Answer :
Sure, let's work through the problems step-by-step.
### Part a
Given:
[tex]\[ x = 7^{a - b} \][/tex]
[tex]\[ y = 7^{b - a} \][/tex]
To show that [tex]\(x \cdot y = 1\)[/tex]:
[tex]\[ x \cdot y = 7^{a-b} \cdot 7^{b-a} \][/tex]
Using properties of exponents:
[tex]\[ 7^{a-b} \cdot 7^{b-a} = 7^{(a-b) + (b-a)} \][/tex]
Simplifying the exponent:
[tex]\[ 7^{(a-b) + (b-a)} = 7^0 \][/tex]
Since any non-zero number raised to the power of 0 is 1:
[tex]\[ 7^0 = 1 \][/tex]
Thus:
[tex]\[ x \cdot y = 1 \][/tex]
### Part b
Given:
[tex]\[ x + y = x \][/tex]
We need to show that [tex]\( x^y = 1 \)[/tex].
Since [tex]\( x + y = x \)[/tex], subtract [tex]\( x \)[/tex] from both sides:
[tex]\[ y = 0 \][/tex]
Any non-zero number raised to the power of 0 is 1:
[tex]\[ x^y = x^0 = 1 \][/tex]
Thus:
[tex]\[ x^y = 1 \][/tex]
### Part c
Given:
[tex]\[ a = 2 \][/tex]
[tex]\[ b = 1 \][/tex]
We need to show that:
[tex]\[ (a+b)^2 = a^2 + 2ab + b^2 \][/tex]
First, calculate [tex]\( (a+b)^2 \)[/tex]:
[tex]\[ (2 + 1)^2 = 3^2 = 9 \][/tex]
Now calculate the right-hand side:
[tex]\[ a^2 + 2ab + b^2 \][/tex]
[tex]\[ 2^2 + 2(2)(1) + 1^2 \][/tex]
[tex]\[ 4 + 4 + 1 = 9 \][/tex]
Both sides are equal:
[tex]\[ 9 = 9 \][/tex]
Thus:
[tex]\[ (a+b)^2 = a^2 + 2ab + b^2 \][/tex]
### Part d
Given:
[tex]\[ a = 5 \][/tex]
[tex]\[ b = 3 \][/tex]
We need to show that:
[tex]\[ (a-b)^2 = a^2 - 2ab + b^2 \][/tex]
First, calculate [tex]\( (a-b)^2 \)[/tex]:
[tex]\[ (5 - 3)^2 = 2^2 = 4 \][/tex]
Now calculate the right-hand side:
[tex]\[ a^2 - 2ab + b^2 \][/tex]
[tex]\[ 5^2 - 2(5)(3) + 3^2 \][/tex]
[tex]\[ 25 - 30 + 9 = 4 \][/tex]
Both sides are equal:
[tex]\[ 4 = 4 \][/tex]
Thus:
[tex]\[ (a-b)^2 = a^2 - 2ab + b^2 \][/tex]
### Part e
Given:
[tex]\[ p + q + r = 0 \][/tex]
We need to show that the expression [tex]\[ x^{p+q} \times x^{p-q} \times x^{q+r} \times x^{q-r} \times x^{r+p} \][/tex]
simplifies to 1.
Using properties of exponents:
[tex]\[ x^{p+q} \times x^{p-q} \times x^{q+r} \times x^{q-r} \times x^{r+p} = x^{(p+q) + (p-q) + (q+r) + (q-r) + (r+p)} \][/tex]
Simplifying the exponent:
[tex]\[ (p+q) + (p-q) + (q+r) + (q-r) + (r+p) \][/tex]
Combining like terms:
[tex]\[ p + q + p - q + q + r + q - r + r + p \][/tex]
Grouping all similar terms together:
[tex]\[ (p + p + p) + (q + q - q + q) + (r + r - r) \][/tex]
[tex]\[ 3p + 2q - q + q + 2r - r \][/tex]
[tex]\[ 3p + q + r \][/tex]
Since [tex]\( p + q + r = 0 \)[/tex]:
[tex]\[ 3p + q + r = 3(p + q + r) \][/tex]
[tex]\[ 3 \times 0 = 0 \][/tex]
Thus:
[tex]\[ x^{0} = 1 \][/tex]
Therefore, we have shown that:
[tex]\[ x^{p+q} \times x^{p-q} \times x^{q+r} \times x^{q-r} \times x^{r+p} = 1 \][/tex]
This covers all parts of the given problem statement.
