The value of x which satisfy the equation `5^(1+x)+5^(1-x) = 26`

Options

A. -1,1
B. 0,1
C. 1,2
D. -1,0

Answer

To solve the equation \( 5^{1+x} + 5^{1-x} = 26 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 5^{1+x} + 5^{1-x} = 26 \]
We can express \( 5^{1+x} \) and \( 5^{1-x} \) as:
\[ 5^{1+x} = 5 \cdot 5^x \quad \text{and} \quad 5^{1-x} = 5 \cdot 5^{-x} \]
Thus, we can rewrite the equation as:
\[ 5 \cdot 5^x + 5 \cdot 5^{-x} = 26 \]

### Step 2: Factor out the common term
Now, we can factor out the common term \( 5 \):
\[ 5(5^x + 5^{-x}) = 26 \]
Dividing both sides by 5 gives:
\[ 5^x + 5^{-x} = \frac{26}{5} \]

### Step 3: Substitute for simplification
Let \( y = 5^x \). Then \( 5^{-x} = \frac{1}{y} \). Substituting these into the equation gives:
\[ y + \frac{1}{y} = \frac{26}{5} \]

### Step 4: Multiply through by \( y \)
To eliminate the fraction, multiply through by \( y \):
\[ y^2 + 1 = \frac{26}{5}y \]
Rearranging gives:
\[ y^2 - \frac{26}{5}y + 1 = 0 \]

### Step 5: Solve the quadratic equation
Now we can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -\frac{26}{5}, c = 1 \):
\[ b^2 - 4ac = \left(-\frac{26}{5}\right)^2 - 4 \cdot 1 \cdot 1 = \frac{676}{25} - 4 = \frac{676}{25} - \frac{100}{25} = \frac{576}{25} \]
Thus,
\[ y = \frac{\frac{26}{5} \pm \sqrt{\frac{576}{25}}}{2} \]
Calculating the square root:
\[ \sqrt{\frac{576}{25}} = \frac{24}{5} \]
So we have:
\[ y = \frac{\frac{26}{5} \pm \frac{24}{5}}{2} \]
This gives us two potential solutions:
\[ y = \frac{50/5}{2} = 5 \quad \text{and} \quad y = \frac{2/5}{2} = \frac{1}{5} \]

### Step 6: Back substitute for \( x \)
Recall that \( y = 5^x \):
1. For \( y = 5 \):
\[ 5^x = 5 \implies x = 1 \]
2. For \( y = \frac{1}{5} \):
\[ 5^x = \frac{1}{5} \implies x = -1 \]

### Conclusion
The values of \( x \) that satisfy the equation are:
\[ x = 1 \quad \text{and} \quad x = -1 \]