Practice · Multiply top and bottom to use the given power · Sensim Math

Variant 1 easy

Use the rule $a^{-n}=\frac{1}{a^n}$ for every natural number $n$ . If $a^{-1}=3$ , find the value of $\frac{a-a^{-1}}{a+a^{-1}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know $a^{-1}=3$ , where $a^{-n}$ means $\frac{1}{a^n}$ . We want the value of the fraction $\frac{a-a^{-1}}{a+a^{-1}}$ .

Givens

$a^{-n}=\frac{1}{a^n}$ for natural numbers $n$ .
$a^{-1}=3$ .

Unknowns

The value of $\frac{a-a^{-1}}{a+a^{-1}}$ .

Constraints

$a\neq0$ , so multiplying the top and bottom by a power of $a$ does not change the fraction.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways

We cannot easily find $a$ itself, but we know $a^{-1}$ . Multiply the top and bottom by $a^{-1}$ so that every term becomes either $1$ or $a^{-2}$ , which is a power of $a^{-1}$ .

3 · Execute3 carry out the plan

1Turn the given into $a^{-2}$

#15 Organize Information in More Ways 8.EE.A.1

Square the given value:

a^{-2}=\frac{1}{a^2}=\left(\frac{1}{a}\right)^2

.

a^{-2}=(a^{-1})^2=3^2=9

Raising a power to a power multiplies the exponents.

2Multiply top and bottom by $a^{-1}$

#13 Convert to Algebra 8.EE.A.1

This keeps the value of the fraction. Each

a\times a^{-1}

becomes

a^0=1

, and each

a^{-1}\times a^{-1}

becomes

a^{-2}

.

\frac{a-a^{-1}}{a+a^{-1}}=\frac{1-a^{-2}}{1+a^{-2}}

a\times\frac{1}{a}=1

.

3Substitute and simplify

#13 Convert to Algebra 7.NS.A.2

Put

a^{-2}=9

into the new fraction.

\frac{1-9}{1+9}=\frac{-8}{10}=-\frac{4}{5}

A negative number divided by a positive number is negative.

Answer:

-\frac{4}{5}

4 · Reviewdoes it hold up?

Here $a^{-2}=9>1$ , so in $\frac{1-a^{-2}}{1+a^{-2}}$ the top is negative and the bottom is positive and bigger in size. The answer should be negative and between $-1$ and $0$ . $-\frac{4}{5}$ fits.

Another way: Work with positive powers instead: $a^2=\frac{1}{a^{-2}}=\frac{1}{9}$ . Multiply the top and bottom by $a$ : $\frac{a^2-1}{a^2+1}=\frac{\frac{1}{9}-1}{\frac{1}{9}+1}=-\frac{4}{5}$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Using $a^{-n}=\frac{1}{a^n}$, the product rule, and the power rule with negative exponents.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Dividing a negative number by a positive number.

💡Takeaway. If you know one power of

a

, reshape the expression so only that power appears.

Variant 2 easy

Use the rule $a^{-n}=\frac{1}{a^n}$ for every natural number $n$ . If $a^{-1}=2$ , find the value of $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know $a^{-1}=2$ , where $a^{-n}$ means $\frac{1}{a^n}$ . We want the value of the fraction $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Givens

$a^{-n}=\frac{1}{a^n}$ for natural numbers $n$ .
$a^{-1}=2$ .

Unknowns

The value of $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Constraints

$a\neq0$ , so multiplying the top and bottom by a power of $a$ does not change the fraction.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways

We cannot easily find $a$ itself, but we know $a^{-1}$ . Multiply the top and bottom by $a^{-2}$ so that every term becomes either $1$ or $a^{-4}$ , which is a power of $a^{-1}$ .

3 · Execute3 carry out the plan

1Turn the given into $a^{-4}$

#15 Organize Information in More Ways 8.EE.A.1

Raise the given value to the power

4

:

a^{-4}=\frac{1}{a^4}=\left(\frac{1}{a}\right)^4

.

a^{-4}=(a^{-1})^4=2^4=16

Raising a power to a power multiplies the exponents.

2Multiply top and bottom by $a^{-2}$

#13 Convert to Algebra 8.EE.A.1

This keeps the value of the fraction. Each

a^2\times a^{-2}

becomes

a^0=1

, and each

a^{-2}\times a^{-2}

becomes

a^{-4}

.

