Question

$(2^{5})^{2} \cdot 2^{0} = 2^{5} \cdot 2^{5} \cdot 0$
$(2^{2})^{3} \cdot 2^{-1} = 2^{6} \cdot \frac{1}{2}$
$2^{-9} \cdot 2^{3} \cdot \frac{1}{2^{4}} = \frac{1}{2^{10}}$
$2 \cdot \frac{1}{2^{9}} \cdot 2^{2} = 2^{(1 + 9 + 2)}$
$(2^{3})^{4} = 2^{7}$

Explanation:

To determine the correctness of each equation, we use the properties of exponents: \( (a^m)^n = a^{mn} \), \( a^m \cdot a^n = a^{m + n} \), \( a^{-n} = \frac{1}{a^n} \), and \( a^0 = 1 \) (for \( a
eq 0 \)).

1. \( \boldsymbol{(2^5)^2 \cdot 2^0 = 2^5 \cdot 2^5 \cdot 0} \)

• Left - hand side (LHS):

Using \( (a^m)^n = a^{mn} \), \( (2^5)^2 = 2^{5\times2}=2^{10} \).
Using \( a^0 = 1 \), \( 2^0 = 1 \).
So, \( (2^5)^2 \cdot 2^0=2^{10}\cdot1 = 2^{10} \).

• Right - hand side (RHS):

\( 2^5\cdot2^5\cdot0=(2^{5 + 5})\cdot0 = 2^{10}\cdot0 = 0 \).
Since \( 2^{10}
eq0 \), the equation is False.

2. \( \boldsymbol{(2^2)^3 \cdot 2^{-1} = 2^6 \cdot \frac{1}{2}} \)

• LHS:

Using \( (a^m)^n = a^{mn} \), \( (2^2)^3 = 2^{2\times3}=2^6 \).
Using \( a^{-n}=\frac{1}{a^n} \), \( 2^{-1}=\frac{1}{2} \).
So, \( (2^2)^3\cdot2^{-1}=2^6\cdot\frac{1}{2} \), which matches the RHS.
The equation is True (the original selection of “False” is incorrect).

3. \( \boldsymbol{2^{-9} \cdot 2^3 \cdot \frac{1}{2^4} = \frac{1}{2^{10}}} \)

• LHS:

Using \( a^m\cdot a^n = a^{m + n} \) and \( \frac{1}{a^n}=a^{-n} \), we rewrite \( \frac{1}{2^4} \) as \( 2^{-4} \).
Then, \( 2^{-9}\cdot2^3\cdot2^{-4}=2^{-9 + 3-4}=2^{-10} \).
Using \( a^{-n}=\frac{1}{a^n} \), \( 2^{-10}=\frac{1}{2^{10}} \), which matches the RHS.
The equation is True (the original selection of “False” is incorrect).

4. \( \boldsymbol{2 \cdot \frac{1}{2^9} \cdot 2^2 = 2^{(1 + 9 + 2)}} \)

• LHS:

Rewrite \( 2 \) as \( 2^1 \) and \( \frac{1}{2^9} \) as \( 2^{-9} \).
Then, \( 2^1\cdot2^{-9}\cdot2^2=2^{1-9 + 2}=2^{-6}=\frac{1}{2^6} \).

• RHS:

\( 2^{(1 + 9+2)}=2^{12} \).
Since \( \frac{1}{2^6}
eq2^{12} \), the equation is False (the original selection of “True” is incorrect).

5. \( \boldsymbol{(2^3)^4 = 2^7} \)

• LHS:

Using \( (a^m)^n = a^{mn} \), \( (2^3)^4 = 2^{3\times4}=2^{12} \).

• RHS: \( 2^7 \).

Since \( 2^{12}
eq2^7 \), the equation is False.

Correcting the Selections:

1. \( (2^5)^2 \cdot 2^0 = 2^5 \cdot 2^5 \cdot 0 \): False (correct as marked).
2. \( (2^2)^3 \cdot 2^{-1} = 2^6 \cdot \frac{1}{2} \): True (should be marked “True”).
3. \( 2^{-9} \cdot 2^3 \cdot \frac{1}{2^4} = \frac{1}{2^{10}} \): True (should be marked “True”).
4. \( 2 \cdot \frac{1}{2^9} \cdot 2^2 = 2^{(1 + 9 + 2)} \): False (should be marked “False”).
5. \( (2^3)^4 = 2^7 \): False (correct as marked).

