Practice · Rewrite in base 2 and factor the sum · Sensim Math

Variant 1 easy

When $\left(\frac{4^6+8^3}{8^6+8^5}\right)^3$ is simplified, it equals $\left(\frac{1}{a}\right)^b$ , where $a$ is a prime number and $b$ is a natural number. Find the value of $b^a$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Simplify $\left(\frac{4^6+8^3}{8^6+8^5}\right)^3$ into the form $\left(\frac{1}{a}\right)^b$ with $a$ prime and $b$ a natural number, then compute $b^a$ .

Givens

$\left(\frac{4^6+8^3}{8^6+8^5}\right)^3=\left(\frac{1}{a}\right)^b$
$a$ is prime; $b$ is a natural number.

Unknowns

The value of $b^a$ .

Constraints

Exponent laws do not split sums, so the sums must be handled by factoring, not term by term.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#13 Convert to Algebra

$4$ and $8$ are both powers of $2$ , so write everything in base $2$ . The denominator then turns out to be a power of $2$ times the numerator, and the fraction collapses.

3 · Execute5 carry out the plan

1Write every power in base $2$

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

Use

4=2^2

and

8=2^3

with the power-of-a-power rule.

4^6=2^{12},\ 8^3=2^9,\ 8^6=2^{18},\ 8^5=2^{15}

(2^2)^6=2^{2\times6}=2^{12}

, and so on.

2Factor the denominator

#7 Identify Subproblems 6.EE.A.3

Each denominator term is

2^6

times a numerator term.

2^{18}+2^{15}=2^6\times2^{12}+2^6\times2^9=2^6(2^{12}+2^9)

This is the distributive property run backwards.

3Cancel the common sum

#7 Identify Subproblems 8.EE.A.1

The same sum

2^{12}+2^9

appears on top and bottom.

\frac{2^{12}+2^9}{2^6(2^{12}+2^9)}=\frac{1}{2^6}

Whatever the sum is, it divides out.

4Cube the result

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

Cube

\frac{1}{2^6}

.

\left(\frac{1}{2^6}\right)^3=\frac{1}{2^{18}}=\left(\frac{1}{2}\right)^{18}

Cubing triples the exponent.

5Read off $a$ and $b$

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

\left(\frac{1}{a}\right)^b=\left(\frac{1}{2}\right)^{18}

means

a^b=2^{18}

; since

a

is prime it must be

2

, so

b=18

.

b^a=18^2=324

Only the prime

2

divides

2^{18}

.

Answer:

324

4 · Reviewdoes it hold up?

Check numerically: $4^6+8^3=4096+512=4608$ and $8^6+8^5=262144+32768=294912$ . Then $294912\div4608=64=2^6$ , so the fraction really is $\frac{1}{2^6}$ .

Another way: Factor both parts by $2^9$ : the top is $2^9(2^3+1)$ and the bottom is $2^{15}(2^3+1)$ , so the fraction is $\frac{2^9}{2^{15}}=\frac{1}{2^6}$ and the rest is the same.

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Rewriting in base $2$, cancelling, and cubing powers.
6.EE.A.3 Apply properties to generate equivalent expressions — Factoring a common power out of a sum.

💡Takeaway. Exponent rules don't work on sums -- so factor the sum first, then the rules take over.

Variant 2 easy

When $\left(\frac{8^4+4^4}{8^6+4^7}\right)^2$ is simplified, it equals $\left(\frac{1}{a}\right)^b$ , where $a$ is a prime number and $b$ is a natural number. Find the value of $b^a$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Simplify $\left(\frac{8^4+4^4}{8^6+4^7}\right)^2$ into the form $\left(\frac{1}{a}\right)^b$ with $a$ prime and $b$ a natural number, then compute $b^a$ .

Givens

$\left(\frac{8^4+4^4}{8^6+4^7}\right)^2=\left(\frac{1}{a}\right)^b$
$a$ is prime; $b$ is a natural number.

Unknowns

The value of $b^a$ .

Constraints

Exponent laws do not split sums, so the sums must be handled by factoring, not term by term.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#13 Convert to Algebra

$8$ and $4$ are both powers of $2$ , so write everything in base $2$ . The denominator then turns out to be a power of $2$ times the numerator, and the fraction collapses.

3 · Execute5 carry out the plan

1Write every power in base $2$

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

Use

8=2^3

and

4=2^2

with the power-of-a-power rule.

8^4=2^{12},\ 4^4=2^8,\ 8^6=2^{18},\ 4^7=2^{14}

(2^3)^4=2^{3\times4}=2^{12}

, and so on.