\frac{a^2-a^{-2}}{a^2+a^{-2}}=\frac{1-a^{-4}}{1+a^{-4}}

a^2\times\frac{1}{a^2}=1

.

3Substitute and simplify

#13 Convert to Algebra 7.NS.A.2

Put

a^{-4}=16

into the new fraction.

\frac{1-16}{1+16}=\frac{-15}{17}=-\frac{15}{17}

A negative number divided by a positive number is negative.

Answer:

-\frac{15}{17}

4 · Reviewdoes it hold up?

Here $a^{-4}=16>1$ , so in $\frac{1-a^{-4}}{1+a^{-4}}$ the top is negative and the bottom is positive and bigger in size. The answer should be negative and between $-1$ and $0$ . $-\frac{15}{17}$ fits.

Another way: Work with positive powers instead: $a^4=\frac{1}{a^{-4}}=\frac{1}{16}$ . Multiply the top and bottom by $a^2$ : $\frac{a^4-1}{a^4+1}=\frac{\frac{1}{16}-1}{\frac{1}{16}+1}=-\frac{15}{17}$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Using $a^{-n}=\frac{1}{a^n}$, the product rule, and the power rule with negative exponents.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Dividing a negative number by a positive number.

💡Takeaway. If you know one power of

a

, reshape the expression so only that power appears.

Variant 3 easy

Use the rule $a^{-n}=\frac{1}{a^n}$ for every natural number $n$ . If $a^{-2}=2$ , find the value of $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know $a^{-2}=2$ , where $a^{-n}$ means $\frac{1}{a^n}$ . We want the value of the fraction $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Givens

$a^{-n}=\frac{1}{a^n}$ for natural numbers $n$ .
$a^{-2}=2$ .

Unknowns

The value of $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Constraints

$a\neq0$ , so multiplying the top and bottom by a power of $a$ does not change the fraction.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways

We cannot easily find $a$ itself, but we know $a^{-2}$ . Multiply the top and bottom by $a^{-2}$ so that every term becomes either $1$ or $a^{-4}$ , which is a power of $a^{-2}$ .

3 · Execute3 carry out the plan

1Turn the given into $a^{-4}$

#15 Organize Information in More Ways 8.EE.A.1

Square the given value:

a^{-4}=\frac{1}{a^4}=\left(\frac{1}{a^2}\right)^2

.

a^{-4}=(a^{-2})^2=2^2=4

Raising a power to a power multiplies the exponents.

2Multiply top and bottom by $a^{-2}$

#13 Convert to Algebra 8.EE.A.1

This keeps the value of the fraction. Each

a^2\times a^{-2}

becomes

a^0=1

, and each

a^{-2}\times a^{-2}

becomes

a^{-4}

.

\frac{a^2-a^{-2}}{a^2+a^{-2}}=\frac{1-a^{-4}}{1+a^{-4}}

a^2\times\frac{1}{a^2}=1

.

3Substitute and simplify

#13 Convert to Algebra 7.NS.A.2

Put

a^{-4}=4

into the new fraction.

\frac{1-4}{1+4}=\frac{-3}{5}=-\frac{3}{5}

A negative number divided by a positive number is negative.

Answer:

-\frac{3}{5}

4 · Reviewdoes it hold up?

Here $a^{-4}=4>1$ , so in $\frac{1-a^{-4}}{1+a^{-4}}$ the top is negative and the bottom is positive and bigger in size. The answer should be negative and between $-1$ and $0$ . $-\frac{3}{5}$ fits.

Another way: Work with positive powers instead: $a^4=\frac{1}{a^{-4}}=\frac{1}{4}$ . Multiply the top and bottom by $a^2$ : $\frac{a^4-1}{a^4+1}=\frac{\frac{1}{4}-1}{\frac{1}{4}+1}=-\frac{3}{5}$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Using $a^{-n}=\frac{1}{a^n}$, the product rule, and the power rule with negative exponents.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Dividing a negative number by a positive number.

💡Takeaway. If you know one power of

a

, reshape the expression so only that power appears.

Variant 4 medium

Use the rule $a^{-n}=\frac{1}{a^n}$ for every natural number $n$ . If $a^{-2}=3$ , find the value of $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know $a^{-2}=3$ , where $a^{-n}$ means $\frac{1}{a^n}$ . We want the value of the fraction $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Givens

$a^{-n}=\frac{1}{a^n}$ for natural numbers $n$ .
$a^{-2}=3$ .

