Advertisements
Advertisements
प्रश्न
If `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m),` prove that `x^my^nz^l = x^ny^lz^m`
Advertisements
उत्तर
Given `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m)`
Putting the values of x, y and z in `x^my^nz^l,` we get
`x^my^nz^l`
= `(a^(m + n))^m(a^(n + l))^n(a^(l + m))^l`
= `(a^(m^2 + nm))(a^(n^2 + ln))(a^(l^2 + lm))`
= `a^(m^2 + n^2 + l^2 + nm + ln + lm)`
Putting the values of x, y and z in `x^ny^lz^m,` we get
`x^ny^lz^m`
= `(a^(m + n))^n(a^(n + l))^l(a^(l + m))^m`
= `(a^(mn + n^2))(a^(nl + l^2))(a^(lm + m^2))`
= `a^(mn+n^2 + nl + l^2 + lm + m^2)`
So, `x^my^nz^l = x^ny^lz^m`.
APPEARS IN
संबंधित प्रश्न
Find:-
`9^(3/2)`
Solve the following equation for x:
`2^(3x-7)=256`
Assuming that x, y, z are positive real numbers, simplify the following:
`(x^-4/y^-10)^(5/4)`
Simplify:
`((25)^(3/2)xx(243)^(3/5))/((16)^(5/4)xx(8)^(4/3))`
Prove that:
`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`
If a, b, c are positive real numbers, then \[\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}\] is equal to
If \[\sqrt{5^n} = 125\] then `5nsqrt64`=
\[\frac{1}{\sqrt{9} - \sqrt{8}}\] is equal to
Simplify:
`11^(1/2)/11^(1/4)`
Simplify:
`(3/5)^4 (8/5)^-12 (32/5)^6`
