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4^x + 4^-x = 2, x = ?

Expert replies
Source: — Data Sufficiency |

by GMATGuruNY » Wed Aug 28, 2013 3:01 am
gmattesttaker2 wrote:
If 4^x + 4^(-x) = 2, what is the value of x?

-1
-1/2
0
1/2
1
We can plug in the answers, which represent the value of x.
A quick scan shows that only C will work:

Answer choice C: x=0
4� + 4� = 2
1 + 1 = 2
2 = 2.
Success!

The correct answer is C.

Here's one way to solve algebraically:

Let 4^x = a.
The equation becomes:
a + a¯¹ = 2.

Solving the rephrased equation, we get:
a + 1/a = 2
(a² + 1)/a = 2
a² + 1 = 2a
a² - 2a + 1 = 0
(a-1)² = 0
a = 1.

Since a = 4^x, we get:
4^x = 1
x = 0.

Plugging in the answers seems easier and faster.
Last edited by GMATGuruNY on Wed Aug 28, 2013 7:40 am, edited 1 time in total.
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by Brent@GMATPrepNow » Wed Aug 28, 2013 5:59 am
gmattesttaker2 wrote:
If 4^x + 4^(-x) = 2, what is the value of x?

A) -1
B) -1/2
C) 0
D) 1/2
E) 1
As you can see, solving this question algebraically can take a lot of time.
Mitch pointed out that checking the answer choices is the best route, and I thought I'd quickly mention that we can quickly eliminate some of the answer choices.

Notice that, if b^k is an integer, then b^(-k) will not be an integer, and vice versa. The only time when this is not true is when k=0, or when b = 0, 1 or -1. Since the base (b) equals 4 in this question, the second part of that proviso doesn't count here.

So, for example, when we examine answer choice A (x=-1), we can see that 4^(-x) = 4^1 = 4. Since 4^(-x) is an integer, we immediately know that 4^x is not an integer. Since there's no way that an integer plus a non-integer can equal 2, we can eliminate A.
Likewise, without performing any calculations, we can eliminate E.

The same applies to answer choices B and D, HOWEVER, if we're checking answer choices, we should be checking the easiest options first. So, once we eliminate 1 and -1 (answers A and E), we should probably check C (x=0) before we start checking fractions.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by gmattesttaker2 » Wed Aug 28, 2013 6:54 pm
GMATGuruNY wrote:
gmattesttaker2 wrote:
If 4^x + 4^(-x) = 2, what is the value of x?

-1
-1/2
0
1/2
1
We can plug in the answers, which represent the value of x.
A quick scan shows that only C will work:

Answer choice C: x=0
4� + 4� = 2
1 + 1 = 2
2 = 2.
Success!

The correct answer is C.

Here's one way to solve algebraically:

Let 4^x = a.
The equation becomes:
a + a¯¹ = 2.

Solving the rephrased equation, we get:
a + 1/a = 2
(a² + 1)/a = 2
a² + 1 = 2a
a² - 2a + 1 = 0
(a-1)² = 0
a = 1.

Since a = 4^x, we get:
4^x = 1
x = 0.

Plugging in the answers seems easier and faster.
Hello Mitch,

Thank you very much for both the approaches to this question. Thanks again for all your help.

Best Regards,
Sri
Join the discussion

by gmattesttaker2 » Wed Aug 28, 2013 6:56 pm
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote:
If 4^x + 4^(-x) = 2, what is the value of x?

A) -1
B) -1/2
C) 0
D) 1/2
E) 1
As you can see, solving this question algebraically can take a lot of time.
Mitch pointed out that checking the answer choices is the best route, and I thought I'd quickly mention that we can quickly eliminate some of the answer choices.

Notice that, if b^k is an integer, then b^(-k) will not be an integer, and vice versa. The only time when this is not true is when k=0, or when b = 0, 1 or -1. Since the base (b) equals 4 in this question, the second part of that proviso doesn't count here.

So, for example, when we examine answer choice A (x=-1), we can see that 4^(-x) = 4^1 = 4. Since 4^(-x) is an integer, we immediately know that 4^x is not an integer. Since there's no way that an integer plus a non-integer can equal 2, we can eliminate A.
Likewise, without performing any calculations, we can eliminate E.

The same applies to answer choices B and D, HOWEVER, if we're checking answer choices, we should be checking the easiest options first. So, once we eliminate 1 and -1 (answers A and E), we should probably check C (x=0) before we start checking fractions.

Cheers,
Brent

Hello Brent,

Thank you very much for explaining this technique. Thanks again for all your help.

Best Regards,
Sri
Join the discussion