2Factor the denominator

#7 Identify Subproblems 6.EE.A.3

Each denominator term is

2^6

times a numerator term.

2^{18}+2^{14}=2^6\times2^{12}+2^6\times2^8=2^6(2^{12}+2^8)

This is the distributive property run backwards.

3Cancel the common sum

#7 Identify Subproblems 8.EE.A.1

The same sum

2^{12}+2^8

appears on top and bottom.

\frac{2^{12}+2^8}{2^6(2^{12}+2^8)}=\frac{1}{2^6}

Whatever the sum is, it divides out.

4Square the result

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

Square

\frac{1}{2^6}

.

\left(\frac{1}{2^6}\right)^2=\frac{1}{2^{12}}=\left(\frac{1}{2}\right)^{12}

Squaring doubles the exponent.

5Read off $a$ and $b$

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

\left(\frac{1}{a}\right)^b=\left(\frac{1}{2}\right)^{12}

means

a^b=2^{12}

; since

a

is prime it must be

2

, so

b=12

.

b^a=12^2=144

Only the prime

2

divides

2^{12}

.

Answer:

144

4 · Reviewdoes it hold up?

Check numerically: $8^4+4^4=4096+256=4352$ and $8^6+4^7=262144+16384=278528$ . Then $278528\div4352=64=2^6$ , so the fraction really is $\frac{1}{2^6}$ .

Another way: Factor both parts by $2^8$ : the top is $2^8(2^4+1)$ and the bottom is $2^{14}(2^4+1)$ , so the fraction is $\frac{2^8}{2^{14}}=\frac{1}{2^6}$ and the rest is the same.

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Rewriting in base $2$, cancelling, and squaring powers.
6.EE.A.3 Apply properties to generate equivalent expressions — Factoring a common power out of a sum.

💡Takeaway. Exponent rules don't work on sums -- so factor the sum first, then the rules take over.

Variant 3 easy

When $\left(\frac{9^3+3^3}{9^4+3^5}\right)^3$ is simplified, it equals $\left(\frac{1}{a}\right)^b$ , where $a$ is a prime number and $b$ is a natural number. Find the value of $b^a$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Simplify $\left(\frac{9^3+3^3}{9^4+3^5}\right)^3$ into the form $\left(\frac{1}{a}\right)^b$ with $a$ prime and $b$ a natural number, then compute $b^a$ .

Givens

$\left(\frac{9^3+3^3}{9^4+3^5}\right)^3=\left(\frac{1}{a}\right)^b$
$a$ is prime; $b$ is a natural number.

Unknowns

The value of $b^a$ .

Constraints

Exponent laws do not split sums, so the sums must be handled by factoring, not term by term.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#13 Convert to Algebra

$9$ is a power of $3$ , so write everything in base $3$ . The denominator then turns out to be a power of $3$ times the numerator, and the fraction collapses.

3 · Execute5 carry out the plan

1Write every power in base $3$

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

Use

9=3^2

with the power-of-a-power rule.

9^3=3^6,\ 9^4=3^8

(3^2)^3=3^{2\times3}=3^6

, and so on.

2Factor the denominator

#7 Identify Subproblems 6.EE.A.3

Each denominator term is

3^2

times a numerator term.

3^8+3^5=3^2\times3^6+3^2\times3^3=3^2(3^6+3^3)

This is the distributive property run backwards.

3Cancel the common sum

#7 Identify Subproblems 8.EE.A.1

The same sum

3^6+3^3

appears on top and bottom.

\frac{3^6+3^3}{3^2(3^6+3^3)}=\frac{1}{3^2}

Whatever the sum is, it divides out.

4Cube the result

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

Cube

\frac{1}{3^2}

.

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

Cubing triples the exponent.

5Read off $a$ and $b$

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

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

means

a^b=3^6

; since

a

is prime it must be

3

, so

b=6

.

b^a=6^3=216

Only the prime

3

divides

3^6

.

Answer:

216

4 · Reviewdoes it hold up?

Check numerically: $9^3+3^3=729+27=756$ and $9^4+3^5=6561+243=6804$ . Then $6804\div756=9=3^2$ , so the fraction really is $\frac{1}{3^2}$ .

Another way: Factor both parts by $3^3$ : the top is $3^3(3^3+1)$ and the bottom is $3^5(3^3+1)$ , so the fraction is $\frac{3^3}{3^5}=\frac{1}{3^2}$ and the rest is the same.

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Rewriting in base $3$, cancelling, and cubing powers.
6.EE.A.3 Apply properties to generate equivalent expressions — Factoring a common power out of a sum.

💡Takeaway. Exponent rules don't work on sums -- so factor the sum first, then the rules take over.

