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Negative and Zero 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 A bracketed value

Find the value of $\bigl((-0.25)^2\bigr)^0$ .
1. Hint 1
  The zero exponent applies to the whole value inside the outer parentheses.
2. Hint 2
  Check whether that base is zero before deciding what a zero exponent means.
Answer

$1$ .

Full solution

The base of the outer power is a square of a nonzero number, so it is not zero.
$(-0.25)^2=0.0625$

A zero power of this nonzero base equals one.
$0.0625^0=1$

Answer

$1$ .

Key idea

A zero exponent gives one when its entire base is nonzero.
Problem 1: Mark complete
Problem 2 A negative base

Write $(-6)^{-3}$ as a fraction with no powers.
1. Hint 1
  A negative exponent asks for the reciprocal of the matching positive power.
2. Hint 2
  Find the sign of the cube before taking its reciprocal.
Answer

$-\frac{1}{216}$ .

Full solution

The negative exponent calls for the reciprocal of the matching positive power.
$(-6)^{-3}=\frac{1}{(-6)^3}$

Three negative factors give a negative cube.
$(-6)^3=-216$

Its reciprocal has the same sign.
$\frac{1}{-216}=-\frac{1}{216}$

As a check, multiplying $(-6)^3$ by this fraction gives one, as a reciprocal must.
$(-216)\times\left(-\frac{1}{216}\right)=1$

Answer

$-\frac{1}{216}$ .

Key idea

A negative exponent takes a reciprocal, and the sign comes from the negative base and the odd power.
Problem 2: Mark complete
Problem 3 A missing numerator

A positive whole number $k$ satisfies $\left(\frac35\right)^{-2}=\left(\frac{k}{3}\right)^2$ . Find $k$ .
1. Hint 1
  A fraction to a negative power equals another fraction to a positive power; find which one.
2. Hint 2
  Turn the fraction on the left over before squaring, then compare the two squares with their shared denominator.
Answer

$k=5$ .

Full solution

The left side is the square of the reciprocal of $3/5$ .
$\left(\frac35\right)^{-2}=\left(\frac53\right)^2$

This value is $25/9$ .
The right side is $k^2/9$ , so its numerator must be $25$ .
$k^2=25$

The positive whole number whose square is $25$ is $5$ , so $k=5$ .
Substituting it makes both sides $25/9$ .

Answer

$k=5$ .

Key idea

Rewriting a negative power as a positive power of the reciprocal can reveal a missing value.
Problem 3: Mark complete
Problem 4 A three-step calculation

Start with $72$ , multiply it by $3^{-2}$ , and then subtract $(-2)^0$ . What value do you get?
1. Hint 1
  Work out what each power represents before following the steps.
2. Hint 2
  The multiplier is a reciprocal square, and the quantity subtracted is a zero power of a nonzero base.
Answer

$7$ .

Full solution

The multiplier is the reciprocal of nine.
$3^{-2}=\frac19$

Multiply the starting value by this fraction.
$72\times\frac19=8$

The quantity subtracted is $(-2)^0=1$ , so the final value is
$8-1=7$

Answer

$7$ .

Key idea

Evaluate each power with its full base before using it in a sequence of operations.
Problem 4: Mark complete
Problem 5 A layered fraction

Find the value of $\frac{\bigl((-2)^{-1}\bigr)^3}{(-2)^{-5}}$ .
1. Hint 1
  The base $-2$ is nonzero, so the power and quotient rules apply.
2. Hint 2
  First combine the exponents in the numerator.
3. Hint 3
  Subtracting the denominator exponent means subtracting a negative number.
Answer

$4$ .

Full solution

The numerator is a power raised to a power.
$\bigl((-2)^{-1}\bigr)^3=(-2)^{-3}$

Subtract the denominator exponent.
$-3-(-5)=2$

The whole expression is therefore $(-2)^2$ , which equals $4$ .
Check directly: the numerator is $-\frac18$ and the denominator is $-\frac1{32}$ .
$\left(-\frac18\right)\div\left(-\frac1{32}\right)=4$

Answer

$4$ .

Key idea

The quotient rule subtracts signed exponents in the same order as positive exponents.
Problem 5: Mark complete
Problem 6 A product with a reciprocal

For $t\ne0$ , write $t^4\times\frac{1}{t^9}\times(t^2)^0$ in the form $\frac{1}{t^n}$ with $n$ a positive whole number.
1. Hint 1
  A reciprocal of a power can be rewritten with a negative exponent.
2. Hint 2
  A nonzero base to the zero power is one, and $t^2$ is nonzero here.
3. Hint 3
  Add the exponents, then move the negative power into a denominator.
Answer

$\frac{1}{t^5}$ , so $n=5$ .

Full solution

Rewrite the reciprocal power with a negative exponent.
$\frac{1}{t^9}=t^{-9}$

Since $t\ne0$ , the base $t^2$ is nonzero, so its zero power is one.
$(t^2)^0=1$

Add the exponents of the two remaining powers of $t$ .
$t^4\times t^{-9}=t^{-5}$

A negative exponent moves the power into the denominator, so $n=5$ .
$t^{-5}=\frac{1}{t^5}$

As a check at $t=2$ , the expression is $16\times\frac{1}{512}\times1$ .
$16\times\frac{1}{512}=\frac{1}{32}$

This matches $\frac{1}{2^5}$ , since $2^5=32$ .

Answer

$\frac{1}{t^5}$ , so $n=5$ .