Answer:

To determine the correctness of each equation, we use the properties of exponents: \( (a^m)^n = a^{mn} \), \( a^m \cdot a^n = a^{m + n} \), \( a^{-n} = \frac{1}{a^n} \), and \( a^0 = 1 \) (for \( a
eq 0 \)).

1. \( \boldsymbol{(2^5)^2 \cdot 2^0 = 2^5 \cdot 2^5 \cdot 0} \)

• Left - hand side (LHS):

Using \( (a^m)^n = a^{mn} \), \( (2^5)^2 = 2^{5\times2}=2^{10} \).
Using \( a^0 = 1 \), \( 2^0 = 1 \).
So, \( (2^5)^2 \cdot 2^0=2^{10}\cdot1 = 2^{10} \).

• Right - hand side (RHS):

\( 2^5\cdot2^5\cdot0=(2^{5 + 5})\cdot0 = 2^{10}\cdot0 = 0 \).
Since \( 2^{10}
eq0 \), the equation is False.

2. \( \boldsymbol{(2^2)^3 \cdot 2^{-1} = 2^6 \cdot \frac{1}{2}} \)

• LHS:

Using \( (a^m)^n = a^{mn} \), \( (2^2)^3 = 2^{2\times3}=2^6 \).
Using \( a^{-n}=\frac{1}{a^n} \), \( 2^{-1}=\frac{1}{2} \).
So, \( (2^2)^3\cdot2^{-1}=2^6\cdot\frac{1}{2} \), which matches the RHS.
The equation is True (the original selection of “False” is incorrect).

3. \( \boldsymbol{2^{-9} \cdot 2^3 \cdot \frac{1}{2^4} = \frac{1}{2^{10}}} \)

• LHS:

Using \( a^m\cdot a^n = a^{m + n} \) and \( \frac{1}{a^n}=a^{-n} \), we rewrite \( \frac{1}{2^4} \) as \( 2^{-4} \).
Then, \( 2^{-9}\cdot2^3\cdot2^{-4}=2^{-9 + 3-4}=2^{-10} \).
Using \( a^{-n}=\frac{1}{a^n} \), \( 2^{-10}=\frac{1}{2^{10}} \), which matches the RHS.
The equation is True (the original selection of “False” is incorrect).

4. \( \boldsymbol{2 \cdot \frac{1}{2^9} \cdot 2^2 = 2^{(1 + 9 + 2)}} \)

• LHS:

Rewrite \( 2 \) as \( 2^1 \) and \( \frac{1}{2^9} \) as \( 2^{-9} \).
Then, \( 2^1\cdot2^{-9}\cdot2^2=2^{1-9 + 2}=2^{-6}=\frac{1}{2^6} \).

• RHS:

\( 2^{(1 + 9+2)}=2^{12} \).
Since \( \frac{1}{2^6}
eq2^{12} \), the equation is False (the original selection of “True” is incorrect).

5. \( \boldsymbol{(2^3)^4 = 2^7} \)

• LHS:

Using \( (a^m)^n = a^{mn} \), \( (2^3)^4 = 2^{3\times4}=2^{12} \).

• RHS: \( 2^7 \).

Since \( 2^{12}
eq2^7 \), the equation is False.

Correcting the Selections:

1. \( (2^5)^2 \cdot 2^0 = 2^5 \cdot 2^5 \cdot 0 \): False (correct as marked).
2. \( (2^2)^3 \cdot 2^{-1} = 2^6 \cdot \frac{1}{2} \): True (should be marked “True”).
3. \( 2^{-9} \cdot 2^3 \cdot \frac{1}{2^4} = \frac{1}{2^{10}} \): True (should be marked “True”).
4. \( 2 \cdot \frac{1}{2^9} \cdot 2^2 = 2^{(1 + 9 + 2)} \): False (should be marked “False”).
5. \( (2^3)^4 = 2^7 \): False (correct as marked).