The value of x satisfying the equation `2log_(10)x - log_(10) (2x-75) = 2` is

Options

A. 150
B. 50
C. 200
D. 250

Answer

To solve the equation \( 2\log_{10}x - \log_{10}(2x - 75) = 2 \), we will follow these steps:

### Step 1: Apply the logarithmic property
Using the property of logarithms that states \( b \log_a c = \log_a(c^b) \), we can rewrite the first term:
\[ 2\log_{10}x = \log_{10}(x^2) \]
So, the equation becomes:
\[ \log_{10}(x^2) - \log_{10}(2x - 75) = 2 \]

### Step 2: Use the quotient rule of logarithms
Using the property of logarithms that states \( \log_a b - \log_a c = \log_a\left(\frac{b}{c}\right) \), we can combine the logarithmic terms:
\[ \log_{10}\left(\frac{x^2}{2x - 75}\right) = 2 \]

### Step 3: Convert the logarithmic equation to exponential form
The equation \( \log_{10}\left(\frac{x^2}{2x - 75}\right) = 2 \) can be rewritten in exponential form:
\[ \frac{x^2}{2x - 75} = 10^2 \]
This simplifies to:
\[ \frac{x^2}{2x - 75} = 100 \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ x^2 = 100(2x - 75) \]
Expanding the right side:
\[ x^2 = 200x - 7500 \]

### Step 5: Rearrange the equation
Rearranging the equation leads to:
\[ x^2 - 200x + 7500 = 0 \]

### Step 6: Factor the quadratic equation
We need to factor the quadratic equation. We look for two numbers that multiply to \( 7500 \) and add to \( -200 \). The numbers \( -150 \) and \( -50 \) work:
\[ (x - 150)(x - 50) = 0 \]

### Step 7: Solve for \( x \)
Setting each factor to zero gives us the possible solutions:
\[ x - 150 = 0 \quad \Rightarrow \quad x = 150 \]
\[ x - 50 = 0 \quad \Rightarrow \quad x = 50 \]

### Step 8: Check the validity of the solutions
We need to ensure that both solutions satisfy the original logarithmic conditions:
1. \( x > 0 \)
2. \( 2x - 75 > 0 \)

For \( x = 150 \):
- \( 2(150) - 75 = 300 - 75 = 225 > 0 \) (valid)

For \( x = 50 \):
- \( 2(50) - 75 = 100 - 75 = 25 > 0 \) (valid)

Both solutions are valid.

### Final Answer
The values of \( x \) satisfying the equation are \( x = 150 \) and \( x = 50 \).

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