### Part a
Given:
[tex]\[ x = 7^{a - b} \][/tex]
[tex]\[ y = 7^{b - a} \][/tex]
To show that [tex]\(x \cdot y = 1\)[/tex]:
[tex]\[ x \cdot y = 7^{a-b} \cdot 7^{b-a} \][/tex]
Using properties of exponents:
[tex]\[ 7^{a-b} \cdot 7^{b-a} = 7^{(a-b) + (b-a)} \][/tex]
Simplifying the exponent:
[tex]\[ 7^{(a-b) + (b-a)} = 7^0 \][/tex]
Since any non-zero number raised to the power of 0 is 1:
[tex]\[ 7^0 = 1 \][/tex]
Thus:
[tex]\[ x \cdot y = 1 \][/tex]
### Part b
Given:
[tex]\[ x + y = x \][/tex]
We need to show that [tex]\( x^y = 1 \)[/tex].
Since [tex]\( x + y = x \)[/tex], subtract [tex]\( x \)[/tex] from both sides:
[tex]\[ y = 0 \][/tex]
Any non-zero number raised to the power of 0 is 1:
[tex]\[ x^y = x^0 = 1 \][/tex]
Thus:
[tex]\[ x^y = 1 \][/tex]
### Part c
Given:
[tex]\[ a = 2 \][/tex]
[tex]\[ b = 1 \][/tex]
We need to show that:
[tex]\[ (a+b)^2 = a^2 + 2ab + b^2 \][/tex]
First, calculate [tex]\( (a+b)^2 \)[/tex]:
[tex]\[ (2 + 1)^2 = 3^2 = 9 \][/tex]
Now calculate the right-hand side:
[tex]\[ a^2 + 2ab + b^2 \][/tex]
[tex]\[ 2^2 + 2(2)(1) + 1^2 \][/tex]
[tex]\[ 4 + 4 + 1 = 9 \][/tex]
Both sides are equal:
[tex]\[ 9 = 9 \][/tex]
Thus:
[tex]\[ (a+b)^2 = a^2 + 2ab + b^2 \][/tex]
### Part d
Given:
[tex]\[ a = 5 \][/tex]
[tex]\[ b = 3 \][/tex]
We need to show that:
[tex]\[ (a-b)^2 = a^2 - 2ab + b^2 \][/tex]
First, calculate [tex]\( (a-b)^2 \)[/tex]:
[tex]\[ (5 - 3)^2 = 2^2 = 4 \][/tex]
Now calculate the right-hand side:
[tex]\[ a^2 - 2ab + b^2 \][/tex]
[tex]\[ 5^2 - 2(5)(3) + 3^2 \][/tex]
[tex]\[ 25 - 30 + 9 = 4 \][/tex]
Both sides are equal:
[tex]\[ 4 = 4 \][/tex]
Thus:
[tex]\[ (a-b)^2 = a^2 - 2ab + b^2 \][/tex]
### Part e
Given:
[tex]\[ p + q + r = 0 \][/tex]
We need to show that the expression [tex]\[ x^{p+q} \times x^{p-q} \times x^{q+r} \times x^{q-r} \times x^{r+p} \][/tex]
simplifies to 1.
Using properties of exponents:
[tex]\[ x^{p+q} \times x^{p-q} \times x^{q+r} \times x^{q-r} \times x^{r+p} = x^{(p+q) + (p-q) + (q+r) + (q-r) + (r+p)} \][/tex]
Simplifying the exponent:
[tex]\[ (p+q) + (p-q) + (q+r) + (q-r) + (r+p) \][/tex]
Combining like terms:
[tex]\[ p + q + p - q + q + r + q - r + r + p \][/tex]
Grouping all similar terms together:
[tex]\[ (p + p + p) + (q + q - q + q) + (r + r - r) \][/tex]
[tex]\[ 3p + 2q - q + q + 2r - r \][/tex]
[tex]\[ 3p + q + r \][/tex]
Since [tex]\( p + q + r = 0 \)[/tex]:
[tex]\[ 3p + q + r = 3(p + q + r) \][/tex]
[tex]\[ 3 \times 0 = 0 \][/tex]
Thus:
[tex]\[ x^{0} = 1 \][/tex]
Therefore, we have shown that:
[tex]\[ x^{p+q} \times x^{p-q} \times x^{q+r} \times x^{q-r} \times x^{r+p} = 1 \][/tex]
This covers all parts of the given problem statement.
