Question:

Step 1: Solve the logarithmic equation. The given equation is $x^{\log_2 x - 4} = 32$. For $\log_2 x$ to be defined, $x > 0$. Taking $\log_2$ of both sides of the equation: $$ \log_2 \left( x^{\log_2 x - 4} \right) = \log_2 32 $$ Using the logarithm property $\log_b(a^c) = c \log_b a$ and knowing $\log_2 32 = 5$: $$ (\log_2 x - 4) \log_2 x = 5 $$ Step 2: Solve for $\log_2 x$. Let $t = \log_2 x$. Substituting $t$ into the equation yields a quadratic equation: $$ t(t - 4) = 5 $$ $$ t^2 - 4t - 5 = 0 $$ Factoring the quadratic equation: $$ (t - 5)(t + 1) = 0 $$ This gives two solutions for $t$: $t = 5$ or $t = -1$. Step 3: Determine the values of $x_1$ and $x_2$. Since $t = \log_2 x$: If $t = 5$, then $\log_2 x = 5 \implies x = 2^5 = 32$. If $t = -1$, then $\log_2 x = -1 \implies x = 2^{-1} = \frac{1}{2}$. Given that $x_1 > x_2$, we identify the solutions as $x_1 = 32$ and $x_2 = \frac{1}{2}$. Step 4: Calculate the product $x_1 \cdot x_2$. $$ x_1 \cdot x_2 = 32 \cdot \frac{1}{2} = 16 $$ Alternatively, using the properties of logarithms, $\log_2(x_1 x_2) = \log_2 x_1 + \log_2 x_2 = t_1 + t_2$. From the quadratic equation $t^2 - 4t - 5 = 0$, the sum of the roots is $t_1 + t_2 = -(-4)/1 = 4$. Thus, $\log_2(x_1 x_2) = 4$, which implies $x_1 x_2 = 2^4 = 16$. Step 5: Calculate the sum $x_1 + x_2$. $$ x_1 + x_2 = 32 + \frac{1}{2} = \frac{64}{2} + \frac{1}{2} = \frac{65}{2} $$ Step 6: Determine the characteristic of $\log_3(x_1)$. $$ \log_3(x_1) = \log_3 32 $$ Since $3^3 = 27$ and $3^4 = 81$, it follows that $3^3 < 32 < 3^4$. Therefore, $3 < \log_3 32 < 4$. The characteristic of $\log_3(x_1)$ is the integer part of $\log_3 32$, which is 3. Step 7: Determine the mantissa of $\log_2(x_2)$. $$ \log_2(x_2) = \log_2 \left( \frac{1}{2} \right) = \log_2 (2^{-1}) = -1 $$ The mantissa of a number is the non-negative fractional part when the number is expressed as the sum of an integer (characteristic) and a non-negative fractional part. For $-1$, it can be written as $-1 + 0$. Thus, the mantissa of $\log_2(x_2)$ is 0.