Key Concepts and Formulas

• Logarithm Properties: $k \log_b M = \log_b M^k$, $\log_b M - \log_b N = \log_b (M/N)$, $\log_b M + \log_b N = \log_b (MN)$, $\log_b M = 0 \iff M=1$, $\log_b a = \frac{\log_c a}{\log_c b}$.
• Vieta's Formulas: For a quadratic equation of the form $ay^2 + by + c = 0$, the product of its roots ($y_1, y_2$) is given by $y_1 y_2 = c/a$.
• Exponential and Logarithmic Forms: $a^x = b \iff x = \log_a b$.

Step-by-Step Solution

Step 1: State the equation and analyze the domain.

The equation is: $x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$ For the logarithms to be defined, we need:

1. $3 + 2^x > 0$, which is always true since $2^x > 0$ for all $x$.
2. $10 - 2^{-x} > 0 \implies 10 > 2^{-x} \implies 10 > \frac{1}{2^x} \implies 2^x > \frac{1}{10}$. This implies $x > \log_2(\frac{1}{10}) = -\log_2(10)$.

Step 2: Rewrite the equation using logarithm properties.

We want to simplify the equation using properties of logarithms. First, rewrite $x+1$ as a base-2 logarithm: $x+1 = \log_2(2^{x+1})$. Also, change the base of the second logarithm from 4 to 2 using $\log_4(a) = \frac{\log_2(a)}{\log_2(4)} = \frac{\log_2(a)}{2}$.

$\log_2(2^{x+1}) - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$ $\log_2(2^{x+1}) - 2{\log _2}(3 + {2^x}) + 2\frac{{\log _2}(10 - {2^{ - x}})}{2} = 0$ $\log_2(2^{x+1}) - 2{\log _2}(3 + {2^x}) + {\log _2}(10 - {2^{ - x}}) = 0$

Step 3: Combine the logarithms.

Use the power rule of logarithms ($a \log_b c = \log_b c^a$) and the product/quotient rules ($\log_b a + \log_b c = \log_b(ac)$ and $\log_b a - \log_b c = \log_b(a/c)$):

$\log_2(2^{x+1}) - {\log _2}((3 + {2^x})^2) + {\log _2}(10 - {2^{ - x}}) = 0$ $\log_2\left(\frac{2^{x+1}(10 - 2^{-x})}{(3+2^x)^2}\right) = 0$

Step 4: Remove the logarithm.

Since $\log_2(A) = 0$ implies $A = 1$, we have:

$\frac{2^{x+1}(10 - 2^{-x})}{(3+2^x)^2} = 1$ $2^{x+1}(10 - 2^{-x}) = (3+2^x)^2$

Step 5: Simplify and substitute.

Let $y = 2^x$. Then $2^{x+1} = 2 \cdot 2^x = 2y$ and $2^{-x} = \frac{1}{2^x} = \frac{1}{y}$. Substituting these into the equation gives:

$2y(10 - \frac{1}{y}) = (3+y)^2$ $20y - 2 = 9 + 6y + y^2$ $y^2 - 14y + 11 = 0$

Step 6: Apply Vieta's formulas.

Let $y_1$ and $y_2$ be the roots of the quadratic equation $y^2 - 14y + 11 = 0$. By Vieta's formulas, $y_1 y_2 = \frac{11}{1} = 11$.

Step 7: Relate the roots of the quadratic to the roots of the original equation.

Since $y = 2^x$, let $y_1 = 2^{x_1}$ and $y_2 = 2^{x_2}$, where $x_1$ and $x_2$ are the roots of the original equation. Then $y_1 y_2 = 2^{x_1} 2^{x_2} = 2^{x_1 + x_2} = 11$. Taking the logarithm base 2 of both sides, we get: $\log_2(2^{x_1 + x_2}) = \log_2(11)$ $x_1 + x_2 = \log_2(11)$

Common Mistakes & Tips

• Remember to check the domain of the logarithmic functions. Discard any solutions that make the arguments of the logarithms non-positive.
• When simplifying logarithmic expressions, be careful to apply the logarithm properties correctly.
• Make sure to relate the roots of the transformed equation back to the original equation.

Summary

By applying logarithm properties, changing the base of logarithms, and substituting $y = 2^x$, the given logarithmic equation was transformed into a quadratic equation $y^2 - 14y + 11 = 0$. Using Vieta's formulas, the product of the roots of the quadratic equation was found to be 11. This led to the sum of the roots of the original equation being $\log_2(11)$.

Final Answer

The final answer is \boxed{log 2 11}, which corresponds to option (B).