Unknowns

The value of $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Constraints

$a\neq0$ , so multiplying the top and bottom by a power of $a$ does not change the fraction.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways

We cannot easily find $a$ itself, but we know $a^{-2}$ . Multiply the top and bottom by $a^{-2}$ so that every term becomes either $1$ or $a^{-4}$ , which is a power of $a^{-2}$ .

3 · Execute3 carry out the plan

1Turn the given into $a^{-4}$

#15 Organize Information in More Ways 8.EE.A.1

Square the given value:

a^{-4}=\frac{1}{a^4}=\left(\frac{1}{a^2}\right)^2

.

a^{-4}=(a^{-2})^2=3^2=9

Raising a power to a power multiplies the exponents.

2Multiply top and bottom by $a^{-2}$

#13 Convert to Algebra 8.EE.A.1

This keeps the value of the fraction. Each

a^2\times a^{-2}

becomes

a^0=1

, and each

a^{-2}\times a^{-2}

becomes

a^{-4}

.

\frac{a^2-a^{-2}}{a^2+a^{-2}}=\frac{1-a^{-4}}{1+a^{-4}}

a^2\times\frac{1}{a^2}=1

.

3Substitute and simplify

#13 Convert to Algebra 7.NS.A.2

Put

a^{-4}=9

into the new fraction.

\frac{1-9}{1+9}=\frac{-8}{10}=-\frac{4}{5}

A negative number divided by a positive number is negative.

Answer:

-\frac{4}{5}

4 · Reviewdoes it hold up?

Here $a^{-4}=9>1$ , so in $\frac{1-a^{-4}}{1+a^{-4}}$ the top is negative and the bottom is positive and bigger in size. The answer should be negative and between $-1$ and $0$ . $-\frac{4}{5}$ fits.

Another way: Work with positive powers instead: $a^4=\frac{1}{a^{-4}}=\frac{1}{9}$ . Multiply the top and bottom by $a^2$ : $\frac{a^4-1}{a^4+1}=\frac{\frac{1}{9}-1}{\frac{1}{9}+1}=-\frac{4}{5}$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Using $a^{-n}=\frac{1}{a^n}$, the product rule, and the power rule with negative exponents.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Dividing a negative number by a positive number.

💡Takeaway. If you know one power of

a

, reshape the expression so only that power appears.

Variant 5 easy

Use the rule $a^{-n}=\frac{1}{a^n}$ for every natural number $n$ . If $a^{-2}=3$ , find the value of $\frac{a^3-a^{-3}}{a^3+a^{-3}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know $a^{-2}=3$ , where $a^{-n}$ means $\frac{1}{a^n}$ . We want the value of the fraction $\frac{a^3-a^{-3}}{a^3+a^{-3}}$ .

Givens

$a^{-n}=\frac{1}{a^n}$ for natural numbers $n$ .
$a^{-2}=3$ .

Unknowns

The value of $\frac{a^3-a^{-3}}{a^3+a^{-3}}$ .

Constraints

$a\neq0$ , so multiplying the top and bottom by a power of $a$ does not change the fraction.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways

We cannot easily find $a$ itself, but we know $a^{-2}$ . Multiply the top and bottom by $a^{-3}$ so that every term becomes either $1$ or $a^{-6}$ , which is a power of $a^{-2}$ .

3 · Execute3 carry out the plan

1Turn the given into $a^{-6}$

#15 Organize Information in More Ways 8.EE.A.1

Cube the given value:

a^{-6}=\frac{1}{a^6}=\left(\frac{1}{a^2}\right)^3

.

a^{-6}=(a^{-2})^3=3^3=27

Raising a power to a power multiplies the exponents.

2Multiply top and bottom by $a^{-3}$

#13 Convert to Algebra 8.EE.A.1

This keeps the value of the fraction. Each

a^3\times a^{-3}

becomes

a^0=1

, and each

a^{-3}\times a^{-3}

becomes

a^{-6}

.

\frac{a^3-a^{-3}}{a^3+a^{-3}}=\frac{1-a^{-6}}{1+a^{-6}}

a^3\times\frac{1}{a^3}=1

.

3Substitute and simplify

#13 Convert to Algebra 7.NS.A.2

Put

a^{-6}=27

into the new fraction.

