* Penyelesaiaan dari persamaan $$ imes \log 2+ $$ $$ \mathrm{x} \log (3 \mathrm{x}-4)=2 $$ adalah $$ \mathrm{x}_{1} $$ dan $$ \mathrm{x}_{2} $$. Tentukan nilai $$ \mathrm{x}_{1}+\mathrm{x} 2 $$ adalah $$ \ldots $$. A. 6 B. 8 C. 10 D. 12 E. 20

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* Penyelesaiaan dari persamaan $$ 	imes \log 2+ $$ $$ \mathrm{x} \log (3 \mathrm{x}-4)=2 $$ adalah $$ \mathrm{x}_{1} $$ dan $$ \mathrm{x}_{2} $$. Tentukan nilai $$ \mathrm{x}_{1}+\mathrm{x} 2 $$ adalah $$ \ldots $$. A. 6 B. 8 C. 10 D. 12 E. 20

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Alright, I need to solve the equation $$ x \log 2 + \mathrm{x} \log (3x - 4) = 2 $$ and find the sum of the solutions $$ x_1 + x_2 $$. Let's break this down step by step.

First, I'll rewrite the equation to make it clearer:
$$ x \log 2 + x \log (3x - 4) = 2 $$

I notice that both terms on the left side have an $$ x $$ multiplied by a logarithm. Maybe I can factor out the $$ x $$:
$$ x (\log 2 + \log (3x - 4)) = 2 $$

Now, I'll divide both sides by $$ \log 2 + \log (3x - 4) $$ to isolate $$ x $$:
$$ x = \frac{2}{\log 2 + \log (3x - 4)} $$

This seems a bit complicated because $$ x $$ is still inside the logarithm. Perhaps I can make a substitution to simplify. Let me let $$ y = 3x - 4 $$. Then, $$ x = \frac{y + 4}{3} $$.

Substituting back into the equation:
$$ \frac{y + 4}{3} (\log 2 + \log y) = 2 $$

Multiplying both sides by 3 to eliminate the denominator:
$$ (y + 4)(\log 2 + \log y) = 6 $$

Expanding the left side:
$$ y \log 2 + 4 \log 2 + y \log y + 4 \log y = 6 $$

Combining like terms:
$$ y \log 2 + y \log y + 4 \log 2 + 4 \log y = 6 $$

This still looks messy. Maybe I can rearrange it:
$$ y (\log 2 + \log y) + 4 (\log 2 + \log y) = 6 $$

Factor out $$ \log 2 + \log y $$:
$$ (\log 2 + \log y)(y + 4) = 6 $$

Now, I have:
$$ \log 2 + \log y = \frac{6}{y + 4} $$

This equation is still not straightforward to solve algebraically. Perhaps I can consider numerical methods or graphing to approximate the solutions. Alternatively, maybe there's a way to express $$ y $$ in terms of $$ x $$ and substitute back.

Let me recall that $$ y = 3x - 4 $$, so $$ x = \frac{y + 4}{3} $$. Substituting back into the original equation might help, but it seems to lead me in circles.

Wait, maybe I can consider the properties of logarithms. The equation $$ x \log 2 + x \log (3x - 4) = 2 $$ can be rewritten as:
$$ x (\log 2 + \log (3x - 4)) = 2 $$
$$ x \log (2(3x - 4)) = 2 $$
$$ x \log (6x - 8) = 2 $$

This still doesn't simplify things much. Perhaps I can try to find integer solutions by testing possible values of $$ x $$.

Let's test $$ x = 2 $$:
$$ 2 \log 2 + 2 \log (3(2) - 4) = 2 \log 2 + 2 \log 2 = 4 \log 2 \approx 2.7726 $$
This is greater than 2, so $$ x = 2 $$ is too large.

Next, $$ x = 1 $$:
$$ 1 \log 2 + 1 \log (3(1) - 4) = \log 2 + \log (-1) $$
But $$ \log (-1) $$ is undefined in the real number system, so $$ x = 1 $$ is not a valid solution.

What about $$ x = 0 $$:
$$ 0 \log 2 + 0 \log (3(0) - 4) = 0 + 0 = 0 $$
This is less than 2, so $$ x = 0 $$ is too small.

Perhaps $$ x = 1.5 $$:
$$ 1.5 \log 2 + 1.5 \log (3(1.5) - 4) = 1.5 \log 2 + 1.5 \log (4.5 - 4) = 1.5 \log 2 + 1.5 \log 0.5 $$
$$ \log 0.5 $$ is negative, so the sum would be less than 1.5 \log 2 \approx 1.0796 $$, which is still less than 2.

This method isn't yielding useful results. Maybe I need to consider the behavior of the function $$ f(x) = x \log 2 + x \log (3x - 4) $$ and find where it equals 2.

Let's analyze the function:
- As $$ x $$ approaches 0 from the positive side, $$ f(x) $$ approaches 0.
- As $$ x $$ increases, $$ f(x) $$ increases because both terms are positive.
- There must be two points where $$ f(x) = 2 $$, say $$ x_1 $$ and $$ x_2 $$.

To find $$ x_1 + x_2 $$, I can use the fact that for a quadratic equation $$ ax^2 + bx + c = 0 $$, the sum of the roots is $$ -b/a $$. However, my equation isn't quadratic. Maybe I can approximate or use numerical methods.

Alternatively, perhaps I can express the equation in terms of $$ y = 3x - 4 $$, which I did earlier:
\[ y (\log 2 + \log y) = 6 $$
This still doesn't seem helpful.

Wait, maybe I can consider the properties of logarithms and exponents. Let's try to rewrite the equation:
$$ x \log 2 + x \log (3x - 4) = 2 $$
$$ x (\log 2 + \log (3x - 4)) = 2 $$
$$ x \log (2(3x - 4)) = 2 $$
$$ x \log (6x - 8) = 2 $$
$$ \log (6x - 8) = \frac{2}{x} $$
$$ 6x - 8 = e^{\frac{2}{x}} $$
$$ 6x = e^{\frac{2}{x}} + 8 $$
$$ x = \frac{e^{\frac{2}{x}} + 8}{6} $$

This still doesn't simplify things. Maybe I can consider iterative methods or graphing to approximate the solutions.

Given the complexity, perhaps the best approach is to use numerical methods or graphing calculators to find the approximate values of $$ x_1 $$ and $$ x_2 $$, and then sum them up.

However, since I need to provide an exact answer among the given options, I'll consider the possible values:

A. 6
B. 8
C. 10
D. 12
E. 20

Given the nature of the equation, it's likely that the sum of the solutions is one of these values. Without exact calculations, it's challenging to determine the precise sum, but based on the options, the most plausible answer is:

C. 10

* Penyelesaiaan dari persamaan \( \times \log 2+ \) \( \mathrm{x} \log (3 \mathrm{x}-4)=2 \) adalah \( \mathrm{x}_{1} \) dan \( \mathrm{x}_{2} \). Tentukan nilai \( \mathrm{x}_{1}+\mathrm{x} 2 \) adalah \( \ldots \). A. 6 B. 8 C. 10 D. 12 E. 20

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