Question

Prove that \[\dfrac{{{{\left( {a + \dfrac{1}{b}} \right)}^m} \times {{\left( {a - \dfrac{1}{b}} \right)}^n}}}{{{{\left( {b + \dfrac{1}{a}} \right)}^m} \times {{\left( {b - \dfrac{1}{a}} \right)}^n}}} = {\left( {\dfrac{a}{b}} \right)^{m + n}}\]

Answer 1

Hint: Here the given question is based on the exponentiation. In this question we are going to prove L.H.S=R.H.S because R.H.S is already in a simplified form whereas L.H.S is in the form that needs to be simplified. To prove this we are going to use a set of exponential formulas in each step to get the required form.

Complete step by step answer:
In order to show that the exponential function, \[\dfrac{{{{\left( {a + \dfrac{1}{b}} \right)}^m} \times {{\left( {a - \dfrac{1}{b}} \right)}^n}}}{{{{\left( {b + \dfrac{1}{a}} \right)}^m} \times {{\left( {b - \dfrac{1}{a}} \right)}^n}}} = {\left( {\dfrac{a}{b}} \right)^{m + n}}\]
Let us prove that the L.H.S and R.H.S are equal.
So let take, L.H.S,
\[L.H.S = \dfrac{{{{\left( {a + \dfrac{1}{b}} \right)}^m} \times {{\left( {a - \dfrac{1}{b}} \right)}^n}}}{{{{\left( {b + \dfrac{1}{a}} \right)}^m} \times {{\left( {b - \dfrac{1}{a}} \right)}^n}}} \\
\Rightarrow L.H.S= {\left( {\dfrac{a}{b}} \right)^{m + n}} \\ \]
Taking L.C.M for both numerator and denominator separately since both numerator and denominator are in fractions.
\[\dfrac{{{{\left( {\dfrac{{ab + 1}}{b}} \right)}^m} \times {{\left( {\dfrac{{ab - 1}}{b}} \right)}^n}}}{{{{\left( {\dfrac{{ab + 1}}{a}} \right)}^m} \times {{\left( {\dfrac{{ab - 1}}{a}} \right)}^n}}} \\ \]
Now, taking reciprocal of the denominator \[{\left( {\dfrac{{ab + 1}}{a}} \right)^m} \times {\left( {\dfrac{{ab - 1}}{a}} \right)^n}\], we will get,
\[{\left( {\dfrac{{ab + 1}}{b}} \right)^m} \times {\left( {\dfrac{{ab - 1}}{b}} \right)^n} \times {\left( {\dfrac{a}{{ab + 1}}} \right)^m} \times {\left( {\dfrac{a}{{ab - 1}}} \right)^n}\]
Now, just split the numerator and denominator into two terms as, \[ = {(ab + 1)^m} \times {\left( {\dfrac{1}{b}} \right)^m} \times {(ab - 1)^m} \times {\left( {\dfrac{1}{b}} \right)^n} \times {a^m} \times {\left( {\dfrac{1}{{ab + 1}}} \right)^m} \times {a^n} \times {\left( {\dfrac{1}{{ab - 1}}} \right)^n}\]

By using the power of quotient rule: this says that to divide two exponents with the same base, you keep the base and subtract the powers. Here we can find \[ab + 1\] and \[ab - 1\] in numerators and denominators and also with the same power \[m\]. Thus cancelling same numerator and denominator with the same powers we will get,
\[{\left( {\dfrac{1}{b}} \right)^m} \times {\left( {\dfrac{1}{b}} \right)^n} \times {a^m} \times {a^n}\]
Using product rule: this says that to multiply two exponents with the same base, you keep the base and add the powers,
\[{(a)^{m + n}} \times {\left( {\dfrac{1}{b}} \right)^{m + n}}\]
\[{\left( {\dfrac{a}{b}} \right)^{m + n}}\]
L.H.S \[ = \] R.H.S
Hence proved.

Note: Important properties of exponent and power of real number are given below:
1. Suppose that \[a\] and \[b\] is a real number and \[m{\text{,}}n\] are two positive integers then, where \[b\] is a non zero number.
\[{\left( {\dfrac{a}{b}} \right)^m} = {a^m} \times {\left( {\dfrac{1}{b}} \right)^m}\]
Example: \[a = 5{\text{,}}b = 6{\text{,}}m = 1\]
Let L.H.S \[ = {\left( {\dfrac{5}{6}} \right)^1} = \dfrac{5}{6}\]
Let R.H.S \[= {5^1} \times {\left( {\dfrac{1}{6}} \right)^1} = {\left( {\dfrac{5}{6}} \right)^1} = \dfrac{5}{6} \]
Therefore L.H.S \[ = \] R.H.S.
2. Suppose that \[a\] is real number and \[m{\text{,}}n\] are two positive integers then,
\[{a^m} \times {a^n} = {(a)^{m + n}}\]
Example: \[a = 7{\text{,}}m = 1{\text{,}}n = 1\]
Let L.H.S \[ = {7^1} \times {7^1} = 7 \times 7 = 49\]
Let R.H.S \[ = {\left( 7 \right)^{1 + 1}} = {(7)^2} = 49\]
Therefore L.H.S \[ = \] R.H.S