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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
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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
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