Variant 4 easy

When $\left(\frac{4^5+8^2}{8^5+2^{11}}\right)^2$ is simplified, it equals $\left(\frac{1}{a}\right)^b$ , where $a$ is a prime number and $b$ is a natural number. Find the value of $b^a$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Simplify $\left(\frac{4^5+8^2}{8^5+2^{11}}\right)^2$ into the form $\left(\frac{1}{a}\right)^b$ with $a$ prime and $b$ a natural number, then compute $b^a$ .

Givens

$\left(\frac{4^5+8^2}{8^5+2^{11}}\right)^2=\left(\frac{1}{a}\right)^b$
$a$ is prime; $b$ is a natural number.

Unknowns

The value of $b^a$ .

Constraints

Exponent laws do not split sums, so the sums must be handled by factoring, not term by term.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#13 Convert to Algebra

$4$ and $8$ are both powers of $2$ , so write everything in base $2$ . The denominator then turns out to be a power of $2$ times the numerator, and the fraction collapses.

3 · Execute5 carry out the plan

1Write every power in base $2$

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

Use

4=2^2

and

8=2^3

with the power-of-a-power rule.

4^5=2^{10},\ 8^2=2^6,\ 8^5=2^{15}

(2^2)^5=2^{2\times5}=2^{10}

, and so on.

2Factor the denominator

#7 Identify Subproblems 6.EE.A.3

Each denominator term is

2^5

times a numerator term.

2^{15}+2^{11}=2^5\times2^{10}+2^5\times2^6=2^5(2^{10}+2^6)

This is the distributive property run backwards.

3Cancel the common sum

#7 Identify Subproblems 8.EE.A.1

The same sum

2^{10}+2^6

appears on top and bottom.

\frac{2^{10}+2^6}{2^5(2^{10}+2^6)}=\frac{1}{2^5}

Whatever the sum is, it divides out.

4Square the result

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

Square

\frac{1}{2^5}

.

\left(\frac{1}{2^5}\right)^2=\frac{1}{2^{10}}=\left(\frac{1}{2}\right)^{10}

Squaring doubles the exponent.

5Read off $a$ and $b$

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

\left(\frac{1}{a}\right)^b=\left(\frac{1}{2}\right)^{10}

means

a^b=2^{10}

; since

a

is prime it must be

2

, so

b=10

.

b^a=10^2=100

Only the prime

2

divides

2^{10}

.

Answer:

100

4 · Reviewdoes it hold up?

Check numerically: $4^5+8^2=1024+64=1088$ and $8^5+2^{11}=32768+2048=34816$ . Then $34816\div1088=32=2^5$ , so the fraction really is $\frac{1}{2^5}$ .

Another way: Factor both parts by $2^6$ : the top is $2^6(2^4+1)$ and the bottom is $2^{11}(2^4+1)$ , so the fraction is $\frac{2^6}{2^{11}}=\frac{1}{2^5}$ and the rest is the same.

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Rewriting in base $2$, cancelling, and squaring powers.
6.EE.A.3 Apply properties to generate equivalent expressions — Factoring a common power out of a sum.

💡Takeaway. Exponent rules don't work on sums -- so factor the sum first, then the rules take over.

Variant 5 medium

When $\left(\frac{8^5+4^3}{4^{11}+2^{13}}\right)^2$ is simplified, it equals $\left(\frac{1}{a}\right)^b$ , where $a$ is a prime number and $b$ is a natural number. Find the value of $b^a$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Simplify $\left(\frac{8^5+4^3}{4^{11}+2^{13}}\right)^2$ into the form $\left(\frac{1}{a}\right)^b$ with $a$ prime and $b$ a natural number, then compute $b^a$ .

Givens

$\left(\frac{8^5+4^3}{4^{11}+2^{13}}\right)^2=\left(\frac{1}{a}\right)^b$
$a$ is prime; $b$ is a natural number.

Unknowns

The value of $b^a$ .

Constraints

Exponent laws do not split sums, so the sums must be handled by factoring, not term by term.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#13 Convert to Algebra

$8$ and $4$ are both powers of $2$ , so write everything in base $2$ . The denominator then turns out to be a power of $2$ times the numerator, and the fraction collapses.

3 · Execute5 carry out the plan

1Write every power in base $2$

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

Use

8=2^3

and

4=2^2

with the power-of-a-power rule.

8^5=2^{15},\ 4^3=2^6,\ 4^{11}=2^{22}

(2^3)^5=2^{3\times5}=2^{15}

, and so on.

2Factor the denominator

#7 Identify Subproblems 6.EE.A.3

Each denominator term is

2^7

times a numerator term.

