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4^x + 4^-x = 2 What is the value of x?

a) -1
b) -1/2
c) 0
d) 1/2
e) 1

Both X and -X are present in the equation as exponents. Hence if there is any solution to the equation, there would be another solution to the equation with the same absolute value and opposite polarity and so there cannot be a unique solution. So hypothetically if -1 is the answer then +1 would also be a solution and the same for -1/2 and +1/2. '0' is the only number without polarity. Hence answer is C
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Can someone please let me know if this method is a correct way to solve this problem?

My work:
4^x+4^-x=2
2^2x+2^-2x=2^1
2^2x (1+1)=2^1
2^2x(2^1)=2^1
(Same bases, so remove them)
2x+1=1
x=0

Please let me know if you can or CANNOT do this! Thanks!
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onsameline
Can someone please let me know if this method is a correct way to solve this problem?

My work:
4^x+4^-x=2
2^2x+2^-2x=2^1
2^2x (1+1)=2^1
2^2x(2^1)=2^1
(Same bases, so remove them)
2x+1=1
x=0

Please let me know if you can or CANNOT do this! Thanks!

The red part is not correct:

\(2^{2x}+2^{-2x}=2^1\) --> \(2^{2x}+\frac{1}{2^{2x}}=2^1\) --> you cannot factor out 2^2x from LHS the way you did.

Hope it's clear.
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Exponents: math-number-theory-88376.html

All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39
All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60

Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html


Alternate Solution:

4^x + 4 ^-x = 2

Can be written as

(2^x) ^ 2 - 2 + 1/(2^x) ^ 2 = 0

(2^x - 1/2^x ) ^ 2 = 0 (The above LHS equation is a perfect square)

2^x = 1/2^x = 2^-x

So x = -x = 0 Answer = C
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nimc2012
4^x + 4^-x = 2 What is the value of x?

A. -1
B. -1/2
C. 0
D. 1/2
E. 1

If we plug from the answer options we get x = 0 as 4^0 + 4^0 = 2
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joaolucas193
If 4^x + 4^(-x) = 2, which of the following is the value of x?

A. -1
B. -1/2
C. 0
D. 1/2
E. 1



could someone explain me step by step how to solve this kind of question? pls

it's easy, but the logic did't enter in my mind

\(4^x + 4^{-x} = 2\)

Or, \(2^{2x} + 2^{-2x} = 2^1\)

Or, \(2^{2x} + \frac{1}{2^{2x}} = 2^1\)

Or, \(2^{4x} + 1 = 2\)

Or, \(2^{4x} = 1\)

Or, \(2^{4x} = 2^0\)

So, \(4x = 0\)

Or, x = 0, Answer must be (C) 0
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joaolucas193
If 4^x + 4^(-x) = 2, which of the following is the value of x?

A. -1
B. -1/2
C. 0
D. 1/2
E. 1



could someone explain me step by step how to solve this kind of question? pls

it's easy, but the logic did't enter in my mind

Yes. Only step is go by putting the values in options. When You put 0 you will get Your answer.
Or follow ZONEF preference method.
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Abhishek009
How did you get from step 3 to step 4?
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nimc2012
4^x + 4^-x = 2 What is the value of x?

A. -1
B. -1/2
C. 0
D. 1/2
E. 1

Use the options
A. 4^-1 + 4 is not 2
B. 4^-1/2 + 4^1/2 is not correct
C. 4^0 + 4^0 = 2 Correct
D. 4^1/2 + 4^-1/2 Incorrect
E. 4 + 4^-1 Incorrect.

C is correct
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This is a very good question on exponents. However, if you just stick to the concepts on exponents, you may take longer to solve this question, compared to someone who relies on logic and some common mathematical sense.

Remember that \(x^{-n}\) = \(\frac{1}{x^n}\). This means that \(4^{-x}\) can be rewritten as \(¼^x\).

Remember that \(x^0\) = 1 (when x≠0).

You need to understand that it is only \(4^0\) that will give you a value less than 4. All other powers of 4 will give you a value greater than or equal to 4.

Similarly, all other powers of \(¼^x\) will give us fractions; it’s only \(¼^0\) which will give us 1.

If you observe carefully, this is the only possible case which satisfies the above equation. In all other cases, the first term will be an integer greater than or equal to 4, while the second term will be a fraction less than or equal to ¼. The addition of these two can never give you 2.

Therefore, the value of x HAS to be 0.

The correct answer option is C.
Hope this helps!
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nimc2012
4^x + 4^-x = 2 What is the value of x?

A. -1
B. -1/2
C. 0
D. 1/2
E. 1

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

Or, \(2^{2x} + \frac{1}{2^{2x}} = 2^1\)

Or, \(2^{4x + 1} = 2^{2x + 1}\)

Or, \(4x + 1 = 2x + 1\)

Or, \(2x = 0\)

Or, \(x = 0\), Answer must be (C)
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How did you do 2^4x +1 = 2^(4x+1) ?

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nimc2012
4^x + 4^-x = 2 What is the value of x?

A. -1
B. -1/2
C. 0
D. 1/2
E. 1


As x is a real number, we can say that 4^x will always be a positive real number (as the base is positive). Additionally, we know that the arithmetic mean (AM) for any 2 positive real numbers is greater than or equal to the geometric mean (GM) of the same numbers.

Knowing this, the numbers, 4^x and 4^(-x) will follow suit. Hence we can say that 4^x+4^(-x) will always be greater than equal to 2.
In question, we know the value to be 2, which is only possible when the 2 numbers are equal to each other. Hence the value of x must be equal to 0.

P.S. This question is easily solvable without the concept of AM GM (through options), but some questions out there might use this concept.
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If I try to solve it I get:

(2^2)^X + (2^2)^(-X) = 2^1
(2)^2X + (2)^(-2X) = 2^1
2x - 2x = 1

which obviously makes no sense.
Can someone help me to understand what's wrong in my solution?
Thanks
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How did you do 2^4x +1 = 2^(4x+1) ?

Posted from my mobile device

It can not be solved that way. We have to plug in the options to get the answer.
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