Is 7^(x + 2)/49 1 ?

Question

Is  \frac{7^{(x+2)}}{49} > 1  ?

(1)  7^{(x - 2)} > \frac{1}{49}

(2)  7^{(x - 1)} > \frac{1}{49}

chetan2u · Answer

Slightly tricky q..

\frac{7^{(x+2)}}{49} > 1..........7^x>1  MEANS --- Is x>0?

lets see the statements
(1)  7^{(x - 2)} > \frac{1}{49}...........7^{(x-2+2)}>1...........7^x>1 
Sufficient

(2)  7^{(x - 1)} > \frac{1}{49}...........7^{(x-1+2)}>1.......7^{(x+1)}>0 
so x+1>0....x>-1
if x is between -1 and 0 , ans is NO
otherwise Yes
insuff

A

exc4libur · Answer

Simplify:  \frac{7^{(x+2)}}{49} > 1 = \frac{7^{(x+2)}}{(7^2)} > 1 =7^{(x+2)-2}> 1.

Question: Is  x > 0?

(1)  7^{(x - 2)} > \frac{1}{49} = 7^{(x - 2)} > \frac{1}{7^2} = 7^{(x - 2)} > 7^{(-2)} = x-2>-2 = x > 0.  Sufficient.

(2)  7^{(x - 1)} > \frac{1}{49} = 7^{(x - 1)} > \frac{1}{7^2} = 7^{(x - 1)} > 7^{(-2)} = x-1>-2 = x-1+2 > 0 = x > -1.  Insufficient.

(A) is the answer.

philipssonicare · Answer

It's mentioned in both responses that the question is asking is x>0? But what if x is a fraction, e.g. 7^1/100, still less than 1?

firas92 · Answer

7^0 is exactly 1. So 7 raised to any power greater than 0 will always be greater than 1.

As a specific example, in the expression 7^n, as the value of the power (n) progresses from 0 to 1, the value of the expression (7^n) will progress from 1 to 7

Hope this is clear!

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Is 7^(x + 2)/49 > 1 ?

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