2^{22}+2^{13}=2^7\times2^{15}+2^7\times2^6=2^7(2^{15}+2^6)

This is the distributive property run backwards.

3Cancel the common sum

#7 Identify Subproblems 8.EE.A.1

The same sum

2^{15}+2^6

appears on top and bottom.

\frac{2^{15}+2^6}{2^7(2^{15}+2^6)}=\frac{1}{2^7}

Whatever the sum is, it divides out.

4Square the result

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

Square

\frac{1}{2^7}

.

\left(\frac{1}{2^7}\right)^2=\frac{1}{2^{14}}=\left(\frac{1}{2}\right)^{14}

Squaring doubles the exponent.

5Read off $a$ and $b$

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

\left(\frac{1}{a}\right)^b=\left(\frac{1}{2}\right)^{14}

means

a^b=2^{14}

; since

a

is prime it must be

2

, so

b=14

.

b^a=14^2=196

Only the prime

2

divides

2^{14}

.

Answer:

196

4 · Reviewdoes it hold up?

Check numerically: $8^5+4^3=32768+64=32832$ and $4^{11}+2^{13}=4194304+8192=4202496$ . Then $4202496\div32832=128=2^7$ , so the fraction really is $\frac{1}{2^7}$ .

Another way: Factor both parts by $2^6$ : the top is $2^6(2^9+1)$ and the bottom is $2^{13}(2^9+1)$ , so the fraction is $\frac{2^6}{2^{13}}=\frac{1}{2^7}$ and the rest is the same.

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Rewriting in base $2$, cancelling, and squaring powers.
6.EE.A.3 Apply properties to generate equivalent expressions — Factoring a common power out of a sum.

💡Takeaway. Exponent rules don't work on sums -- so factor the sum first, then the rules take over.

Variant 6 medium

When $\left(\frac{16^2+4^2}{2^{13}+8^3}\right)^3$ is simplified, it equals $\left(\frac{1}{a}\right)^b$ , where $a$ is a prime number and $b$ is a natural number. Find the value of $b^a$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Simplify $\left(\frac{16^2+4^2}{2^{13}+8^3}\right)^3$ into the form $\left(\frac{1}{a}\right)^b$ with $a$ prime and $b$ a natural number, then compute $b^a$ .

Givens

$\left(\frac{16^2+4^2}{2^{13}+8^3}\right)^3=\left(\frac{1}{a}\right)^b$
$a$ is prime; $b$ is a natural number.

Unknowns

The value of $b^a$ .

Constraints

Exponent laws do not split sums, so the sums must be handled by factoring, not term by term.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#13 Convert to Algebra

$16$ and $4$ and $8$ are all powers of $2$ , so write everything in base $2$ . The denominator then turns out to be a power of $2$ times the numerator, and the fraction collapses.

3 · Execute5 carry out the plan

1Write every power in base $2$

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

Use

16=2^4

and

4=2^2

and

8=2^3

with the power-of-a-power rule.

16^2=2^8,\ 4^2=2^4,\ 8^3=2^9

(2^4)^2=2^{4\times2}=2^8

, and so on.

2Factor the denominator

#7 Identify Subproblems 6.EE.A.3

Each denominator term is

2^5

times a numerator term.

2^{13}+2^9=2^5\times2^8+2^5\times2^4=2^5(2^8+2^4)

This is the distributive property run backwards.

3Cancel the common sum

#7 Identify Subproblems 8.EE.A.1

The same sum

2^8+2^4

appears on top and bottom.

\frac{2^8+2^4}{2^5(2^8+2^4)}=\frac{1}{2^5}

Whatever the sum is, it divides out.

4Cube the result

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

Cube

\frac{1}{2^5}

.

\left(\frac{1}{2^5}\right)^3=\frac{1}{2^{15}}=\left(\frac{1}{2}\right)^{15}

Cubing triples the exponent.

5Read off $a$ and $b$

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

\left(\frac{1}{a}\right)^b=\left(\frac{1}{2}\right)^{15}

means

a^b=2^{15}

; since

a

is prime it must be

2

, so

b=15

.

b^a=15^2=225

Only the prime

2

divides

2^{15}

.

Answer:

225

4 · Reviewdoes it hold up?

Check numerically: $16^2+4^2=256+16=272$ and $2^{13}+8^3=8192+512=8704$ . Then $8704\div272=32=2^5$ , so the fraction really is $\frac{1}{2^5}$ .

Another way: Factor both parts by $2^4$ : the top is $2^4(2^4+1)$ and the bottom is $2^9(2^4+1)$ , so the fraction is $\frac{2^4}{2^9}=\frac{1}{2^5}$ and the rest is the same.

