If AM of the number `5^(1+x) and 5^(1-x)` is 13 then the set of possible real values of x is -

Options

A. `5,(1)/(5)`
B. {-1,1}
C. {0,1}
D. none of these

Answer

To solve the problem step by step, we need to find the set of possible real values of \( x \) given that the arithmetic mean (AM) of the numbers \( 5^{(1+x)} \) and \( 5^{(1-x)} \) is equal to 13.

### Step 1: Write the expression for the AM
The arithmetic mean of two numbers \( a \) and \( b \) is given by:
\[ AM = \frac{a + b}{2} \]
In our case, we have:
\[ AM = \frac{5^{(1+x)} + 5^{(1-x)}}{2} = 13 \]

### Step 2: Set up the equation
From the AM expression, we can set up the equation:
\[ \frac{5^{(1+x)} + 5^{(1-x)}}{2} = 13 \]
Multiplying both sides by 2 gives:
\[ 5^{(1+x)} + 5^{(1-x)} = 26 \]

### Step 3: Simplify the expression
Using the property of exponents, we can rewrite \( 5^{(1+x)} \) and \( 5^{(1-x)} \):
\[ 5^{(1+x)} = 5 \cdot 5^x \quad \text{and} \quad 5^{(1-x)} = 5 \cdot 5^{-x} \]
Substituting these into the equation gives:
\[ 5 \cdot 5^x + 5 \cdot 5^{-x} = 26 \]
Dividing the entire equation by 5 results in:
\[ 5^x + 5^{-x} = \frac{26}{5} \]

### Step 4: Let \( y = 5^x \)
Let \( y = 5^x \). Then \( 5^{-x} = \frac{1}{y} \). Substituting these into the equation gives:
\[ y + \frac{1}{y} = \frac{26}{5} \]

### Step 5: Multiply through by \( y \)
To eliminate the fraction, multiply through by \( y \):
\[ y^2 + 1 = \frac{26}{5}y \]
Rearranging gives:
\[ 5y^2 - 26y + 5 = 0 \]

### Step 6: Solve the quadratic equation
Now we will solve the quadratic equation \( 5y^2 - 26y + 5 = 0 \) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 5 \), \( b = -26 \), and \( c = 5 \):
\[ y = \frac{26 \pm \sqrt{(-26)^2 - 4 \cdot 5 \cdot 5}}{2 \cdot 5} \]
Calculating the discriminant:
\[ (-26)^2 - 4 \cdot 5 \cdot 5 = 676 - 100 = 576 \]
Thus:
\[ y = \frac{26 \pm \sqrt{576}}{10} = \frac{26 \pm 24}{10} \]
Calculating the two possible values for \( y \):
1. \( y = \frac{50}{10} = 5 \)
2. \( y = \frac{2}{10} = \frac{1}{5} \)

### Step 7: Find \( x \) from \( y \)
Recall that \( y = 5^x \):
1. If \( y = 5 \), then \( 5^x = 5 \) implies \( x = 1 \).
2. If \( y = \frac{1}{5} \), then \( 5^x = \frac{1}{5} \) implies \( x = -1 \).

### Conclusion
The set of possible real values of \( x \) is:
\[ \{ -1, 1 \} \]