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
Simplify the following:
`(5^(n+3)-6xx5^(n+1))/(9xx5^x-2^2xx5^n)`
Solve the following equation for x:
`2^(5x+3)=8^(x+3)`
Solve the following equation:
`3^(x+1)=27xx3^4`
If a and b are distinct primes such that `root3 (a^6b^-4)=a^xb^(2y),` find x and y.
State the product law of exponents.
State the power law of exponents.
\[\frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^n - 2 \times 5^{n + 1}}\] is equal to
If x= \[\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\] and y = \[\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\] , then x2 + y +y2 =
If \[x = \sqrt{6} + \sqrt{5}\],then \[x^2 + \frac{1}{x^2} - 2 =\]
Find:-
`32^(1/5)`
