Question

A sequence is 6^(n-1)     B sequence is 50(3^(n-1))

Set them equal and solve for n

 

6^(n-1) = 50(3 ^(n-1))

 Take log of both sides

 

(n-1)Log 6 = Log50 +(n-1)log3 

(n-1)(log6-log3) = log 50

n-1 = log50 /(log6-log3)

n = log50/(log6-log3) + 1

n= 6.64    This is where they would EQUAL,,,,the nex integer is where A>B           = 7

the integer after which A>B is 6

Sequence A:  a_n = 6^(n-1)  or  a_n = 2^(n-1)*3^(n-1)

 

Sequence B:  b_n = 50*3^(n-1)

 

For A to be larger than B:   2^(n-1)*3^(n-1)  >  50*3^(n-1)   

 

or   2^(n-1)  >  50  

 

The smallest value of n for which 2^(n-1)  >  50  is n = 7  (as 2⁵ = 32 and 2⁶ = 64) 

 

Hence after 7 terms A is larger than B.