laws apply and the divisor is not zero.

Key idea

A quotient of a sum can be simplified one term at a time on its original domain.
Problem 7: Mark complete
Problem 8 A proposed output

A machine accepts any positive number $a$ and returns a positive number $u$ whose sixth power is $a$ . A designer labels the output $a^{1/6}$ . Is that label consistent with the power law? Explain why its sign matters.
1. Hint 1
  Ask what happens when the proposed output is raised to the sixth power.
2. Hint 2
  Compare the machine's output requirement with the definition of $a^{1/6}$ .
Answer

Yes; the positive principal sixth root is required.

Full solution

The power law requires the sixth power of the labeled output to be the original input.
$(a^{1/6})^6=a$

The positive output with that sixth power is precisely $\sqrt[6]{a}$ .
There is also a negative number with the same sixth power, but positive bases are assigned positive rational powers.
The sign condition selects the principal root.

Answer

Yes; the positive principal sixth root is required.

Key idea

The power law supplies a root condition, and positivity selects the principal root.
Problem 8: Mark complete
Problem 9 Two predictions

A student says $(-3125)^{1/5}$ and $(-3125)^{2/5}$ are both negative. Decide whether the claim is correct and give both exact values.
1. Hint 1
  An odd root retains the sign of its negative input.
2. Hint 2
  The second exponent squares that odd root.
Answer

Incorrect; $(-3125)^{1/5}=-5$ and $(-3125)^{2/5}=25$ .

Full solution

Both reduced denominators are odd, so the real odd-root convention applies.
$(-3125)^{1/5}=-5$

For the other expression, square the fifth root.
$(-3125)^{2/5}=25$

The first output is negative but its square is positive.
The claim fails for the second value.

Answer

Incorrect; $(-3125)^{1/5}=-5$ and $(-3125)^{2/5}=25$ .

Key idea

A negative base with an odd reduced root index can give a positive result when the numerator exponent is even.
Problem 9: Mark complete
Problem 10 Choosing an index

An integer $n$ is chosen from $2$ through $8$ . For which choices does $(-19)^{2/n}$ have a real value? Explain how you decided.
1. Hint 1
  Reduce each exponent before inspecting its denominator.
2. Hint 2
  A negative base has a real rational power exactly when the reduced denominator is odd.
Answer

$n=2,3,5,6,7$ .

Full solution

For $n=2,3,4,5,6,7,8$ , the reduced exponents are respectively $1$ , $2/3$ , $1/2$ , $2/5$ , $1/3$ , $2/7$ , and $1/4$ .

The even reduced denominators occur at $n=4$ and $n=8$ .
Those two choices have no real value.
Every remaining reduced denominator is odd.
For example, at $n=6$ ,
$\frac2n=\frac13$

This is a cube root of a negative number, so it is real even though the original written denominator was even.

Answer

$n=2,3,5,6,7$ .

Key idea

The reduced denominator determines whether a negative base has a real rational power.
Problem 10: Mark complete

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