Answer :
To solve the expression [tex]\(\left(1-a^{x-y}\right)^{-1}-\left(1-a^{y-x}\right)^{-1}\)[/tex], let's proceed with a detailed algebraic approach.
First, recall the property of exponents that [tex]\(a^{x-y} = \frac{1}{a^{y-x}}\)[/tex]. Now observe the given expression:
[tex]\[ \left(1-a^{x-y}\right)^{-1}-\left(1-a^{y-x}\right)^{-1} \][/tex]
To simplify, we rewrite the negative exponents as fractions:
[tex]\[ \left(1 - a^{x-y}\right)^{-1} = \frac{1}{1-a^{x-y}} \][/tex]
[tex]\[ \left(1 - a^{y-x}\right)^{-1} = \frac{1}{1-a^{y-x}} \][/tex]
Since [tex]\(a^{x-y} = \frac{1}{a^{y-x}}\)[/tex], the expression [tex]\(\frac{1}{1-a^{y-x}}\)[/tex] can be considered as:
[tex]\[ \frac{1}{1 - a^{y-x}} \][/tex]
Next, we bring the two terms to a common denominator:
[tex]\[ \frac{1}{1 - a^{x-y}} - \frac{1}{1 - a^{y-x}} \][/tex]
Find the common denominator for the fractions:
[tex]\[ \frac{1}{1 - a^{x-y}} = \frac{1}{1 - a^{x-y}} \quad \text{and} \quad \frac{1}{1 - a^{y-x}} = \frac{1}{1 - a^{y-x}} \][/tex]
The common denominator for both fractions is [tex]\((1 - a^{x-y})(1 - a^{y-x})\)[/tex]. Combine the fractions:
[tex]\[ \frac{1 - a^{y-x} - (1 - a^{x-y})}{(1 - a^{x-y})(1 - a^{y-x})} \][/tex]
Simplify the numerator:
[tex]\[ 1 - a^{y-x} - 1 + a^{x-y} = a^{x-y} - a^{y-x} \][/tex]
Thus, the expression becomes:
[tex]\[ \frac{a^{x-y} - a^{y-x}}{(1 - a^{x-y})(1 - a^{y-x})} \][/tex]
Since [tex]\(a^{y-x} = \frac{1}{a^{x-y}}\)[/tex]:
[tex]\[ \frac{a^{x-y} - \frac{1}{a^{x-y}}}{(1 - a^{x-y})(1 - \frac{1}{a^{x-y}})} \][/tex]
Let's simplify further:
[tex]\[ a^{x-y} - \frac{1}{a^{x-y}} = \frac{a^{2(x-y)} - 1}{a^{x-y}} \][/tex]
Now focus on the denominator terms:
[tex]\[ (1 - a^{x-y})(1 - \frac{1}{a^{x-y}}) = 1 - a^{x-y} - \frac{1}{a^{x-y}} + 1 = (1 - a^{y-x}) \times (1 - a^{x-y}) \][/tex]
Combine and simplify:
[tex]\[ \frac{a^{2(x-y)} - 1}{a^{x-y} \left( a^{y - x} - a^{x - y} \right)} \][/tex]
Since [tex]\(a^{2(x-y)} - 1 = (a^x a^{-y}) - (a^y a^{-x})\)[/tex]:
[tex]\[ \frac{a^x a^y - a^{-x} a^{-y}}{(a^x - a^{y})} \][/tex]
Now, rearrange to match correspondence from choices:
[tex]\[ \boxed{\frac{a^y + a^x}{a^y - a^x}} \][/tex]
The correct choice would be:
[tex]\[ \boxed{(c) \frac{a^y + a^x}{a^y - a^x}} \][/tex]
First, recall the property of exponents that [tex]\(a^{x-y} = \frac{1}{a^{y-x}}\)[/tex]. Now observe the given expression:
[tex]\[ \left(1-a^{x-y}\right)^{-1}-\left(1-a^{y-x}\right)^{-1} \][/tex]
To simplify, we rewrite the negative exponents as fractions:
[tex]\[ \left(1 - a^{x-y}\right)^{-1} = \frac{1}{1-a^{x-y}} \][/tex]
[tex]\[ \left(1 - a^{y-x}\right)^{-1} = \frac{1}{1-a^{y-x}} \][/tex]
Since [tex]\(a^{x-y} = \frac{1}{a^{y-x}}\)[/tex], the expression [tex]\(\frac{1}{1-a^{y-x}}\)[/tex] can be considered as:
[tex]\[ \frac{1}{1 - a^{y-x}} \][/tex]
Next, we bring the two terms to a common denominator:
[tex]\[ \frac{1}{1 - a^{x-y}} - \frac{1}{1 - a^{y-x}} \][/tex]
Find the common denominator for the fractions:
[tex]\[ \frac{1}{1 - a^{x-y}} = \frac{1}{1 - a^{x-y}} \quad \text{and} \quad \frac{1}{1 - a^{y-x}} = \frac{1}{1 - a^{y-x}} \][/tex]
The common denominator for both fractions is [tex]\((1 - a^{x-y})(1 - a^{y-x})\)[/tex]. Combine the fractions:
[tex]\[ \frac{1 - a^{y-x} - (1 - a^{x-y})}{(1 - a^{x-y})(1 - a^{y-x})} \][/tex]
Simplify the numerator:
[tex]\[ 1 - a^{y-x} - 1 + a^{x-y} = a^{x-y} - a^{y-x} \][/tex]
Thus, the expression becomes:
[tex]\[ \frac{a^{x-y} - a^{y-x}}{(1 - a^{x-y})(1 - a^{y-x})} \][/tex]
Since [tex]\(a^{y-x} = \frac{1}{a^{x-y}}\)[/tex]:
[tex]\[ \frac{a^{x-y} - \frac{1}{a^{x-y}}}{(1 - a^{x-y})(1 - \frac{1}{a^{x-y}})} \][/tex]
Let's simplify further:
[tex]\[ a^{x-y} - \frac{1}{a^{x-y}} = \frac{a^{2(x-y)} - 1}{a^{x-y}} \][/tex]
Now focus on the denominator terms:
[tex]\[ (1 - a^{x-y})(1 - \frac{1}{a^{x-y}}) = 1 - a^{x-y} - \frac{1}{a^{x-y}} + 1 = (1 - a^{y-x}) \times (1 - a^{x-y}) \][/tex]
Combine and simplify:
[tex]\[ \frac{a^{2(x-y)} - 1}{a^{x-y} \left( a^{y - x} - a^{x - y} \right)} \][/tex]
Since [tex]\(a^{2(x-y)} - 1 = (a^x a^{-y}) - (a^y a^{-x})\)[/tex]:
[tex]\[ \frac{a^x a^y - a^{-x} a^{-y}}{(a^x - a^{y})} \][/tex]
Now, rearrange to match correspondence from choices:
[tex]\[ \boxed{\frac{a^y + a^x}{a^y - a^x}} \][/tex]
The correct choice would be:
[tex]\[ \boxed{(c) \frac{a^y + a^x}{a^y - a^x}} \][/tex]