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Rewriting in base $2$, cancelling, and cubing powers.
6.EE.A.3 Apply properties to generate equivalent expressions — Factoring a common power out of a sum.

💡Takeaway. Exponent rules don't work on sums -- so factor the sum first, then the rules take over.

Variant 7 medium

When $\left(\frac{8^4+2^5}{16^4+8^3}\right)^2$ is simplified, it equals $\left(\frac{1}{a}\right)^b$ , where $a$ is a prime number and $b$ is a natural number. Find the value of $b^a$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Simplify $\left(\frac{8^4+2^5}{16^4+8^3}\right)^2$ into the form $\left(\frac{1}{a}\right)^b$ with $a$ prime and $b$ a natural number, then compute $b^a$ .

Givens

$\left(\frac{8^4+2^5}{16^4+8^3}\right)^2=\left(\frac{1}{a}\right)^b$
$a$ is prime; $b$ is a natural number.

Unknowns

The value of $b^a$ .

Constraints

Exponent laws do not split sums, so the sums must be handled by factoring, not term by term.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#13 Convert to Algebra

$8$ and $16$ are both powers of $2$ , so write everything in base $2$ . The denominator then turns out to be a power of $2$ times the numerator, and the fraction collapses.

3 · Execute5 carry out the plan

1Write every power in base $2$

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

Use

8=2^3

and

16=2^4

with the power-of-a-power rule.

8^4=2^{12},\ 16^4=2^{16},\ 8^3=2^9

(2^3)^4=2^{3\times4}=2^{12}

, and so on.

2Factor the denominator

#7 Identify Subproblems 6.EE.A.3

Each denominator term is

2^4

times a numerator term.

2^{16}+2^9=2^4\times2^{12}+2^4\times2^5=2^4(2^{12}+2^5)

This is the distributive property run backwards.

3Cancel the common sum

#7 Identify Subproblems 8.EE.A.1

The same sum

2^{12}+2^5

appears on top and bottom.

\frac{2^{12}+2^5}{2^4(2^{12}+2^5)}=\frac{1}{2^4}

Whatever the sum is, it divides out.

4Square the result

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

Square

\frac{1}{2^4}

.

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

Squaring doubles the exponent.

5Read off $a$ and $b$

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

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

means

a^b=2^8

; since

a

is prime it must be

2

, so

b=8

.

b^a=8^2=64

Only the prime

2

divides

2^8

.

Answer:

64

4 · Reviewdoes it hold up?

Check numerically: $8^4+2^5=4096+32=4128$ and $16^4+8^3=65536+512=66048$ . Then $66048\div4128=16=2^4$ , so the fraction really is $\frac{1}{2^4}$ .

Another way: Factor both parts by $2^5$ : the top is $2^5(2^7+1)$ and the bottom is $2^9(2^7+1)$ , so the fraction is $\frac{2^5}{2^9}=\frac{1}{2^4}$ and the rest is the same.

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Rewriting in base $2$, cancelling, and squaring powers.
6.EE.A.3 Apply properties to generate equivalent expressions — Factoring a common power out of a sum.

💡Takeaway. Exponent rules don't work on sums -- so factor the sum first, then the rules take over.

Variant 8 medium

When $\left(\frac{16^3+8^2}{16^4+4^5}\right)^2$ is simplified, it equals $\left(\frac{1}{a}\right)^b$ , where $a$ is a prime number and $b$ is a natural number. Find the value of $b^a$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Simplify $\left(\frac{16^3+8^2}{16^4+4^5}\right)^2$ into the form $\left(\frac{1}{a}\right)^b$ with $a$ prime and $b$ a natural number, then compute $b^a$ .

Givens

$\left(\frac{16^3+8^2}{16^4+4^5}\right)^2=\left(\frac{1}{a}\right)^b$
$a$ is prime; $b$ is a natural number.

Unknowns

The value of $b^a$ .

Constraints

Exponent laws do not split sums, so the sums must be handled by factoring, not term by term.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#13 Convert to Algebra

$16$ and $8$ and $4$ are all powers of $2$ , so write everything in base $2$ . The denominator then turns out to be a power of $2$ times the numerator, and the fraction collapses.

3 · Execute5 carry out the plan

1Write every power in base $2$

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

Use

16=2^4

and

8=2^3

and

4=2^2

with the power-of-a-power rule.

16^3=2^{12},\ 8^2=2^6,\ 16^4=2^{16},\ 4^5=2^{10}

(2^4)^3=2^{4\times3}=2^{12}

, and so on.