\frac{1-27}{1+27}=\frac{-26}{28}=-\frac{13}{14}

A negative number divided by a positive number is negative.

Answer:

-\frac{13}{14}

4 · Reviewdoes it hold up?

Here $a^{-6}=27>1$ , so in $\frac{1-a^{-6}}{1+a^{-6}}$ the top is negative and the bottom is positive and bigger in size. The answer should be negative and between $-1$ and $0$ . $-\frac{13}{14}$ fits.

Another way: Work with positive powers instead: $a^6=\frac{1}{a^{-6}}=\frac{1}{27}$ . Multiply the top and bottom by $a^3$ : $\frac{a^6-1}{a^6+1}=\frac{\frac{1}{27}-1}{\frac{1}{27}+1}=-\frac{13}{14}$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Using $a^{-n}=\frac{1}{a^n}$, the product rule, and the power rule with negative exponents.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Dividing a negative number by a positive number.

💡Takeaway. If you know one power of

a

, reshape the expression so only that power appears.

Variant 6 medium

Use the rule $a^{-n}=\frac{1}{a^n}$ for every natural number $n$ . If $a^{-2}=2$ , find the value of $\frac{a^3-a^{-3}}{a^3+a^{-3}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know $a^{-2}=2$ , where $a^{-n}$ means $\frac{1}{a^n}$ . We want the value of the fraction $\frac{a^3-a^{-3}}{a^3+a^{-3}}$ .

Givens

$a^{-n}=\frac{1}{a^n}$ for natural numbers $n$ .
$a^{-2}=2$ .

Unknowns

The value of $\frac{a^3-a^{-3}}{a^3+a^{-3}}$ .

Constraints

$a\neq0$ , so multiplying the top and bottom by a power of $a$ does not change the fraction.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways

We cannot easily find $a$ itself, but we know $a^{-2}$ . Multiply the top and bottom by $a^{-3}$ so that every term becomes either $1$ or $a^{-6}$ , which is a power of $a^{-2}$ .

3 · Execute3 carry out the plan

1Turn the given into $a^{-6}$

#15 Organize Information in More Ways 8.EE.A.1

Cube the given value:

a^{-6}=\frac{1}{a^6}=\left(\frac{1}{a^2}\right)^3

.

a^{-6}=(a^{-2})^3=2^3=8

Raising a power to a power multiplies the exponents.

2Multiply top and bottom by $a^{-3}$

#13 Convert to Algebra 8.EE.A.1

This keeps the value of the fraction. Each

a^3\times a^{-3}

becomes

a^0=1

, and each

a^{-3}\times a^{-3}

becomes

a^{-6}

.

\frac{a^3-a^{-3}}{a^3+a^{-3}}=\frac{1-a^{-6}}{1+a^{-6}}

a^3\times\frac{1}{a^3}=1

.

3Substitute and simplify

#13 Convert to Algebra 7.NS.A.2

Put

a^{-6}=8

into the new fraction.

\frac{1-8}{1+8}=\frac{-7}{9}=-\frac{7}{9}

A negative number divided by a positive number is negative.

Answer:

-\frac{7}{9}

4 · Reviewdoes it hold up?

Here $a^{-6}=8>1$ , so in $\frac{1-a^{-6}}{1+a^{-6}}$ the top is negative and the bottom is positive and bigger in size. The answer should be negative and between $-1$ and $0$ . $-\frac{7}{9}$ fits.

Another way: Work with positive powers instead: $a^6=\frac{1}{a^{-6}}=\frac{1}{8}$ . Multiply the top and bottom by $a^3$ : $\frac{a^6-1}{a^6+1}=\frac{\frac{1}{8}-1}{\frac{1}{8}+1}=-\frac{7}{9}$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Using $a^{-n}=\frac{1}{a^n}$, the product rule, and the power rule with negative exponents.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Dividing a negative number by a positive number.

💡Takeaway. If you know one power of

a

, reshape the expression so only that power appears.

Variant 7 medium

Use the rule $a^{-n}=\frac{1}{a^n}$ for every natural number $n$ . If $a^{-4}=2$ , find the value of $\frac{a^4-a^{-4}}{a^4+a^{-4}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know $a^{-4}=2$ , where $a^{-n}$ means $\frac{1}{a^n}$ . We want the value of the fraction $\frac{a^4-a^{-4}}{a^4+a^{-4}}$ .

