(i) If $a = x^{m + n} y^{l}, b = x^{n + l} y^{m} and c = x^{l + m} y^{n}$ , prove that $a^{m - n} b^{n - l} c^{l - m} = 1$ .

(ii) If $x = a^{m + n}, y = a^{n + l} and z = a^{l + m}$ , prove that $x^{m} y^{n} z^{l} = x^{n} y^{l} z^{m}$ .

Open in App

Solution

(i) Given: $a = x^{m + n} y^{l}, b = x^{n + l} y^{m} and c = x^{l + m} y^{n}$
Putting the values of a, b and c in $a^{m - n} b^{n - l} c^{l - m}$ , we get
$a^{m - n} b^{n - l} c^{l - m} = {(x^{m + n} y^{l})}^{m - n} {(x^{n + l} y^{m})}^{n - l} {(x^{l + m} y^{n})}^{l - m} = [x^{(m + n) (m - n)} y^{l (m - n)}] [x^{(n + l) (n - l)} y^{m (n - l)}] [x^{(l + m) (l - m)} y^{n (l - m)}] = x^{(m^{2} - n^{2})} x^{(n^{2} - l^{2})} x^{(l^{2} - m^{2})} y^{l m - l n} y^{m n - m l} y^{n l - n m} = x^{m^{2} - n^{2} + n^{2} - l^{2} + l^{2} - m^{2}} y^{l m - l n + m n - m l + n l - n m} = x^{0} y^{0} = 1$

(ii) Given: $x = a^{m + n}, y = a^{n + l} and z = a^{l + m}$
Putting the values of x, y and z in $x^{m} y^{n} z^{l}$ , we get
$x^{m} y^{n} z^{l} = {(a^{m + n})}^{m} {(a^{n + l})}^{n} {(a^{l + m})}^{l} = (a^{m^{2} + n m}) (a^{n^{2} + l n}) (a^{l^{2} + l m}) = a^{m^{2} + n^{2} + l^{2} + n m + l n + l m}$
Putting the values of x, y and z in $x^{n} y^{l} z^{m}$ , we get
$x^{n} y^{l} z^{m} = {(a^{m + n})}^{n} {(a^{n + l})}^{l} {(a^{l + m})}^{m} = (a^{m n + n^{2}}) (a^{n l + l^{2}}) (a^{l m + m^{2}}) = a^{m n + n^{2} + n l + l^{2} + l m + m^{2}}$
So, $x^{m} y^{n} z^{l}$ = $x^{n} y^{l} z^{m}$

Suggest Corrections

Similar questions

a = x^{m + n} . y^{l}; b = x^{n + l} . y^{m}

c = x^{l + m} . y^{n}

a^{m - n} . b^{n - l} . c^{l - m} = 1

a = x^{m + n} \cdot y^{1}; b = x^{n + 1} \cdot y^{m} a n d c = x^{1 + m} \cdot y^{n}

a^{m - n} \cdot b^{n - 1} \cdot c^{1 - m} = 1

\frac{x^{m + n} \times x^{n + 1} \times x^{l + m}}{{(x^{m} \times x^{n} \times x^{l})}^{2}} = 1

\frac{1}{m} log (1 + a^{m}) < \frac{1}{n} log (1 + a^{n})

m > n

{(\frac{x^{m}}{x^{n}})}^{m + n} \times {(\frac{x^{n}}{x^{l}})}^{n + l} \times {(\frac{x^{l}}{x^{m}})}^{l + m}

Join BYJU'S Learning Program