2Factor the denominator

#7 Identify Subproblems 6.EE.A.3

Each denominator term is

2^4

times a numerator term.

2^{16}+2^{10}=2^4\times2^{12}+2^4\times2^6=2^4(2^{12}+2^6)

This is the distributive property run backwards.

3Cancel the common sum

#7 Identify Subproblems 8.EE.A.1

The same sum

2^{12}+2^6

appears on top and bottom.

\frac{2^{12}+2^6}{2^4(2^{12}+2^6)}=\frac{1}{2^4}

Whatever the sum is, it divides out.

4Square the result

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

Square

\frac{1}{2^4}

.

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

Squaring doubles the exponent.

5Read off $a$ and $b$

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

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

means

a^b=2^8

; since

a

is prime it must be

2

, so

b=8

.

b^a=8^2=64

Only the prime

2

divides

2^8

.

Answer:

64

4 · Reviewdoes it hold up?

Check numerically: $16^3+8^2=4096+64=4160$ and $16^4+4^5=65536+1024=66560$ . Then $66560\div4160=16=2^4$ , so the fraction really is $\frac{1}{2^4}$ .

Another way: Factor both parts by $2^6$ : the top is $2^6(2^6+1)$ and the bottom is $2^{10}(2^6+1)$ , so the fraction is $\frac{2^6}{2^{10}}=\frac{1}{2^4}$ and the rest is the same.

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Rewriting in base $2$, cancelling, and squaring powers.
6.EE.A.3 Apply properties to generate equivalent expressions — Factoring a common power out of a sum.

💡Takeaway. Exponent rules don't work on sums -- so factor the sum first, then the rules take over.

Variant 9 hard

When $\left(\frac{4^5+16^2}{2^{13}+2^{11}}\right)^2$ is simplified, it equals $\left(\frac{1}{a}\right)^b$ , where $a$ is a prime number and $b$ is a natural number. Find the value of $b^a$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Simplify $\left(\frac{4^5+16^2}{2^{13}+2^{11}}\right)^2$ into the form $\left(\frac{1}{a}\right)^b$ with $a$ prime and $b$ a natural number, then compute $b^a$ .

Givens

$\left(\frac{4^5+16^2}{2^{13}+2^{11}}\right)^2=\left(\frac{1}{a}\right)^b$
$a$ is prime; $b$ is a natural number.

Unknowns

The value of $b^a$ .

Constraints

Exponent laws do not split sums, so the sums must be handled by factoring, not term by term.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#13 Convert to Algebra

$4$ and $16$ are both powers of $2$ , so write everything in base $2$ . The denominator then turns out to be a power of $2$ times the numerator, and the fraction collapses.

3 · Execute5 carry out the plan

1Write every power in base $2$

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

Use

4=2^2

and

16=2^4

with the power-of-a-power rule.

4^5=2^{10},\ 16^2=2^8

(2^2)^5=2^{2\times5}=2^{10}

, and so on.

2Factor the denominator

#7 Identify Subproblems 6.EE.A.3

Each denominator term is

2^3

times a numerator term.

2^{13}+2^{11}=2^3\times2^{10}+2^3\times2^8=2^3(2^{10}+2^8)

This is the distributive property run backwards.

3Cancel the common sum

#7 Identify Subproblems 8.EE.A.1

The same sum

2^{10}+2^8

appears on top and bottom.

\frac{2^{10}+2^8}{2^3(2^{10}+2^8)}=\frac{1}{2^3}

Whatever the sum is, it divides out.

4Square the result

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

Square

\frac{1}{2^3}

.

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

Squaring doubles the exponent.

5Read off $a$ and $b$

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

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

means

a^b=2^6

; since

a

is prime it must be

2

, so

b=6

.

b^a=6^2=36

Only the prime

2

divides

2^6

.

Answer:

36

4 · Reviewdoes it hold up?

Check numerically: $4^5+16^2=1024+256=1280$ and $2^{13}+2^{11}=8192+2048=10240$ . Then $10240\div1280=8=2^3$ , so the fraction really is $\frac{1}{2^3}$ .

Another way: Factor both parts by $2^8$ : the top is $2^8(2^2+1)$ and the bottom is $2^{11}(2^2+1)$ , so the fraction is $\frac{2^8}{2^{11}}=\frac{1}{2^3}$ and the rest is the same.

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Rewriting in base $2$, cancelling, and squaring powers.
6.EE.A.3 Apply properties to generate equivalent expressions — Factoring a common power out of a sum.

💡Takeaway. Exponent rules don't work on sums -- so factor the sum first, then the rules take over.