Givens

$a^{-n}=\frac{1}{a^n}$ for natural numbers $n$ .
$a^{-4}=2$ .

Unknowns

The value of $\frac{a^4-a^{-4}}{a^4+a^{-4}}$ .

Constraints

$a\neq0$ , so multiplying the top and bottom by a power of $a$ does not change the fraction.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways

We cannot easily find $a$ itself, but we know $a^{-4}$ . Multiply the top and bottom by $a^{-4}$ so that every term becomes either $1$ or $a^{-8}$ , which is a power of $a^{-4}$ .

3 · Execute3 carry out the plan

1Turn the given into $a^{-8}$

#15 Organize Information in More Ways 8.EE.A.1

Square the given value:

a^{-8}=\frac{1}{a^8}=\left(\frac{1}{a^4}\right)^2

.

a^{-8}=(a^{-4})^2=2^2=4

Raising a power to a power multiplies the exponents.

2Multiply top and bottom by $a^{-4}$

#13 Convert to Algebra 8.EE.A.1

This keeps the value of the fraction. Each

a^4\times a^{-4}

becomes

a^0=1

, and each

a^{-4}\times a^{-4}

becomes

a^{-8}

.

\frac{a^4-a^{-4}}{a^4+a^{-4}}=\frac{1-a^{-8}}{1+a^{-8}}

a^4\times\frac{1}{a^4}=1

.

3Substitute and simplify

#13 Convert to Algebra 7.NS.A.2

Put

a^{-8}=4

into the new fraction.

\frac{1-4}{1+4}=\frac{-3}{5}=-\frac{3}{5}

A negative number divided by a positive number is negative.

Answer:

-\frac{3}{5}

4 · Reviewdoes it hold up?

Here $a^{-8}=4>1$ , so in $\frac{1-a^{-8}}{1+a^{-8}}$ the top is negative and the bottom is positive and bigger in size. The answer should be negative and between $-1$ and $0$ . $-\frac{3}{5}$ fits.

Another way: Work with positive powers instead: $a^8=\frac{1}{a^{-8}}=\frac{1}{4}$ . Multiply the top and bottom by $a^4$ : $\frac{a^8-1}{a^8+1}=\frac{\frac{1}{4}-1}{\frac{1}{4}+1}=-\frac{3}{5}$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Using $a^{-n}=\frac{1}{a^n}$, the product rule, and the power rule with negative exponents.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Dividing a negative number by a positive number.

💡Takeaway. If you know one power of

a

, reshape the expression so only that power appears.

Variant 8 medium

Use the rule $a^{-n}=\frac{1}{a^n}$ for every natural number $n$ . If $a^{-2}=4$ , find the value of $\frac{a^3-a^{-3}}{a^3+a^{-3}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know $a^{-2}=4$ , where $a^{-n}$ means $\frac{1}{a^n}$ . We want the value of the fraction $\frac{a^3-a^{-3}}{a^3+a^{-3}}$ .

Givens

$a^{-n}=\frac{1}{a^n}$ for natural numbers $n$ .
$a^{-2}=4$ .

Unknowns

The value of $\frac{a^3-a^{-3}}{a^3+a^{-3}}$ .

Constraints

$a\neq0$ , so multiplying the top and bottom by a power of $a$ does not change the fraction.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways

We cannot easily find $a$ itself, but we know $a^{-2}$ . Multiply the top and bottom by $a^{-3}$ so that every term becomes either $1$ or $a^{-6}$ , which is a power of $a^{-2}$ .

3 · Execute3 carry out the plan

1Turn the given into $a^{-6}$

#15 Organize Information in More Ways 8.EE.A.1

Cube the given value:

a^{-6}=\frac{1}{a^6}=\left(\frac{1}{a^2}\right)^3

.

a^{-6}=(a^{-2})^3=4^3=64

Raising a power to a power multiplies the exponents.

2Multiply top and bottom by $a^{-3}$

#13 Convert to Algebra 8.EE.A.1

This keeps the value of the fraction. Each

a^3\times a^{-3}

becomes

a^0=1

, and each

a^{-3}\times a^{-3}

becomes

a^{-6}

.

\frac{a^3-a^{-3}}{a^3+a^{-3}}=\frac{1-a^{-6}}{1+a^{-6}}

a^3\times\frac{1}{a^3}=1

.

