The value of x for which the equation `5*3^(log_3x)-2^(1-log_2x)-3=0`

Options

A. 1
B. 2
C. `(2)/(5)`
D. 7

Explanation

To solve the equation \( 5 \cdot 3^{\log_3 x} - 2^{1 - \log_2 x} - 3 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 5 \cdot 3^{\log_3 x} - 2^{1 - \log_2 x} - 3 = 0 \]

### Step 2: Apply the logarithmic property
Using the property \( a^{\log_a b} = b \), we can rewrite \( 3^{\log_3 x} \) as \( x \):
\[ 5x - 2^{1 - \log_2 x} - 3 = 0 \]

### Step 3: Simplify the second term
Next, we simplify \( 2^{1 - \log_2 x} \):
\[ 2^{1 - \log_2 x} = 2^1 \cdot 2^{-\log_2 x} = 2 \cdot \frac{1}{x} = \frac{2}{x} \]
So, substituting this back into the equation gives:
\[ 5x - \frac{2}{x} - 3 = 0 \]

### Step 4: Clear the fraction
To eliminate the fraction, multiply the entire equation by \( x \) (assuming \( x \neq 0 \)):
\[ 5x^2 - 2 - 3x = 0 \]
Rearranging gives:
\[ 5x^2 - 3x - 2 = 0 \]

### Step 5: Factor the quadratic equation
Now, we will factor the quadratic equation:
\[ 5x^2 - 5x + 2x - 2 = 0 \]
Grouping the terms:
\[ (5x^2 - 5x) + (2x - 2) = 0 \]
Factoring out common terms:
\[ 5x(x - 1) + 2(x - 1) = 0 \]
Factoring further gives:
\[ (5x + 2)(x - 1) = 0 \]

### Step 6: Solve for \( x \)
Setting each factor to zero:
1. \( 5x + 2 = 0 \) leads to \( x = -\frac{2}{5} \)
2. \( x - 1 = 0 \) leads to \( x = 1 \)

### Step 7: Determine valid solutions
Since \( x \) must be greater than 0 (as logarithm is not defined for non-positive values), we discard \( x = -\frac{2}{5} \). Thus, the only valid solution is:
\[ \boxed{1} \]