Variant 10 hard

When $\left(\frac{9^4+27^2}{27^4+9^5}\right)^2$ is simplified, it equals $\left(\frac{1}{a}\right)^b$ , where $a$ is a prime number and $b$ is a natural number. Find the value of $b^a$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Simplify $\left(\frac{9^4+27^2}{27^4+9^5}\right)^2$ into the form $\left(\frac{1}{a}\right)^b$ with $a$ prime and $b$ a natural number, then compute $b^a$ .

Givens

$\left(\frac{9^4+27^2}{27^4+9^5}\right)^2=\left(\frac{1}{a}\right)^b$
$a$ is prime; $b$ is a natural number.

Unknowns

The value of $b^a$ .

Constraints

Exponent laws do not split sums, so the sums must be handled by factoring, not term by term.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#13 Convert to Algebra

$9$ and $27$ are both powers of $3$ , so write everything in base $3$ . The denominator then turns out to be a power of $3$ times the numerator, and the fraction collapses.

3 · Execute5 carry out the plan

1Write every power in base $3$

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

Use

9=3^2

and

27=3^3

with the power-of-a-power rule.

9^4=3^8,\ 27^2=3^6,\ 27^4=3^{12},\ 9^5=3^{10}

(3^2)^4=3^{2\times4}=3^8

, and so on.

2Factor the denominator

#7 Identify Subproblems 6.EE.A.3

Each denominator term is

3^4

times a numerator term.

3^{12}+3^{10}=3^4\times3^8+3^4\times3^6=3^4(3^8+3^6)

This is the distributive property run backwards.

3Cancel the common sum

#7 Identify Subproblems 8.EE.A.1

The same sum

3^8+3^6

appears on top and bottom.

\frac{3^8+3^6}{3^4(3^8+3^6)}=\frac{1}{3^4}

Whatever the sum is, it divides out.

4Square the result

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

Square

\frac{1}{3^4}

.

\left(\frac{1}{3^4}\right)^2=\frac{1}{3^8}=\left(\frac{1}{3}\right)^{8}

Squaring doubles the exponent.

5Read off $a$ and $b$

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

\left(\frac{1}{a}\right)^b=\left(\frac{1}{3}\right)^{8}

means

a^b=3^8

; since

a

is prime it must be

3

, so

b=8

.

b^a=8^3=512

Only the prime

3

divides

3^8

.

Answer:

512

4 · Reviewdoes it hold up?

Check numerically: $9^4+27^2=6561+729=7290$ and $27^4+9^5=531441+59049=590490$ . Then $590490\div7290=81=3^4$ , so the fraction really is $\frac{1}{3^4}$ .

Another way: Factor both parts by $3^6$ : the top is $3^6(3^2+1)$ and the bottom is $3^{10}(3^2+1)$ , so the fraction is $\frac{3^6}{3^{10}}=\frac{1}{3^4}$ and the rest is the same.

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Rewriting in base $3$, cancelling, and squaring powers.
6.EE.A.3 Apply properties to generate equivalent expressions — Factoring a common power out of a sum.

💡Takeaway. Exponent rules don't work on sums -- so factor the sum first, then the rules take over.

Variant 11 hard

When $\left(\frac{27^2+9^2}{27^3+3^7}\right)^2$ is simplified, it equals $\left(\frac{1}{a}\right)^b$ , where $a$ is a prime number and $b$ is a natural number. Find the value of $b^a$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Simplify $\left(\frac{27^2+9^2}{27^3+3^7}\right)^2$ into the form $\left(\frac{1}{a}\right)^b$ with $a$ prime and $b$ a natural number, then compute $b^a$ .

Givens

$\left(\frac{27^2+9^2}{27^3+3^7}\right)^2=\left(\frac{1}{a}\right)^b$
$a$ is prime; $b$ is a natural number.

Unknowns

The value of $b^a$ .

Constraints

Exponent laws do not split sums, so the sums must be handled by factoring, not term by term.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#13 Convert to Algebra

$27$ and $9$ are both powers of $3$ , so write everything in base $3$ . The denominator then turns out to be a power of $3$ times the numerator, and the fraction collapses.

3 · Execute5 carry out the plan

1Write every power in base $3$

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

Use

27=3^3

and

9=3^2

with the power-of-a-power rule.

27^2=3^6,\ 9^2=3^4,\ 27^3=3^9

(3^3)^2=3^{3\times2}=3^6

, and so on.

2Factor the denominator

#7 Identify Subproblems 6.EE.A.3

Each denominator term is

3^3

times a numerator term.

3^9+3^7=3^3\times3^6+3^3\times3^4=3^3(3^6+3^4)

This is the distributive property run backwards.