3Substitute and simplify

#13 Convert to Algebra 7.NS.A.2

Put

a^{-6}=64

into the new fraction.

\frac{1-64}{1+64}=\frac{-63}{65}=-\frac{63}{65}

A negative number divided by a positive number is negative.

Answer:

-\frac{63}{65}

4 · Reviewdoes it hold up?

Here $a^{-6}=64>1$ , so in $\frac{1-a^{-6}}{1+a^{-6}}$ the top is negative and the bottom is positive and bigger in size. The answer should be negative and between $-1$ and $0$ . $-\frac{63}{65}$ fits.

Another way: Work with positive powers instead: $a^6=\frac{1}{a^{-6}}=\frac{1}{64}$ . Multiply the top and bottom by $a^3$ : $\frac{a^6-1}{a^6+1}=\frac{\frac{1}{64}-1}{\frac{1}{64}+1}=-\frac{63}{65}$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Using $a^{-n}=\frac{1}{a^n}$, the product rule, and the power rule with negative exponents.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Dividing a negative number by a positive number.

💡Takeaway. If you know one power of

a

, reshape the expression so only that power appears.

Variant 9 hard

Use the rule $a^{-n}=\frac{1}{a^n}$ for every natural number $n$ . If $a^{-1}=\frac{1}{2}$ , find the value of $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We know $a^{-1}=\frac{1}{2}$ , where $a^{-n}$ means $\frac{1}{a^n}$ . We want the value of the fraction $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Givens

$a^{-n}=\frac{1}{a^n}$ for natural numbers $n$ .
$a^{-1}=\frac{1}{2}$ .

Unknowns

The value of $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Constraints

$a\neq0$ , so multiplying the top and bottom by a power of $a$ does not change the fraction.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways

We cannot easily find $a$ itself, but we know $a^{-1}$ . Multiply the top and bottom by $a^{-2}$ so that every term becomes either $1$ or $a^{-4}$ , which is a power of $a^{-1}$ .

3 · Execute3 carry out the plan

1Turn the given into $a^{-4}$

#15 Organize Information in More Ways 8.EE.A.1

Raise the given value to the power

4

:

a^{-4}=\frac{1}{a^4}=\left(\frac{1}{a}\right)^4

.

a^{-4}=(a^{-1})^4=\left(\frac{1}{2}\right)^4=\frac{1}{16}

Raising a power to a power multiplies the exponents.

2Multiply top and bottom by $a^{-2}$

#13 Convert to Algebra 8.EE.A.1

This keeps the value of the fraction. Each

a^2\times a^{-2}

becomes

a^0=1

, and each

a^{-2}\times a^{-2}

becomes

a^{-4}

.

\frac{a^2-a^{-2}}{a^2+a^{-2}}=\frac{1-a^{-4}}{1+a^{-4}}

a^2\times\frac{1}{a^2}=1

.

3Substitute and simplify

#13 Convert to Algebra 7.NS.A.2

Put

a^{-4}=\frac{1}{16}

into the new fraction.

\frac{1-\frac{1}{16}}{1+\frac{1}{16}}=\frac{\frac{15}{16}}{\frac{17}{16}}=\frac{15}{17}

A positive number divided by a positive number is positive.

Answer:

\frac{15}{17}

4 · Reviewdoes it hold up?

Here $a^{-4}=\frac{1}{16}<1$ , so in $\frac{1-a^{-4}}{1+a^{-4}}$ both the top and the bottom are positive and the top is smaller. The answer should be between $0$ and $1$ . $\frac{15}{17}$ fits.

Another way: Work with positive powers instead: $a^4=\frac{1}{a^{-4}}=16$ . Multiply the top and bottom by $a^2$ : $\frac{a^4-1}{a^4+1}=\frac{16-1}{16+1}=\frac{15}{17}$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Using $a^{-n}=\frac{1}{a^n}$, the product rule, and the power rule with negative exponents.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Dividing one positive fraction by another.

💡Takeaway. If you know one power of

a

, reshape the expression so only that power appears.

Variant 10 hard

Use the rule $a^{-n}=\frac{1}{a^n}$ for every natural number $n$ . If $a^{-2}=5$ , find the value of $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

We know $a^{-2}=5$ , where $a^{-n}$ means $\frac{1}{a^n}$ . We want the value of the fraction $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Givens

$a^{-n}=\frac{1}{a^n}$ for natural numbers $n$ .
$a^{-2}=5$ .

