Is there a formula for #root(x)(a) xx root(y)(a)#? For example, #sqrt(81) xx root4(81)#

2 Answers
Jun 6, 2017

Answer:

#root(x)(a^m)xxroot(y)(a^m)=a^((m(x+y))/(xy))=root(xy)(a^(m(x+y)))#

Explanation:

There is no formula i.e. often used by people to solve such problems. However, mathematics is full of surprises and it does not mean that we cannot have a formula.

Here it is observed that in the example you have square root and fourth roots of #81#, which is itself a power of #3# i.e. #3^4#. Hence we will attempt a formula cosidering #a=b^m# and we attempt

#root(x)(a^m)xxroot(y)(a^m)#

= #(a^m)^(1/x)xx(a^m)^(1/y)#

= #a^(m/x)xxa^(m/y)#

= #a^(m/x+m/y)#

= #a^((m(x+y))/(xy))#

= #root(xy)(a^(m(x+y)))# and that is the formula.

i.e. #root(x)(a^m)xxroot(y)(a^m)=a^((m(x+y))/(xy))=root(xy)(a^(m(x+y)))#

If #a# is not a power than you can use #m=1#

Using this #sqrt(3^4)xxroot(4)(3^4)#

= #3^((4(2+4))/(2xx4))#

= #3^((4xx6)/8)#

= #3^3=27#

Jun 7, 2017

Answer:

#27#

Explanation:

#sqrt81 xx root4(81)#

#:.9 xx root4(3*3*3*3)#

#:.root4(a) xx root4(a) xx root4(a) xx root4(a)=a#

#:.9 xx 3=27#