3Cancel the common sum

#7 Identify Subproblems 8.EE.A.1

The same sum

3^6+3^4

appears on top and bottom.

\frac{3^6+3^4}{3^3(3^6+3^4)}=\frac{1}{3^3}

Whatever the sum is, it divides out.

4Square the result

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

Square

\frac{1}{3^3}

.

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

Squaring doubles the exponent.

5Read off $a$ and $b$

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

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

means

a^b=3^6

; since

a

is prime it must be

3

, so

b=6

.

b^a=6^3=216

Only the prime

3

divides

3^6

.

Answer:

216

4 · Reviewdoes it hold up?

Check numerically: $27^2+9^2=729+81=810$ and $27^3+3^7=19683+2187=21870$ . Then $21870\div810=27=3^3$ , so the fraction really is $\frac{1}{3^3}$ .

Another way: Factor both parts by $3^4$ : the top is $3^4(3^2+1)$ and the bottom is $3^7(3^2+1)$ , so the fraction is $\frac{3^4}{3^7}=\frac{1}{3^3}$ and the rest is the same.

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Rewriting in base $3$, cancelling, and squaring powers.
6.EE.A.3 Apply properties to generate equivalent expressions — Factoring a common power out of a sum.

💡Takeaway. Exponent rules don't work on sums -- so factor the sum first, then the rules take over.

Variant 12 hard

When $\left(\frac{125^2+25^2}{5^7+5^5}\right)^2$ is simplified, it equals $\left(\frac{1}{a}\right)^b$ , where $a$ is a prime number and $b$ is a natural number. Find the value of $b^a$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Simplify $\left(\frac{125^2+25^2}{5^7+5^5}\right)^2$ into the form $\left(\frac{1}{a}\right)^b$ with $a$ prime and $b$ a natural number, then compute $b^a$ .

Givens

$\left(\frac{125^2+25^2}{5^7+5^5}\right)^2=\left(\frac{1}{a}\right)^b$
$a$ is prime; $b$ is a natural number.

Unknowns

The value of $b^a$ .

Constraints

Exponent laws do not split sums, so the sums must be handled by factoring, not term by term.

2 · Planchoose the strategy

#15 Organize Information in More Ways · also uses: #7 Identify Subproblems#13 Convert to Algebra

$125$ and $25$ are both powers of $5$ , so write everything in base $5$ . The denominator then turns out to be a power of $5$ times the numerator, and the fraction collapses.

3 · Execute5 carry out the plan

1Write every power in base $5$

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

Use

125=5^3

and

25=5^2

with the power-of-a-power rule.

125^2=5^6,\ 25^2=5^4

(5^3)^2=5^{3\times2}=5^6

, and so on.

2Factor the denominator

#7 Identify Subproblems 6.EE.A.3

Each denominator term is

5

times a numerator term.

5^7+5^5=5\times5^6+5\times5^4=5(5^6+5^4)

This is the distributive property run backwards.

3Cancel the common sum

#7 Identify Subproblems 8.EE.A.1

The same sum

5^6+5^4

appears on top and bottom.

\frac{5^6+5^4}{5(5^6+5^4)}=\frac{1}{5}

Whatever the sum is, it divides out.

4Square the result

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

Square

\frac{1}{5}

.

\left(\frac{1}{5}\right)^2=\frac{1}{5^2}=\left(\frac{1}{5}\right)^{2}

Squaring doubles the exponent.

5Read off $a$ and $b$

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

\left(\frac{1}{a}\right)^b=\left(\frac{1}{5}\right)^{2}

means

a^b=5^2

; since

a

is prime it must be

5

, so

b=2

.

b^a=2^5=32

Only the prime

5

divides

5^2

.

Answer:

32

4 · Reviewdoes it hold up?

Check numerically: $125^2+25^2=15625+625=16250$ and $5^7+5^5=78125+3125=81250$ . Then $81250\div16250=5$ , so the fraction really is $\frac{1}{5}$ .

Another way: Factor both parts by $5^4$ : the top is $5^4(5^2+1)$ and the bottom is $5^5(5^2+1)$ , so the fraction is $\frac{5^4}{5^5}=\frac{1}{5}$ and the rest is the same.

Standardsmin grade 8

8.EE.A.1 Know and apply the properties of integer exponents — Rewriting in base $5$, cancelling, and squaring powers.
6.EE.A.3 Apply properties to generate equivalent expressions — Factoring a common power out of a sum.

💡Takeaway. Exponent rules don't work on sums -- so factor the sum first, then the rules take over.

Rewrite in base 2 and factor the sum · 12 practice problems

Generated variants — 12

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