Advertisements
Advertisements
Question
Prove that:
`sqrt(3xx5^-3)divroot3(3^-1)sqrt5xxroot6(3xx5^6)=3/5`
Advertisements
Solution
we have to prove that `sqrt(3xx5^-3)/(root3(3^-1)sqrt5)xxroot6(3xx5^6)=3/5`
By using rational exponents `a^-n=1/a^n` we get,
`sqrt(3xx5^-3)/(root3(3^-1)sqrt5)xxroot6(3xx5^6)=sqrt(3xx1/5^3)/(root3(1/3)sqrt5)xxroot6(3xx5^6)`
`=(3^(1/2)xx1/5^(3xx1/2))/(1/3^(1/3)xx5^(1/2))xx3^(1/6)xx5^(6xx1/6)`
`=(3^(1/2)/5^(3/2))/(5^(1/2)/3^(1/3))xx3^(1/6)xx5^1`
`=3^(1/2)/5^(3/2)xx3^(1/3)/5^(1/2)xx3^(1/6)xx5^1`
`=3^(1/2)xx3^(1/3)xx5^(-3/2)xx5^(-1/2)xx3^(1/6)xx5^1`
`=3^(1/2+1/3+1/6)xx5^(-3/2-1/2+1)`
`=3^((1xx3)/(2xx3)+(1xx2)/(3xx2)+1/6)xx5^(-3/2-1/2+(1xx2)/(1xx2))`
`=3^((3+2+1)/6)xx5^((-3-1+2)/2)`
`=3^1xx5^-1`
`=3xx1/5`
`=3/5`
Hence `sqrt(3xx5^-3)/(root3(3^-1)sqrt5)xxroot6(3xx5^6)=3/5`
APPEARS IN
RELATED QUESTIONS
If 2x = 3y = 12z, show that `1/z=1/y+2/x`
Find the value of x in the following:
`2^(x-7)xx5^(x-4)=1250`
If 102y = 25, then 10-y equals
If \[\frac{3^{5x} \times {81}^2 \times 6561}{3^{2x}} = 3^7\] then x =
If \[x = 7 + 4\sqrt{3}\] and xy =1, then \[\frac{1}{x^2} + \frac{1}{y^2} =\]
If \[x + \sqrt{15} = 4,\] then \[x + \frac{1}{x}\] =
\[\frac{1}{\sqrt{9} - \sqrt{8}}\] is equal to
The value of \[\sqrt{3 - 2\sqrt{2}}\] is
The value of \[\sqrt{5 + 2\sqrt{6}}\] is
If `a = 2 + sqrt(3)`, then find the value of `a - 1/a`.
