exponent ds

2 ^ (2m+1) = 2 ^ (n+2)

2^2m.2=2^n.2^2

2^2m=2^n.2

2m=n+1--> equ 1


St 1 2 ^ (3n-1) = 256

2^3n/2 = 2^8
2^3n=2^9

3n = 9
n=3
Substituting in equ 1 you get m = 2---SUFF

St 2 2 ^ (m+2n) = 256

2^m.2^2n = 2^8

m+2n = 8

From equ 1 --> 2m=n+1; m=(n+1)/2

m+2n=8

(n+1)/2+2n=8

n=3, m=2---SUFF

Answer D

Thanks cramya for posting this warm up question. Hope to see more such questions. Thanks logitech for taking the initiative.

mals24 wrote:2 ^ (2m+1) = 2 ^ (n+2)

2^2m.2=2^n.2^2

2^2m=2^n.2

2m=n+1--> equ 1


St 1 2 ^ (3n-1) = 256

2^3n/2 = 2^8
2^3n=2^9

3n = 9
n=3
Substituting in equ 1 you get m = 2---SUFF

St 2 2 ^ (m+2n) = 256

2^m.2^2n = 2^8

m+2n = 8

From equ 1 --> 2m=n+1; m=(n+1)/2

m+2n=8

(n+1)/2+2n=8

n=3, m=2---SUFF

Answer D

Thanks cramya for posting this warm up question. Hope to see more such questions. Thanks logitech for taking the initiative.

Cool !

However, we are not supposed to solve in this detail..

2 variables, 2 linear equations ...sufficient....

We have to be clear in the subtle difference in DS and PS

However, if you get confidence by doing this, do it...however try to speed it up as...we cannot afford more than 2 min

@iamcste

Lolz i just explained it in detail so that people should not have any problem in understanding any step. Its just to make things clear. Thanks for the condensed version btw.

mals24, I enjoy reading and learning from your solutions. Thanks!