how to find k if Sq root 3^7 + Sq root 3^5 + Sq root 3^3 + (42)(Sq Root 3) = 3^k?

Answer 1

Sq root(3^7) = Sq root(3^6)* Sq root(3)= 27*sqrt(3)

Sq root(3^5) = Sq root(3^4)* Sq root(3)=9*sqrt(3)

Sq root(3^3) = Sq root(3^2)* Sq root(3)=3*sqrt(3)

HENCE:

LHS

= Sq root 3^7 + Sq root 3^5 + Sq root 3^3 + (42)(Sq Root 3)

= 27* Sq root (3) + 9* Sq root (3) + 3* Sq root (3) + 42* Sq root (3)

= 81* Sq root(3)

= 3^(4+ ½)


Hence, LHS = RHS gives

3^(4 + 1/2) = 3^k

Or, k = 4+ ½

Hence the required value of k is 4 + 1/2 or 4.5(in decimal notation)

Answer 2

by taking tke logarithm on both sides of the equation and then dividint by log(3)

LOG { Sq root 3^7 + Sq root 3^5 + Sq root 3^3 + (42)(Sq Root 3) }= k LOG(3)

LOG { Sq root 3^7 + Sq root 3^5 + Sq root 3^3 + (42)(Sq Root 3) } / LOG(3)= k