Advertisements
Advertisements
प्रश्न
For any positive real number x, write the value of \[\left\{ \left( x^a \right)^b \right\}^\frac{1}{ab} \left\{ \left( x^b \right)^c \right\}^\frac{1}{bc} \left\{ \left( x^c \right)^a \right\}^\frac{1}{ca}\]
उत्तर
\[\left\{ \left( x^a \right)^b \right\}^\frac{1}{ab} \left\{ \left( x^b \right)^c \right\}^\frac{1}{bc} \left\{ \left( x^c \right)^a \right\}^\frac{1}{ca}\] So,
\[\left\{ \left( x^a \right)^b \right\}^\frac{1}{ab} \left\{ \left( x^b \right)^c \right\}^\frac{1}{bc} \left\{ \left( x^c \right)^a \right\}^\frac{1}{ca}\] = `{x^(ab)}^(1/(ab)) {x^(bc)}^(1/(bc)) {x^(ca)}^(1/(ca))`
=`x^(ab xx 1/(ab)) xx x^(bc xx 1/(bc) xx x^(ac xx 1/(ca)))`
= `x^(ab xx 1/(ab)) xx x^(bc xx 1/(bc) xx x^(ac xx 1/(ca)))`
`= x^1 xx x^1 xx x^1`
By using rational exponents `a^m xx a^n xx a^(m+n), ` we get
\[\left\{ \left( x^a \right)^b \right\}^\frac{1}{ab} \left\{ \left( x^b \right)^c \right\}^\frac{1}{bc} \left\{ \left( x^c \right)^a \right\}^\frac{1}{ca}\] = `x^(1+1+1)`
`=x^3`
Hence the value of
\[\left\{ \left( x^a \right)^b \right\}^\frac{1}{ab} \left\{ \left( x^b \right)^c \right\}^\frac{1}{bc} \left\{ \left( x^c \right)^a \right\}^\frac{1}{ca}\] = `x^(1+1+1)` is `=x^3`
APPEARS IN
संबंधित प्रश्न
Find:-
`9^(3/2)`
Simplify the following:
`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Assuming that x, y, z are positive real numbers, simplify the following:
`(x^-4/y^-10)^(5/4)`
Simplify:
`(sqrt2/5)^8div(sqrt2/5)^13`
Prove that:
`((0.6)^0-(0.1)^-1)/((3/8)^-1(3/2)^3+((-1)/3)^-1)=(-3)/2`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
If 2x = 3y = 6-z, show that `1/x+1/y+1/z=0`
Solve the following equation:
`3^(x+1)=27xx3^4`
If 9x+2 = 240 + 9x, then x =