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555-605 (Medium)|   Algebra|   Exponents|                           
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Hi Karishma, thank you for the below. Can you tell me where this part comes from in the first calculation? I'm struggling to understand the rule: +2∗3x∗3−
KarishmaB
yb
If x is an integer, is 9^x + 9^(-x) = b ?

1) 3^x + 3^(-x) = sqrt(b+2)

2) x>0

Note that 9^x is the square of 3^x so we know that we should try to square stmnt 1.

1) \(3^x + 3^{-x} = \sqrt{(b+2)}\)
\(3^{2x} + 3^{-2x} + 2*3^x*3^{-x} = \sqrt{(b+2)}^2\)
\(9^x + 9^{-x} = b + 2 - 2 = b\)

We answer with 'Yes' and hence this statement alone is sufficient.

2) x > 0
Obviously not sufficient.

Answer (A)
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Hi Karishma, thank you for the below. Can you tell me where this part comes from in the first calculation? I'm struggling to understand the rule: +2∗3x∗3−
KarishmaB
yb
If x is an integer, is 9^x + 9^(-x) = b ?

1) 3^x + 3^(-x) = sqrt(b+2)

2) x>0

Note that 9^x is the square of 3^x so we know that we should try to square stmnt 1.

1) \(3^x + 3^{-x} = \sqrt{(b+2)}\)
\(3^{2x} + 3^{-2x} + 2*3^x*3^{-x} = \sqrt{(b+2)}^2\)
\(9^x + 9^{-x} = b + 2 - 2 = b\)

We answer with 'Yes' and hence this statement alone is sufficient.

2) x > 0
Obviously not sufficient.

Answer (A)

The term \(2 * 3^x * 3^{-x}\) comes from squaring \((3^x + 3^{-x})\) using the identity \((a + b)^2 = a^2 + b^2 + 2ab\)

Here, \(a = 3^x\) and \(b = 3^{-x}\), so:

\((3^x + 3^{-x})^2 = (3^x)^2 + (3^{-x})^2 + 2 * 3^x * 3^{-x} = 9^x + 9^{-x} + 2\)

That’s where the \(2 * 3^x * 3^{-x}\) term comes from.
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So, while solving this question, I misunderstood something

Should it be
\sqrt{} b + 2

(\sqrt{} b )+ 2

The answer remains A in both the cases, but I would like to be accurate in my learning.

Bunuel
SOLUTION

If x is an integer, is \(9^x + 9^{-x} = b\) ?

(1) \(3^x + 3^{-x} = \sqrt{b + 2}\);

Square both sides: \(9^x+2*3^x*\frac{1}{3^x}+9^{-x}=b+2\);

\(9^x + 9^{-x} = b\).

So, the answer to the question is YES. Sufficient.


(2) x > 0. No sufficient.


Answer: A.
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RBgmatPre
So, while solving this question, I misunderstood something

Should it be
\sqrt{} b + 2

(\sqrt{} b )+ 2

The answer remains A in both the cases, but I would like to be accurate in my learning.

Bunuel
SOLUTION

If x is an integer, is \(9^x + 9^{-x} = b\) ?

(1) \(3^x + 3^{-x} = \sqrt{b + 2}\);

Square both sides: \(9^x+2*3^x*\frac{1}{3^x}+9^{-x}=b+2\);

\(9^x + 9^{-x} = b\).

So, the answer to the question is YES. Sufficient.


(2) x > 0. No sufficient.


Answer: A.

It's the square root of (b + 2).
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Bunuel
If x is an integer, is \(9^x + 9^{-x} = b\) ?


(1) \(3^x + 3^{-x} = \sqrt{b + 2}\)

(2) x > 0





Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience

[email protected]
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