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by sanju09 » Fri Feb 27, 2009 5:43 am
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
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by Rag_gmat » Fri Feb 27, 2009 5:51 am
x>2/3
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by mjjking » Fri Feb 27, 2009 7:24 am
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?
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by DanaJ » Fri Feb 27, 2009 7:34 am
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
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by mygmat.2009 » Fri Feb 27, 2009 7:34 am
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)



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by marouan » Fri Feb 27, 2009 7:37 am
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 !!
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by mjjking » Fri Feb 27, 2009 7:52 am
thanks folks, stupid mistake!! :oops:
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by sanju09 » Sat Feb 28, 2009 4:50 am
If 3^x > 3^y, does it always mean x > y?
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by DanaJ » Sat Feb 28, 2009 4:55 am
The exponential function is an injective function, so my answer will be yes.
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by sanju09 » Sat Feb 28, 2009 5:01 am
DanaJ wrote:The exponential function is an injective function, so my answer will be yes.
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by senthil » Sat Feb 28, 2009 8:54 am
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 !!
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