What is the value of 2^x + 2^(-x) ? (1) x < 0 (2) 4^x + 4^(−x) = 23

OA: B
What is the value of \(2^x + 2^{-x}\) ?
or What is value of \(x\)?

(1) : \(x < 0\)
plugging \(x =-1\)
\(2^x + 2^{-x} = 2^{-1}+2^{1}= 0.5 + 2 =2.5\)
plugging \(x =-2\)
\(2^x + 2^{-x} = 2^{-2}+2^{2}= \frac{1}{4} + 4 = 0.25 + 4 = 4.25\)
We are not getting a unique value of \(2^x + 2^{-x}\)
So Statement 1 alone is not sufficient

(2) \(4^x + 4^{−x} = 23\)
Let \(2^x + 2^{-x}=y\)
Squaring both sides, we get
\((2^x + 2^{-x})^2=y^2\)
\((2^x)^2 +(2^{-x})^2 + 2(2^x)(2^{-x}) =y^2\)
\(2^x.2^x +2^{-x}.2^{-x} + 2 =y^2\)
\(4^x + 4^{−x} + 2 =y^2\)
\(23 + 2 =y^2\)
\(y^2=25 ; y=+5,-5\) (\(-5\) rejected as minimum value of \(2^x + 2^{-x}\) is \(2\) at \(x =0\))
using A.M≥ G.M
\(\frac{2^x + 2^{-x}}{2}≥\sqrt{{2^x * 2^{-x}}}\)
\(2^x + 2^{-x}≥2\)
So Statement 2 alone is sufficient

Wouldn't root 25 yeild +-5 as two values making it insufficient ?

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. Even roots have only a positive value on the GMAT. That is, \(\sqrt{25}=5\), NOT +5 or -5.

In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5.

VeritasKarishma

Hi!! Have you found any other way to solve this question? I found it hard to know that I had to elevate 2^x.... to the 2...

It is a standard application of the algebraic identity
\((x + y)^2 = x^2 + y^2 + 2xy\)

\((a + 1/a)^2 = a^2 + 1/a^2 + 2\)

Assume 2^x = a to see how it is applicable.

Since statement 2 gives you square of 2^x and 2^-x, and the original question has 2^x and 2^-x, you should identify the identity. Note that without using the algebraic identities, questions based on those can seem almost impossible with the available resources.
Practice some more question based on algebraic identities.

VeritasKarishma nick1816 GMATBusters

Since we don't have to calculate the value, I used below approach, please tell me if it is correct.

Statement 2 ((2) 4^x + 4^(−x) = 23) has only one variable so we can calculate the value of "x" hence it is sufficient.

Am I missing something here?

bhai use this variable approach in linear equations only and also, make sure the equations you're using are consistent (we have only one equation here;so, we don't bother about consistency part.)

Let's look at this question. We can observe that if some random value of x (let say k) satisfies the equation, '-k' will satisfy it too (since 4^x+4^{-x} is symmetric about y-axis). So, if we have to find the value of x, Statement 2 is insufficient, as we get 2 values of x (k and -k).

Exactly what nick1816 said!

If 4^a + 4^(-a) = 17/4

a can be 1 or -1.

The point here is that 2^x + 2^(-x) will still have a unique value.

Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

The first step of the Variable Approach: The first step and the priority is to modify and recheck the original condition and the question to suit the type of information given in the condition.

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Learn the 3 steps. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find the value of \(2^x + 2^{-x}\).

=> \(2^x + 2^{-x}\) = \(2^x + \frac{1 }{ (2^x)}\)

=> Squaring both the sides, we get: \(2^{2x} + 2^{-2x} + 2 = 4^x + 4^{-x} + 2\)

Condition(1) tells us that x < 0.

=> For various values of 'x' less than '0', we will have various values for the above expression

Since the answer is not unique , condition(1) alone is not sufficient by CMT 2.


Condition(2) tells us that\( 4^x + 4^{-x} = 23\) .

=> \(4^x + 4^{-x} + 2 : 23 + 2 = 25\)

=> \((2^x + 2^{-x})^ 2 = 25 \)

=> \(2^x + 2^{-x} = 5\)

Since the answer is unique , condition(2) alone is sufficient by CMT 2.

Condition(2) alone is sufficient.

So, B is the correct answer.

Answer: B


SAVE TIME: By Variable Approach[MODIFICATION], check the condition quickly and separately and mark answer as A or B.

Already good explanations here (thought I couldn't comprehend them) but another way to look at it is that for any value of x, 2^x OR 2^-x both will always be positive, so their sum also can never be negative and thus -5 can simply be neglected as it isn't possible.

Mathematically

function: 2^x+(1/2^x) can never be negative for any value of x. So you can check this way. This is a more rational and appropriate approach rather than saying on GMAT "the only accepted answer is the positive root."

Not sure what you are trying to say there but what is written in my post is true and is an answer to the question asked by user "is \(\sqrt{25}\) both 5 and -5".


Mathematically, \(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

Hope it helps.

{2^x +2^(-x)}^2 = 25
2^x + 2^(-x) can be either +5 or -5. While squaring them, we get 25. Isn't it?
Confused on this part.. ended up choosing C

Bunuel, avigutman, I understand that we squaring up the expression given in the question stem and then identifying how it relates to second statement is the way to answer this question.

Can't we divide this expression 4^x + 4^(−x) = 23 by 2 on both sides to arrive at 2^x + 2^-x = 23/2?

Please let me know.

Thank you!

Ah, CrushTHYGMAT, dividing 4^x by 2 doesn't get you 2^x.
Think about it this way: 4^x = 4*4*4*4*4*...4*4 [x factors of 4 multiplied by one another]
So if you divide that by 2, you get 4*4*4*4*4*...4*2 [so everything is the same, except for that last factor, which is 2 rather than 4]
In order to get 2^x, you'd have to divide 4^x by 2^x.

Bunuel
for statement 2, why do you first need to factor out (2^x + 2^-x)^2 .... why can't you just take the square root of this to get what I have below
2^x + 2^-x = Squareroot(23)

Because, generally, \(\sqrt{a + b}\neq \sqrt{a }+ \sqrt{b}\). Does \(\sqrt{16 + 9}\) equal to \( \sqrt{16}+ \sqrt{9}\)?