The first problem (number 14) states that $log_2 a = x$, $log_2 b = x+1$, and $log_2 c = 2x+3$. We need to find the value of $log_2(\frac{b^2c}{a^4})$.

1. Problem Description

The first problem (number 14) states that $log_2 a = x$, $log_2 b = x+1$, and $log_2 c = 2x+3$. We need to find the value of $log_2(\frac{b^2c}{a^4})$.

2. Solution Steps

We have:

$log_2 a = x$

$log_2 b = x+1$

$log_2 c = 2x+3$

We want to find $log_2(\frac{b^2c}{a^4})$.

Using the properties of logarithms:

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

$log_2(b^2c) = log_2(b^2) + log_2(c) = 2log_2(b) + log_2(c)$

$log_2(a^4) = 4log_2(a)$

Substituting the given values:

$2log_2(b) + log_2(c) = 2(x+1) + (2x+3) = 2x + 2 + 2x + 3 = 4x+5$

$4log_2(a) = 4x$

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

3. Final Answer

The correct value of $log_2(\frac{b^2c}{a^4})$ is

5. So the answer is D.

Now, let's solve the second problem (number 15).

1. Problem Description

Three IT students P, Q, and R developed an app and had GH¢ 135,000.

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0. P received twice as Q and Q received two-thirds as R. Find the amount received by Q.

2. Solution Steps

Let P, Q, and R be the amounts received by the respective students.

We are given:

P + Q + R = 135,000

P = 2Q

Q = (2/3)R

From Q = (2/3)R, we have R = (3/2)Q

Substituting P and R in terms of Q into the first equation:

2Q + Q + (3/2)Q = 135,000

(4/2)Q + (2/2)Q + (3/2)Q = 135,000

(9/2)Q = 135,000

Q = (2/9) * 135,000

Q = 2 * 15,000

Q = 30,000

3. Final Answer

The amount received by Q is GH¢ 30,000.

0

0. So the answer is A.

Now, let's solve the third problem (number 16).

1. Problem Description

Find the correct value of m in the equation $5^{m+1} = 3 \times 5^m + 250$.

2. Solution Steps

We have the equation $5^{m+1} = 3 \times 5^m + 250$.

$5^{m+1} = 5 \times 5^m$

So the equation becomes $5 \times 5^m = 3 \times 5^m + 250$.

$5 \times 5^m - 3 \times 5^m = 250$

$2 \times 5^m = 250$

$5^m = 125$

$5^m = 5^3$

Therefore, $m = 3$.

3. Final Answer

The correct value of m is

3. So the answer is C.

Now, let's solve the fourth problem (number 17).

1. Problem Description

Given that the coefficient of $x^6$ in the binomial $(m+x)^9$ is $\frac{21}{2}$, find the correct value of m.

2. Solution Steps

The general term in the binomial expansion of $(m+x)^9$ is given by:

$T_{k+1} = \binom{9}{k} m^{9-k} x^k$

We want the term with $x^6$, so $k = 6$.

$T_{6+1} = T_7 = \binom{9}{6} m^{9-6} x^6 = \binom{9}{6} m^3 x^6$

$\binom{9}{6} = \binom{9}{3} = \frac{9!}{6!3!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$

So the coefficient of $x^6$ is $84m^3$.

We are given that the coefficient is $\frac{21}{2}$.

Therefore, $84m^3 = \frac{21}{2}$

$m^3 = \frac{21}{2 \times 84} = \frac{21}{168} = \frac{1}{8}$

$m = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$

3. Final Answer

The correct value of m is $\frac{1}{2}$.

So the answer is C.

Now, let's solve the fifth problem (number 18).

1. Problem Description

Find the correct value of y in the equation $log_3(7y-1) = 3$.

2. Solution Steps

We have $log_3(7y-1) = 3$.

To remove the logarithm, we can write the equation in exponential form:

$7y - 1 = 3^3$

$7y - 1 = 27$

$7y = 28$

$y = 4$

3. Final Answer

The correct value of y is

4. So the answer is D.