The image contains several math problems. Question 4 asks to find the value of $x$ that satisfies the equation $log_3(x-2) - log_3(2x-1) = -1$. Question 5 asks to find the fifth term of the binomial expansion of $(1-x)^{10}$. Questions 6-8 relate to the partial fraction decomposition $\frac{x+24}{x^2-x-12} = \frac{A}{x-4} + \frac{B}{x+3}$. Question 6 asks for the value of $B$. Question 7 asks for the value of $A-B$. Question 8 asks for the value of $2B^2 - 3A$.

1. Problem Description

The image contains several math problems.

Question 4 asks to find the value of $x$ that satisfies the equation $log_3(x-2) - log_3(2x-1) = -1$.

Question 5 asks to find the fifth term of the binomial expansion of $(1-x)^{10}$.

Questions 6-8 relate to the partial fraction decomposition $\frac{x+24}{x^2-x-12} = \frac{A}{x-4} + \frac{B}{x+3}$.

Question 6 asks for the value of $B$.

Question 7 asks for the value of $A-B$.

Question 8 asks for the value of $2B^2 - 3A$.

2. Solution Steps

Question 4:

We are given the equation $log_3(x-2) - log_3(2x-1) = -1$.

Using the logarithm property $log_a(b) - log_a(c) = log_a(\frac{b}{c})$, we have

$log_3(\frac{x-2}{2x-1}) = -1$.

Exponentiating both sides with base 3, we get

$\frac{x-2}{2x-1} = 3^{-1} = \frac{1}{3}$.

Multiplying both sides by $3(2x-1)$, we have

$3(x-2) = 2x-1$.

$3x-6 = 2x-1$.

$x = 5$.

We need to check if $x=5$ is a valid solution by plugging it back into the original equation.

$log_3(5-2) - log_3(2(5)-1) = log_3(3) - log_3(9) = 1 - 2 = -1$.

Thus, $x=5$ is a valid solution.

Question 5:

The binomial theorem states that $(a+b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k}b^k$.

In our case, $a=1$, $b=-x$, and $n=10$.

We are looking for the fifth term, which corresponds to $k=4$.

The fifth term is ${10 \choose 4} (1)^{10-4} (-x)^4 = {10 \choose 4} x^4$.

${10 \choose 4} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210$.

Therefore, the fifth term is $210x^4$.

Questions 6, 7, and 8:

We have $\frac{x+24}{x^2-x-12} = \frac{x+24}{(x-4)(x+3)} = \frac{A}{x-4} + \frac{B}{x+3}$.

Multiplying both sides by $(x-4)(x+3)$, we get

$x+24 = A(x+3) + B(x-4)$.

To find $A$, let $x=4$. Then $4+24 = A(4+3) + B(4-4)$.

$28 = 7A$, so $A = 4$.

To find $B$, let $x=-3$. Then $-3+24 = A(-3+3) + B(-3-4)$.

$21 = -7B$, so $B = -3$.

Question 6:

The value of $B$ is $-3$.

Question 7:

$A - B = 4 - (-3) = 4+3 = 7$.

Question 8:

$2B^2 - 3A = 2(-3)^2 - 3(4) = 2(9) - 12 = 18 - 12 = 6$.

3. Final Answer

Question 4: C. 5

Question 5: B. 210x^4

Question 6: C. -3

Question 7: B. 7

Question 8: D. 6