Unknowns

The value of $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Constraints

$a\neq0$ , so multiplying the top and bottom by a power of $a$ does not change the fraction.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways

We cannot easily find $a$ itself, but we know $a^{-2}$ . Multiply the top and bottom by $a^{-2}$ so that every term becomes either $1$ or $a^{-4}$ , which is a power of $a^{-2}$ .

3 · Execute3 carry out the plan

1Turn the given into $a^{-4}$

#15 Organize Information in More Ways 8.EE.A.1

Square the given value:

a^{-4}=\frac{1}{a^4}=\left(\frac{1}{a^2}\right)^2

.

a^{-4}=(a^{-2})^2=5^2=25

Raising a power to a power multiplies the exponents.

2Multiply top and bottom by $a^{-2}$

#13 Convert to Algebra 8.EE.A.1

This keeps the value of the fraction. Each

a^2\times a^{-2}

becomes

a^0=1

, and each

a^{-2}\times a^{-2}

becomes

a^{-4}

.

\frac{a^2-a^{-2}}{a^2+a^{-2}}=\frac{1-a^{-4}}{1+a^{-4}}

a^2\times\frac{1}{a^2}=1

.

3Substitute and simplify

#13 Convert to Algebra 7.NS.A.2

Put

a^{-4}=25

into the new fraction.

\frac{1-25}{1+25}=\frac{-24}{26}=-\frac{12}{13}

A negative number divided by a positive number is negative.

Answer:

-\frac{12}{13}

4 · Reviewdoes it hold up?

Here $a^{-4}=25>1$ , so in $\frac{1-a^{-4}}{1+a^{-4}}$ the top is negative and the bottom is positive and bigger in size. The answer should be negative and between $-1$ and $0$ . $-\frac{12}{13}$ fits.

Another way: Work with positive powers instead: $a^4=\frac{1}{a^{-4}}=\frac{1}{25}$ . Multiply the top and bottom by $a^2$ : $\frac{a^4-1}{a^4+1}=\frac{\frac{1}{25}-1}{\frac{1}{25}+1}=-\frac{12}{13}$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Using $a^{-n}=\frac{1}{a^n}$, the product rule, and the power rule with negative exponents.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Dividing a negative number by a positive number.

💡Takeaway. If you know one power of

a

, reshape the expression so only that power appears.

Variant 11 hard

Use the rule $a^{-n}=\frac{1}{a^n}$ for every natural number $n$ . If $a^{-2}=\frac{1}{3}$ , find the value of $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We know $a^{-2}=\frac{1}{3}$ , where $a^{-n}$ means $\frac{1}{a^n}$ . We want the value of the fraction $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Givens

$a^{-n}=\frac{1}{a^n}$ for natural numbers $n$ .
$a^{-2}=\frac{1}{3}$ .

Unknowns

The value of $\frac{a^2-a^{-2}}{a^2+a^{-2}}$ .

Constraints

$a\neq0$ , so multiplying the top and bottom by a power of $a$ does not change the fraction.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways

We cannot easily find $a$ itself, but we know $a^{-2}$ . Multiply the top and bottom by $a^{-2}$ so that every term becomes either $1$ or $a^{-4}$ , which is a power of $a^{-2}$ .

3 · Execute3 carry out the plan

1Turn the given into $a^{-4}$

#15 Organize Information in More Ways 8.EE.A.1

Square the given value:

a^{-4}=\frac{1}{a^4}=\left(\frac{1}{a^2}\right)^2

.

a^{-4}=(a^{-2})^2=\left(\frac{1}{3}\right)^2=\frac{1}{9}

Raising a power to a power multiplies the exponents.

2Multiply top and bottom by $a^{-2}$

#13 Convert to Algebra 8.EE.A.1

This keeps the value of the fraction. Each

a^2\times a^{-2}

becomes

a^0=1

, and each

a^{-2}\times a^{-2}

becomes

a^{-4}

.

\frac{a^2-a^{-2}}{a^2+a^{-2}}=\frac{1-a^{-4}}{1+a^{-4}}

a^2\times\frac{1}{a^2}=1

.

3Substitute and simplify

#13 Convert to Algebra 7.NS.A.2

Put

a^{-4}=\frac{1}{9}

into the new fraction.

