Advertisements
Advertisements
Question
Prove that:
`(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))=28sqrt2`
Advertisements
Solution
We have to prove that `(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))=28sqrt2`
Let x = `(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))`
`=(3^-3xx3^2xx2^2xxsqrt(7xx7xx2))/(5^2xxroot3(1/25)xx(15)^-(4/3)xx3^(1/3))`
`=(3^(-3+2)xx2^2xx7sqrt2)/(5^2xx1/5^(2xx1/3)xx5^(-4/3)xx3^(-4/3)xx3^(1/3))`
`=(3^-1xx2^2xx7sqrt2)/(5^2/1xx1/5^(2/3)xx1/5^(4/3)xx1/3^(4/3)xx3^(1/3)/1)`
`=3^-1xx2^2xx7sqrt2xx1/5^2xx5^(2/3)xx5^(4/3)xx3^(4/3)xx1/3^(1/3)`
`=3^-1xx3^(4/3)xx1/3^(1/3)xx4xx7sqrt2xx1/5^2xx5^(2/3)xx5^(4/3)`
`=3^(-1+4/3-1/3)xx4xx7sqrt2xx5^(-2+2/3+4/3)`
`=3^((-1xx3)/(1xx3)+4/3-1/3)xx28sqrt2xx5^((-2xx3)/(1xx3)+2/3+4/3)`
`=3^((-3+4-1)/3)xx28sqrt2xx5^((-6+2+4)/3)`
`=3^0xx28sqrt2xx5^0`
`=1xx28sqrt2xx1`
`=28sqrt2`
Hence, `(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))=28sqrt2`
APPEARS IN
RELATED QUESTIONS
Simplify the following:
`(2x^-2y^3)^3`
Given `4725=3^a5^b7^c,` find
(i) the integral values of a, b and c
(ii) the value of `2^-a3^b7^c`
Prove that:
`(64/125)^(-2/3)+1/(256/625)^(1/4)+(sqrt25/root3 64)=65/16`
If 2x = 3y = 12z, show that `1/z=1/y+2/x`
Find the value of x in the following:
`5^(x-2)xx3^(2x-3)=135`
If a and b are distinct primes such that `root3 (a^6b^-4)=a^xb^(2y),` find x and y.
Which one of the following is not equal to \[\left( \frac{100}{9} \right)^{- 3/2}\]?
If \[2^{- m} \times \frac{1}{2^m} = \frac{1}{4},\] then \[\frac{1}{14}\left\{ ( 4^m )^{1/2} + \left( \frac{1}{5^m} \right)^{- 1} \right\}\] is equal to
If \[\frac{2^{m + n}}{2^{n - m}} = 16\], \[\frac{3^p}{3^n} = 81\] and \[a = 2^{1/10}\],than \[\frac{a^{2m + n - p}}{( a^{m - 2n + 2p} )^{- 1}} =\]
The positive square root of \[7 + \sqrt{48}\] is
