Question

The sum of the roots of the equation, \( x + 1 - 2\log_2(3 + 2^x) + 2\log_4(10 - 2^{-x}) = 0 \), is :

• A. \( \log_2 11 \)
• B. \( \log_2 12 \)
• C. \( \log_2 13 \)
• D. \( \log_2 14 \)

Hint:

When asked for the "sum of roots" \( x_1 + x_2 \) in equations involving \( a^x \), look for the product of roots in terms of \( t = a^x \), because \( t_1 t_2 = a^{x_1} a^{x_2} = a^{x_1+x_2} \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The equation involves logarithmic terms with different bases. We must convert them to a common base (base 2) to simplify.
Step 2: Key Formula or Approach:
Use the property \( \log_{a^k} b = \frac{1}{k} \log_a b \).
Specifically, \( 2\log_4(10 - 2^{-x}) = 2 \cdot \frac{1}{2} \log_2(10 - 2^{-x}) = \log_2(10 - 2^{-x}) \).
Step 3: Detailed Explanation:
Rewrite the equation in base 2:
\[ x + 1 - \log_2(3 + 2^x)^2 + \log_2(10 - 2^{-x}) = 0 \]
Express \( x + 1 \) as \( \log_2 2^{x+1} \):
\[ \log_2 2^{x+1} + \log_2(10 - 2^{-x}) = \log_2(3 + 2^x)^2 \]
Using log properties:
\[ \log_2[2^{x+1}(10 - 2^{-x})] = \log_2(3 + 2^x)^2 \]
Remove logs:
\[ 2 \cdot 2^x (10 - \frac{1}{2^x}) = (3 + 2^x)^2 \]
\[ 20 \cdot 2^x - 2 = 9 + (2^x)^2 + 6 \cdot 2^x \]
Let \( 2^x = t \):
\[ t^2 - 14t + 11 = 0 \]
Let the roots be \( x_1 \) and \( x_2 \). Then \( t_1 = 2^{x_1} \) and \( t_2 = 2^{x_2} \).
From the quadratic equation, the product of roots \( t_1 t_2 = 11 \).
\[ 2^{x_1} \cdot 2^{x_2} = 11 \implies 2^{x_1 + x_2} = 11 \]
Taking log base 2:
\[ x_1 + x_2 = \log_2 11 \]
Step 4: Final Answer:
The sum of roots is \( \log_2 11 \).