one or none

[sanju09]

Solve the inequality 3^(3x-2) > 1.

(A) x > 1
(B) x > 3
(C) x > 2/3
(D) x > 1/3
(E) none of the above

[Rag_gmat]

x>2/3

[mjjking]

x>1 IMO.

to be 3^3x-2>1 we need 3x-2>=1, hence 3x-2 >=1 --> x>=1

Am I wrong?

if x>2/3, x could be 3/4 and 3^(3*3/4-2)<1... right?

[DanaJ]

mjjking: in order to solve this inequality, you need to remember that a^0 = 1, with a being a real number except 0. This means that 3^(3x - 2) > 1 is equivalent to 3^(3x - 2) > 3^0 or that 3x - 2 > 0.

This will in turn provide the answer : 3x > 2, x > 2/3

[mygmat.2009]

if 3x-2 = 0, we get 3^0 = 1, so we want anything greater than that

==> 3x - 2 > 0
==> 3x > 2
==> x > 2/3


IMO
Ans is (C)



Cheers,
Sam

[marouan]

Am I wrong? Ya .. you miss it on that part :
to be 3^3x-2>1 we need 3x-2>=1,
you rather need it to be more than 0 , because 3^0=1 !!!!!!
so here is the solution ( just like you did , but replacing the 1 by 0)

3x-2>0 => x > 2/3 !!

[mjjking]

thanks folks, stupid mistake!!

[sanju09]

If 3^x > 3^y, does it always mean x > y?

[DanaJ]

The exponential function is an injective function, so my answer will be yes.

[sanju09]

The exponential function is an injective function, so my answer will be yes.

Take your time

[senthil]

I'd rather blindly folow it the way I have learnt to solve ...

let a = 3x-2 ;

3^a > 1 ... Take log on both sides

alog3 > log1... log (1) is always zero.

therfore , a>0
3x-2>0
3x>2
x> 2/3......


Hope thats clear !!