\frac{1-\frac{1}{9}}{1+\frac{1}{9}}=\frac{\frac{8}{9}}{\frac{10}{9}}=\frac{4}{5}

A positive number divided by a positive number is positive.

Answer:

\frac{4}{5}

4 · Reviewdoes it hold up?

Here $a^{-4}=\frac{1}{9}<1$ , so in $\frac{1-a^{-4}}{1+a^{-4}}$ both the top and the bottom are positive and the top is smaller. The answer should be between $0$ and $1$ . $\frac{4}{5}$ fits.

Another way: Work with positive powers instead: $a^4=\frac{1}{a^{-4}}=9$ . Multiply the top and bottom by $a^2$ : $\frac{a^4-1}{a^4+1}=\frac{9-1}{9+1}=\frac{4}{5}$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Using $a^{-n}=\frac{1}{a^n}$, the product rule, and the power rule with negative exponents.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Dividing one positive fraction by another.

💡Takeaway. If you know one power of

a

, reshape the expression so only that power appears.

Variant 12 hard

Use the rule $a^{-n}=\frac{1}{a^n}$ for every natural number $n$ . If $a^{-2}=\frac{1}{2}$ , find the value of $\frac{a^3-a^{-3}}{a^3+a^{-3}}$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We know $a^{-2}=\frac{1}{2}$ , where $a^{-n}$ means $\frac{1}{a^n}$ . We want the value of the fraction $\frac{a^3-a^{-3}}{a^3+a^{-3}}$ .

Givens

$a^{-n}=\frac{1}{a^n}$ for natural numbers $n$ .
$a^{-2}=\frac{1}{2}$ .

Unknowns

The value of $\frac{a^3-a^{-3}}{a^3+a^{-3}}$ .

Constraints

$a\neq0$ , so multiplying the top and bottom by a power of $a$ does not change the fraction.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #15 Organize Information in More Ways

We cannot easily find $a$ itself, but we know $a^{-2}$ . Multiply the top and bottom by $a^{-3}$ so that every term becomes either $1$ or $a^{-6}$ , which is a power of $a^{-2}$ .

3 · Execute3 carry out the plan

1Turn the given into $a^{-6}$

#15 Organize Information in More Ways 8.EE.A.1

Cube the given value:

a^{-6}=\frac{1}{a^6}=\left(\frac{1}{a^2}\right)^3

.

a^{-6}=(a^{-2})^3=\left(\frac{1}{2}\right)^3=\frac{1}{8}

Raising a power to a power multiplies the exponents.

2Multiply top and bottom by $a^{-3}$

#13 Convert to Algebra 8.EE.A.1

This keeps the value of the fraction. Each

a^3\times a^{-3}

becomes

a^0=1

, and each

a^{-3}\times a^{-3}

becomes

a^{-6}

.

\frac{a^3-a^{-3}}{a^3+a^{-3}}=\frac{1-a^{-6}}{1+a^{-6}}

a^3\times\frac{1}{a^3}=1

.

3Substitute and simplify

#13 Convert to Algebra 7.NS.A.2

Put

a^{-6}=\frac{1}{8}

into the new fraction.

\frac{1-\frac{1}{8}}{1+\frac{1}{8}}=\frac{\frac{7}{8}}{\frac{9}{8}}=\frac{7}{9}

A positive number divided by a positive number is positive.

Answer:

\frac{7}{9}

4 · Reviewdoes it hold up?

Here $a^{-6}=\frac{1}{8}<1$ , so in $\frac{1-a^{-6}}{1+a^{-6}}$ both the top and the bottom are positive and the top is smaller. The answer should be between $0$ and $1$ . $\frac{7}{9}$ fits.

Another way: Work with positive powers instead: $a^6=\frac{1}{a^{-6}}=8$ . Multiply the top and bottom by $a^3$ : $\frac{a^6-1}{a^6+1}=\frac{8-1}{8+1}=\frac{7}{9}$ .

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Using $a^{-n}=\frac{1}{a^n}$, the product rule, and the power rule with negative exponents.
7.NS.A.2 Apply and extend understanding of multiplication and division of rational numbers — Dividing one positive fraction by another.

💡Takeaway. If you know one power of

a

, reshape the expression so only that power appears.

Multiply top and bottom to use the given power · 12 practice problems

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