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

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

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.

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

If we plug from the answer options we get x = 0 as 4^0 + 4^0 = 2

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

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

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.

Abhishek009
How did you get from step 3 to step 4?

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

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!

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

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)

How did you do 2^4x +1 = 2^(4x+1) ?

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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.

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

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

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