If x is an integer, is 9^x + 9^(-x) = b ? (1) 3^x + 3^(-x) = (b + 2)

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

square both sides

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

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

Sufficient.

2) Tells us nothing about b, but rather that x is a positive number. Insufficient.

This was tough. I spent at least 3 minutes (during the exam is not possible)

We can rephrase the stimulus but I see a lot of options to do. Is not straight (at least for me). Ok better to look at the statements

2) this say nothing about \(=\)between the left part of equation and the right part INSUFF

1) square boot sides so we have \((3^x + 3^{-x})^2\)\(=\)\(\sqrt{b + 2}^2\)

Now we 'd have \(9^x\) that is, is the same of \({3^2}^x\) -------> \(9^x + 9^{-x} + 2 ( 3^x + 3^{-x})\)\(=\) \(b + 2\) ---------> \(9^x + 2 + 9^{-x} = b + 2\)

In the end \(9^x + 9^{-x} = b\) SUFF

A should be the answer

Note: \(2 ( 3^x + 3^{-x})\) is zero because we have \(3^ {x- x}\) . a number power zero is 1 ---> \(2*1 = 2\) for me is more than 600 level

Question please: Can we do anything with the 9^x + 9^{-x} = b or simplify any more than what is given? I tried to do something more but could not find anything proper...

I'd say \(9^x + 9^{-x}\) the simplest way of writing this expression and as you can see from the solution we don't even need to manipulate with it further to answer the question.

why is 3^x * 3^x = 9x? shouldnt it be 9^2x?

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

For more check Number Theory chapter of our Math Book: math-number-theory-88376.html

Hope it helps.

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)

Bunuel , what is x=0 ?

If x = 0, then the questions asks whether b = 2.

(1) gives \(2 = \sqrt{b+2}\) --> b = 2. Sufficient.

Hope it's clear.

Bunuel VeritasKarishma or other experts, is this correct?

Stem:
9^x+9^ -x = b?
(3^2)^x+(3^2)^-x = b?
3^2(1^x+1^-x)
3^2(2) =
18 = b?

1) 3^x+3^-x = √(b+2)
3(1^x+1^-x) = (3^x+3^-x)^2 = b+2
3(2) = 3^2x + 3^0 + 3^0 + 3^-2x = b+2
36 = b+2 3^2x + 3^-2x + 2 = b + 2
34 = b 3^2x+3^-2x = b
Definite yes, Sufficient

2) Irrelevant because base 1 will always be 1 if x is an int.

Edit: Thanks Karishma, I must have been half asleep when I did this one...
Takeaway: Don't confuse multiplication and addition of exponents when factoring out!

No energetics.

Note that we cannot add the exponents when the terms are added. We do that only when the terms are multiples.

3^a * 3^b = 3^(a + b) ---- Correct

3^a + 3^b is not 3^(a + b).

Question : is \(9^x + 9^{-x} = b\)


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

i.e. Squaring both sides \(3^{2x} + 3^{-2x} +2*(3^x*3^{-x} = √(b+2)\)

i.e. \(3^{2x} + 3^{-2x} +2 = (b+2)\)

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

i.e. answer to the question is DEFINITELY YES hence

SUFFICIENT

Statement 2: x > 0

No association of x with b is given to check the question statement's authenticity to answer the question hence

NOT SUFFICIENT

Answer: Option A

When you square sqrt(b+2), doesnt it become |b+2|, and you can get two answers?

You would get 9^x +9^(-x) = b+2 OR = -b-4

No. I think you are mixing squaring and taking the square root.

\(\sqrt{x^2}=|x|\) but \((\sqrt{y})^2=y\).

Does this make sense?

Thanks for the response; I did not know that! I had assumed that since \(\sqrt{x^2}=(\sqrt{x})^2\)), that it would be the same in this case.

I just checked back on some notes, and this is not covered in any material I can find. Is there a "name/term" for this and where I can read up on this weird interaction?

i.e. \(\sqrt{x^2}=|x|\) but \((\sqrt{y})^2=y\), but also \(\sqrt{x^2}=(\sqrt{x})^2\)?

Two points before we move to your actual question:


Now, about \(\sqrt{x^2}=|x|\). Notice that the square root there gives, as it should, non-negative result: |x| (the absolute value of a number is always nonnegative). Consider this, say we have \(\sqrt{x^2}=5\). What is the value of x? Well, x can obviously be 5: \(\sqrt{5^2}=\sqrt{25}=5\) but it can also be -5: \(\sqrt{(-5)^2}=\sqrt{25}=5\). So, as you can see \(\sqrt{x^2}=5\) means that \(\sqrt{x^2}=|x|=5\), which gives x = 5 or x = -5.

Next, \(\sqrt{x^2}\) and \((\sqrt{x})^2\) are not exactly the same. For \((\sqrt{x})^2\) to be defined on the GMAT, x, which is under the square root, must be nonnegative, so if say, it's given that \((\sqrt{x})^2=5\), then x can only be 5. But if it's given that \(\sqrt{x^2}=5\), then x can be 5 or -5.

Hope it's clear.

avigutman

I'd love to know if this question can be solved using a reasoning-based approach? It took me quite some time to arrive at the right answer even with plenty of rephrasing