Is 7^x > 100 ? (1) 7^(x+2) > 9,800 (2) 7^(2x) > 10,000

Target question: Is \(7^x > 100\)

Statement 1: \(7^{(x+2)} > 9,800\)
Rewrite as: \((7^x)(7^2) > 9,800\)
In other words: \((7^x)(49) > 9,800\)
Divide both sides by 49 to get: \((7^x) > 200\)
Since 200 > 100, we can write: \((7^x) > 200 >100\)
So, the answer to the target question is YES, 7^x IS greater than 100
Since we can answer the target question with certainty, statement 1 is SUFFICIENT


Statement 2: \(7^{(2x)} > 10,000\)
Rewrite as: \((7^x)^2 > 100^2\)
This means: \(7^x > 100\)
The answer to the target question is YES, 7^x IS greater than 100
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent
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Hi BrentGMATPrepNow,

Square root of x^2 can be either +x or -x. On the same lines, could you please explain why we ignored the negative possibilities in statement 2? If we were to assume there are 2 possibilities - positive and negative, statement 2 would have been insufficient. Please help.

Thanks,
Harshbir

\(7^x > 100\)..

(1) \(7^{(x+2)} > 9,800........7^x*7^2>9800......7^x*49>9800.......7^x>200>100.......7^x>100\)
Yes and sufficient

(2) \(7^{(2x)} > 10,000....(7^x)^2>100^2........7^x>100\)
Yes and sufficient

D

harshbirsingh, on your query that \(7^x<-100\) should also be checked.
\(7^x\) can never be negative. Even if x is some negative quantity to the tune of say -10000000, \(7^{-10000000}=\frac{1}{7^{10000000}}>0\), so \(7^x<-100\) is NOT valid.

Since the base, 7, is positive, we can be certain that 7^x is positive for all values of x.

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