Solve for x: `5^(log x) + 5x^(log 5) =3 (a>0)`

Options

A. `2^("log"_(a)5)`
B. `2^(-"log"_(a)5)`
C. `2^(-"log"_(5)a)`
D. `2^("log"_(5)a)`

Explanation

To solve the equation \( 5^{\log x} + 5x^{\log 5} = 3 \) (where \( a > 0 \)), we will follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ 5^{\log x} + 5x^{\log 5} = 3 \]

### Step 2: Apply logarithmic properties
Using the property of logarithms that states \( x^{\log y} = y^{\log x} \), we can rewrite \( 5x^{\log 5} \):
\[ 5^{\log x} + 5^{1} \cdot x^{\log 5} = 3 \]
This can be rewritten as:
\[ 5^{\log x} + 5 \cdot 5^{\log x} = 3 \]

### Step 3: Combine like terms
Now, we can factor out \( 5^{\log x} \):
\[ (1 + 5) \cdot 5^{\log x} = 3 \]
This simplifies to:
\[ 6 \cdot 5^{\log x} = 3 \]

### Step 4: Isolate \( 5^{\log x} \)
Dividing both sides by 6 gives:
\[ 5^{\log x} = \frac{3}{6} = \frac{1}{2} \]

### Step 5: Take logarithm on both sides
Taking logarithm base 5 on both sides:
\[ \log x = \log_5\left(\frac{1}{2}\right) \]

### Step 6: Use the change of base formula
Using the change of base formula, we can express this as:
\[ \log x = -\log_5(2) \]

### Step 7: Solve for \( x \)
This implies:
\[ x = 2^{-\log_5(2)} \]
Using the property \( a^{\log_b(c)} = c^{\log_b(a)} \), we can rewrite this as:
\[ x = 2^{-\log_5(2)} = 2^{\log_5(2^{-1})} = 2^{\log_5\left(\frac{1}{2}\right)} \]

### Step 8: Final expression
Thus, we can express \( x \) as:
\[ x = 2^{-\log_5(2)} = 2^{\log_5(2^{-1})} = 5^{-\log_2(2)} = 5^{-1} = \frac{1}{5} \]

### Conclusion
The solution for \( x \) is:
\[ x = \frac{1}{5} \]