Advertisements
Advertisements
Question
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt2/sqrt3)^5(6/7)^2`
Advertisements
Solution
We have to simplify the following, assuming that x, y, z are positive real numbers
Given `(sqrt2/sqrt3)^5(6/7)^2`
`=(sqrt2/sqrt3)^(2+2+1)(6/7)^2`
`=(sqrt2/sqrt3)^2xx(sqrt2/sqrt3)^2xx(sqrt2/sqrt3)^1xx(6/7)^2`
`=(2/3)xx(2/3)xx(sqrt2/sqrt3)^1xx(6/7)^2`
`=(16sqrt2)/(49sqrt3)`
`=sqrt(512/7203)`
`=(512/7203)^(1/2)`
APPEARS IN
RELATED QUESTIONS
Simplify the following:
`(6(8)^(n+1)+16(2)^(3n-2))/(10(2)^(3n+1)-7(8)^n)`
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Simplify:
`((5^-1xx7^2)/(5^2xx7^-4))^(7/2)xx((5^-2xx7^3)/(5^3xx7^-5))^(-5/2)`
Find the value of x in the following:
`5^(2x+3)=1`
Solve the following equation:
`3^(x+1)=27xx3^4`
Solve the following equation:
`4^(2x)=(root3 16)^(-6/y)=(sqrt8)^2`
Write the value of \[\sqrt[3]{125 \times 27}\].
If \[\frac{x}{x^{1 . 5}} = 8 x^{- 1}\] and x > 0, then x =
The simplest rationalising factor of \[2\sqrt{5}-\]\[\sqrt{3}\] is
Simplify:-
`(1/3^3)^7`
