The problem consists of three parts: a. Evaluate two logarithmic expressions: $\log_{\frac{1}{2}}4$ and $\log_{0.1}10$. b. Find the square roots of the complex number $3-4i$. c. Express two rational functions in partial fractions: $\frac{x}{(x+3)(x+2)}$ and $\frac{2x^2+6x+1}{x(x+1)}$.

1. Problem Description

The problem consists of three parts:

a. Evaluate two logarithmic expressions: $\log_{\frac{1}{2}}4$ and $\log_{0.1}10$.

b. Find the square roots of the complex number $3-4i$.

c. Express two rational functions in partial fractions: $\frac{x}{(x+3)(x+2)}$ and $\frac{2x^2+6x+1}{x(x+1)}$.

2. Solution Steps

a.

i. Evaluate $\log_{\frac{1}{2}}4$. Let $y = \log_{\frac{1}{2}}4$. Then, $(\frac{1}{2})^y = 4$. Since $4 = 2^2$ and $\frac{1}{2} = 2^{-1}$, we have $(2^{-1})^y = 2^2$, which means $2^{-y} = 2^2$. Therefore, $-y = 2$, so $y = -2$.

ii. Evaluate $\log_{0.1}10$. Let $y = \log_{0.1}10$. Then $(0.1)^y = 10$. Since $0.1 = \frac{1}{10} = 10^{-1}$, we have $(10^{-1})^y = 10$, which means $10^{-y} = 10^1$. Therefore, $-y = 1$, so $y = -1$.

b. Find the square roots of $3-4i$. Let $\sqrt{3-4i} = a+bi$, where $a$ and $b$ are real numbers. Squaring both sides, we get $3-4i = (a+bi)^2 = a^2 + 2abi - b^2 = (a^2 - b^2) + 2abi$.

Equating the real and imaginary parts, we have:

$a^2 - b^2 = 3$ and $2ab = -4$, which implies $ab = -2$, so $b = -\frac{2}{a}$.

Substituting $b$ into the first equation:

$a^2 - (-\frac{2}{a})^2 = 3$

$a^2 - \frac{4}{a^2} = 3$

$a^4 - 4 = 3a^2$

$a^4 - 3a^2 - 4 = 0$

Let $u = a^2$. Then $u^2 - 3u - 4 = 0$.

$(u-4)(u+1) = 0$.

So $u = 4$ or $u = -1$. Since $a$ is real, $a^2$ must be non-negative. Therefore, $a^2 = 4$, which means $a = 2$ or $a = -2$.

If $a = 2$, then $b = -\frac{2}{2} = -1$.

If $a = -2$, then $b = -\frac{2}{-2} = 1$.

Therefore, the square roots are $2-i$ and $-2+i$.

c.

i. Express $\frac{x}{(x+3)(x+2)}$ in partial fractions.

We can write $\frac{x}{(x+3)(x+2)} = \frac{A}{x+3} + \frac{B}{x+2}$.

Multiplying both sides by $(x+3)(x+2)$, we get $x = A(x+2) + B(x+3)$.

If $x = -3$, then $-3 = A(-3+2) + B(-3+3) = -A$, so $A = 3$.

If $x = -2$, then $-2 = A(-2+2) + B(-2+3) = B$, so $B = -2$.

Therefore, $\frac{x}{(x+3)(x+2)} = \frac{3}{x+3} - \frac{2}{x+2}$.

ii. Express $\frac{2x^2+6x+1}{x(x+1)}$ in partial fractions.

Since the degree of the numerator is equal to the degree of the denominator, we perform polynomial long division first.

$\frac{2x^2+6x+1}{x(x+1)} = \frac{2x^2+6x+1}{x^2+x} = 2 + \frac{4x+1}{x(x+1)}$.

Now we express $\frac{4x+1}{x(x+1)}$ in partial fractions.

$\frac{4x+1}{x(x+1)} = \frac{C}{x} + \frac{D}{x+1}$.

Multiplying both sides by $x(x+1)$, we get $4x+1 = C(x+1) + Dx$.

If $x = 0$, then $1 = C(0+1) + D(0) = C$, so $C = 1$.

If $x = -1$, then $-4+1 = C(-1+1) + D(-1) = -D$, so $-3 = -D$, which means $D = 3$.

Therefore, $\frac{4x+1}{x(x+1)} = \frac{1}{x} + \frac{3}{x+1}$.

So, $\frac{2x^2+6x+1}{x(x+1)} = 2 + \frac{1}{x} + \frac{3}{x+1}$.

3. Final Answer

a.

i. $\log_{\frac{1}{2}}4 = -2$

ii. $\log_{0.1}10 = -1$

b. The square roots of $3-4i$ are $2-i$ and $-2+i$.

c.

i. $\frac{x}{(x+3)(x+2)} = \frac{3}{x+3} - \frac{2}{x+2}$

ii. $\frac{2x^2+6x+1}{x(x+1)} = 2 + \frac{1}{x} + \frac{3}{x+1}$