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The sum of all the values of a satisfyin...

The sum of all the values of a satisfying the equation `|[log_10 a,-1],[log_10(a-1),2]|=log_10 a+log_10 2`

A

0

B

1

C

2

D

none of these

Text Solution

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The correct Answer is:
To solve the equation \( |[\log_{10} a, -1], [\log_{10}(a-1), 2]| = \log_{10} a + \log_{10} 2 \), we will follow these steps: ### Step 1: Calculate the Determinant The determinant of the matrix \( |[\log_{10} a, -1], [\log_{10}(a-1), 2]| \) can be calculated using the formula for the determinant of a 2x2 matrix: \[ \text{det} = ad - bc \] where \( a = \log_{10} a \), \( b = -1 \), \( c = \log_{10}(a-1) \), and \( d = 2 \). Thus, we have: \[ \text{det} = (\log_{10} a)(2) - (-1)(\log_{10}(a-1)) = 2 \log_{10} a + \log_{10}(a-1) \] ### Step 2: Set the Determinant Equal to the Right Side Now we set the determinant equal to the right side of the equation: \[ 2 \log_{10} a + \log_{10}(a-1) = \log_{10} a + \log_{10} 2 \] ### Step 3: Simplify the Equation We can simplify the equation: \[ 2 \log_{10} a + \log_{10}(a-1) - \log_{10} a - \log_{10} 2 = 0 \] This simplifies to: \[ \log_{10} a + \log_{10}(a-1) - \log_{10} 2 = 0 \] ### Step 4: Use Logarithm Properties Using the property of logarithms that states \( \log_b x + \log_b y = \log_b(xy) \), we can rewrite the equation: \[ \log_{10}(a(a-1)) = \log_{10} 2 \] ### Step 5: Exponentiate Both Sides Exponentiating both sides gives: \[ a(a-1) = 2 \] ### Step 6: Rearrange the Equation Rearranging the equation leads to: \[ a^2 - a - 2 = 0 \] ### Step 7: Factor the Quadratic Equation Now we can factor the quadratic: \[ (a - 2)(a + 1) = 0 \] ### Step 8: Solve for \( a \) Setting each factor to zero gives us the solutions: \[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \] \[ a + 1 = 0 \quad \Rightarrow \quad a = -1 \] ### Step 9: Consider Validity of Solutions Since \( a \) must be greater than 1 (as \( \log_{10}(a-1) \) must be defined), we discard \( a = -1 \). Thus, the only valid solution is: \[ a = 2 \] ### Step 10: Find the Sum of All Valid Values of \( a \) Since the only valid value of \( a \) is 2, the sum of all values of \( a \) is: \[ \text{Sum} = 2 \]

To solve the equation \( |[\log_{10} a, -1], [\log_{10}(a-1), 2]| = \log_{10} a + \log_{10} 2 \), we will follow these steps: ### Step 1: Calculate the Determinant The determinant of the matrix \( |[\log_{10} a, -1], [\log_{10}(a-1), 2]| \) can be calculated using the formula for the determinant of a 2x2 matrix: \[ \text{det} = ad - bc \] where \( a = \log_{10} a \), \( b = -1 \), \( c = \log_{10}(a-1) \), and \( d = 2 \). ...
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