Key idea

For a nonzero base, a reciprocal power can be rewritten with a negative exponent so the product rule combines it.
Problem 6: Mark complete
Problem 7 Two model weights

A model has two parts. One part weighs $3\times\left(\frac16\right)^{-2}$ grams, and the other weighs $5\times\left(\frac16\right)^{-1}$ grams. Find the total weight.
1. Hint 1
  The negative exponents turn the fraction bases into powers of their reciprocals.
2. Hint 2
  Find each part weight before adding them.
Answer

$138$ grams.

Full solution

The first power becomes a square of $6$ , the reciprocal of $\frac16$ .
$\left(\frac16\right)^{-2}=36$

The first part therefore weighs $3\times36=108$ grams.
The second power is a single reciprocal.
$\left(\frac16\right)^{-1}=6$

The second part weighs $5\times6=30$ grams.
Add the two weights.
$108+30=138$

The total weight is $138$ grams.

Answer

$138$ grams.

Key idea

A negative exponent on a positive fraction can produce a value greater than one.
Problem 7: Mark complete
Problem 8 A proposed definition

Alex proposes setting $a^0=0$ for every nonzero number $a$ , while keeping $a^2\times a^0=a^2$ . Can both of these hold at once? Explain what value of $a^0$ the equality $a^2\times a^0=a^2$ requires.
1. Hint 1
  Think about which multiplier leaves a nonzero number unchanged.
2. Hint 2
  Try Alex's value in the left side: what is any number times zero?
3. Hint 3
  Compare the multiplier the equality needs with the value Alex proposes.
Answer

No. The equality requires $a^0=1$ .

Full solution

The equality says that multiplying $a^2$ by $a^0$ gives back $a^2$ .
Since $a\ne0$ , the value $a^2$ is nonzero.

The only multiplier that leaves a nonzero number unchanged is $1$ , so the equality requires
$a^0=1$

Alex's value fails, because a multiplier of zero turns the left side into zero, and zero is not the nonzero value $a^2$ .
$a^2\times0=0$

The quotient rule agrees.
It writes the quotient of $a^2$ by itself as a power.
$\frac{a^2}{a^2}=a^{2-2}$

That quotient is $1$ , since a nonzero number over itself is $1$ , and the exponent $2-2$ is $0$ , so again $a^0=1$ .

Answer

No. The equality requires $a^0=1$ .

Key idea

A zero power of a nonzero base must equal one, because one is the multiplier that leaves a nonzero power unchanged.
Problem 8: Mark complete
Problem 9 Lena's two expressions

Lena says $(-7)^{-2}$ and $-7^{-2}$ are the same number. Is she correct? Explain.
1. Hint 1
  Decide what the base is in each expression before evaluating anything.
2. Hint 2
  Without parentheses, the leading minus sign waits until the power is evaluated.
3. Hint 3
  An even number of negative factors gives a positive product, and a reciprocal keeps its sign.
Answer

No. $(-7)^{-2}=\frac{1}{49}$ , but $-7^{-2}=-\frac{1}{49}$ .

Full solution

With the parentheses, the base is $-7$ , and the negative exponent calls for the reciprocal of $(-7)^2$ .
$(-7)^{-2}=\frac{1}{(-7)^2}$

Two negative factors give a positive square, $(-7)^2=49$ , and its reciprocal keeps that sign.
$\frac{1}{(-7)^2}=\frac{1}{49}$

Without parentheses, the base is $7$ , and the leading minus sign waits until the power is evaluated.
$7^{-2}=\frac{1}{49}$

Applying the waiting minus sign gives
$-7^{-2}=-\frac{1}{49}$

One value is positive and the other is negative, so Lena is not correct.

With an odd exponent the two readings would agree: $(-7)^{-3}$ and $-7^{-3}$ both equal $-\frac{1}{343}$ .

Answer

No. $(-7)^{-2}=\frac{1}{49}$ , but $-7^{-2}=-\frac{1}{49}$ .

Key idea

Parentheses decide whether the minus sign belongs to the base, and with an even exponent the two readings have opposite signs.
Problem 9: Mark complete
Problem 10 A proposed ordering

Ria says $a^{-2}<a^0$ for every positive number $a$ . Is she correct? Justify your answer.
1. Hint 1
  A claim about every positive number fails if a single positive number breaks it.
2. Hint 2
  For a base between zero and one, its reciprocal is greater than one.
3. Hint 3
  Compare the square of that reciprocal with the zero power.
Answer

No. Any $a$ with $0<a\le1$ is a counterexample; for example, $a=\frac12$ gives $a^{-2}=4$ and $a^0=1$ .

Full solution

Choose the positive base $a=\frac12$ , which is between zero and one.
Its reciprocal is $2$ .
$\left(\frac12\right)^{-2}=2^2$ $2^2=4$

The same nonzero base to the zero power gives
$\left(\frac12\right)^0=1$

Here $4>1$ , so the claimed ordering fails at $a=\frac12$ , and one failure is enough to show that Ria is not correct.

The same happens for every $a$ between zero and one: the reciprocal of $a$ is greater than one, so its square, $a^{-2}$ , is greater than $1$ , which is $a^0$ .

At $a=1$ both sides equal $1$ , and $1<1$ is false, so the strict inequality fails there too.

Answer

No. Any $a$ with $0<a\le1$ is a counterexample; for example, $a=\frac12$ gives $a^{-2}=4$ and $a^0=1$ .

Key idea

A negative power of a base between zero and one can exceed its zero power.
Problem 10: Mark complete

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