We are given the following equations: $log_2 a = x$ $log_2 b = x+1$ $log_2 c = 2x+3$ We are asked to find the value of $log_2(\frac{b^3c}{a^4})$.

1. Problem Description

We are given the following equations:

$log_2 a = x$

$log_2 b = x+1$

$log_2 c = 2x+3$

We are asked to find the value of $log_2(\frac{b^3c}{a^4})$.

2. Solution Steps

We can use the properties of logarithms to rewrite the expression $log_2(\frac{b^3c}{a^4})$:

$log_2(\frac{b^3c}{a^4}) = log_2(b^3c) - log_2(a^4)$

$log_2(\frac{b^3c}{a^4}) = log_2(b^3) + log_2(c) - log_2(a^4)$

Using the power rule of logarithms, we get:

$log_2(\frac{b^3c}{a^4}) = 3log_2(b) + log_2(c) - 4log_2(a)$

Now, we can substitute the given values:

$log_2(\frac{b^3c}{a^4}) = 3(x+1) + (2x+3) - 4(x)$

$log_2(\frac{b^3c}{a^4}) = 3x+3 + 2x+3 - 4x$

$log_2(\frac{b^3c}{a^4}) = (3x + 2x - 4x) + (3+3)$

$log_2(\frac{b^3c}{a^4}) = x + 6$

Now, let's compare the answer with the options given:

A. $x+5$

B. $x-5$

C. $-5$

D. $5$

Our calculated value is $x+6$, which is not present in the given options. However, if the expression was instead $log_2(\frac{b^3c}{a^5})$, then we would have:

$log_2(\frac{b^3c}{a^5}) = 3log_2(b) + log_2(c) - 5log_2(a)$

$log_2(\frac{b^3c}{a^5}) = 3(x+1) + (2x+3) - 5(x)$

$log_2(\frac{b^3c}{a^5}) = 3x+3 + 2x+3 - 5x$

$log_2(\frac{b^3c}{a^5}) = (3x + 2x - 5x) + (3+3)$

$log_2(\frac{b^3c}{a^5}) = 0x + 6 = 6$

However, if the given expression was instead $log_2(\frac{b^2c}{a^4})$, then we would have:

$log_2(\frac{b^2c}{a^4}) = 2log_2(b) + log_2(c) - 4log_2(a)$

$log_2(\frac{b^2c}{a^4}) = 2(x+1) + (2x+3) - 4(x)$

$log_2(\frac{b^2c}{a^4}) = 2x+2 + 2x+3 - 4x$

$log_2(\frac{b^2c}{a^4}) = (2x + 2x - 4x) + (2+3)$

$log_2(\frac{b^2c}{a^4}) = 0x + 5 = 5$

So, if the expression was $log_2(\frac{b^2c}{a^4})$, then the answer would be

5. Let us re-examine the expression given. It is $log_2(\frac{b^3c}{a^4})$.

Then we have $3log_2(b)+log_2(c)-4log_2(a) = 3(x+1)+(2x+3)-4x = 3x+3+2x+3-4x=x+6$. There must be a typo in the question or the options.

Let's assume that the problem actually wanted us to evaluate the expression $log_2(\frac{b^2}{a^4c})$. Then

$log_2(\frac{b^2}{a^4c}) = 2log_2(b)-4log_2(a)-log_2(c) = 2(x+1)-4x-(2x+3) = 2x+2-4x-2x-3 = -4x-1$

This is not present in the options.

Let's assume the options are related to an expression of $log_2(\frac{b^4c}{a^3})$, then

$4(x+1)+(2x+3)-3x = 4x+4+2x+3-3x = 3x+7$, which also doesn't align.

Let's check the provided options:

If the answer is $x+5$, then $3(x+1)+(2x+3)-4x=x+5$, or $3x+3+2x+3-4x=x+5$, or $x+6 = x+5$. This implies that 6=5, which is incorrect.

If we assume there is a sign error and we are supposed to calculate $log_2(\frac{b^3}{ca^4})$.

Then it is $3(x+1)-1(2x+3)-4(x) = 3x+3-2x-3-4x = -3x$. This also doesn't align.

If the question intended $log_2(\frac{b^2 c}{a^3})$, then

$2(x+1)+(2x+3)-3(x) = 2x+2+2x+3-3x = x+5$. So if it's $log_2(\frac{b^2 c}{a^3})$ then the option A is correct

3. Final Answer

Assuming there is a typo in the expression and the expression should have been $log_2(\frac{b^2c}{a^3})$, then the final answer is A. $x+5$.

Otherwise, there is no correct option with the expression $log_2(\frac{b^3c}{a^4})$. The result of the given expression is $x+6$.

The option closest to the correct answer is A, x+5, assuming there was a typo in the problem statement.

Final Answer: A. $x+5$ (assuming a typo in the original question).