Simplify: 2log2(x²y) - log2(4x) + log2(y³)

The product rule is $\log_b M + \log_b N = \log_b(MN)$.
The quotient rule is $\log_b M - \log_b N = \log_b(\frac{M}{N})$.

$$\log_{2}(x^4y^2) - \log_{2}(4x) + \log_{2}(y^3) = \log_{2}\left(\frac{x^4y^2 \cdot y^3}{4x}\right)$$

$$\log_{2}\left(\frac{x^4y^5}{4x}\right) = \log_{2}\left(\frac{x^{4-1}y^5}{4}\right)$$ $$\log_{2}\left(\frac{x^3y^5}{4}\right)$$

$\boxed{\log_{2}\left(\frac{x^3y^5}{4}\right)}$

b) Solve:

i) $2^{3x-2} = 7$

$$3x-2 = \frac{\log(7)}{\log(2)}$$ $$3x = 2 + \frac{\log(7)}{\log(2)}$$ $$x = \frac{1}{3}\left(2 + \frac{\log(7)}{\log(2)}\right)$$

Using $\log(7) \approx 0.8451$ and $\log(2) \approx 0.3010$:

$$x \approx \frac{1}{3}\left(2 + \frac{0.8451}{0.3010}\right)$$ $$x \approx \frac{1}{3}(2 + 2.8076)$$ $$x \approx \frac{1}{3}(4.8076)$$ $$x \approx 1.6025$$

$\boxed{x = \frac{1}{3}\left(2 + \frac{\log(7)}{\log(2)}\right) \approx 1.6025}$

ii) $\log_{3}(2x+14) - \log_{3}x = 2$

If $\log_b M = N$, then $M = b^N$.

$$\frac{2x+14}{x} = 3^2$$ $$\frac{2x+14}{x} = 9$$

$$2x+14 = 9x$$ $$14 = 9x - 2x$$ $$14 = 7x$$ $$x = \frac{14}{7}$$ $$x = 2$$

Check for valid domain: $2x+14 > 0 \implies 2(2)+14 = 18 > 0$. $x > 0 \implies 2 > 0$. Both conditions are met.
$\boxed{x = 2}$

iii) $4^y - 6(2^y) - 16 = 0$

Substitute $u$ into the equation.

$$u^2 - 6u - 16 = 0$$

Factor the quadratic:

$$(u-8)(u+2) = 0$$

This gives two possible solutions for $u$:

$$u = 8 \quad \text{or} \quad u = -2$$

Case 1: $2^y = 8$

$$2^y = 2^3$$ $$y = 3$$

Case 2: $2^y = -2$
There is no real solution for $y$ since $2^y$ must always be positive.
$\boxed{y = 3}$

QUESTION FOUR

a) Solve the following equations for values

i) $4 \cos \theta = \cot \theta$

$$4 \cos \theta - \frac{\cos \theta}{\sin \theta} = 0$$

Factor out $\cos \theta$:

$$\cos \theta \left(4 - \frac{1}{\sin \theta}\right) = 0$$

This gives two possibilities:
Possibility 1: $\cos \theta = 0$
Possibility 2: $4 - \frac{1}{\sin \theta} = 0$

For $\cos \theta = 0$:
$\theta = 90^\circ + 180^\circ n$, where $n$ is an integer.
In the range $0^\circ \le \theta < 360^\circ$, $\theta = 90^\circ, 270^\circ$.

For $4 - \frac{1}{\sin \theta} = 0$:

$$4 = \frac{1}{\sin \theta}$$ $$\sin \theta = \frac{1}{4}$$

Using a calculator, the principal value is $\theta_1 = \arcsin\left(\frac{1}{4}\right) \approx 14.48^\circ$.
Since $\sin \theta$ is positive, $\theta$ is in the first or second quadrant.
$\theta_2 = 180^\circ - \theta_1 \approx 180^\circ - 14.48^\circ = 165.52^\circ$.
General solutions:
$\theta = 14.48^\circ + 360^\circ n$
$\theta = 165.52^\circ + 360^\circ n$
In the range $0^\circ \le \theta < 360^\circ$, $\theta \approx 14.48^\circ, 165.52^\circ$.

Combining all solutions:
$\boxed{\theta = 90^\circ, 270^\circ, \arcsin\left(\frac{1}{4}\right), 180^\circ - \arcsin\left(\frac{1}{4}\right) \text{ (and their co-terminal angles)}}$
$\boxed{\theta \approx 90^\circ, 270^\circ, 14.48^\circ, 165.52^\circ}$

ii) $\tan^2 \theta - 1 = \frac{2}{3}$

$$\tan^2 \theta = 1 + \frac{2}{3}$$ $$\tan^2 \theta = \frac{3}{3} + \frac{2}{3}$$ $$\tan^2 \theta = \frac{5}{3}$$

$$\tan \theta = \pm\sqrt{\frac{5}{3}}$$ $$\tan \theta = \pm\frac{\sqrt{15}}{3}$$

For $\tan \theta = \sqrt{\frac{5}{3}} \approx 1.291$:
The principal value is $\theta_1 = \arctan\left(\sqrt{\frac{5}{3}}\right) \approx 52.24^\circ$.
Since $\tan \theta$ has a period of $180^\circ$, the general solution is $\theta = 52.24^\circ + 180^\circ n$.
In the range $0^\circ \le \theta < 360^\circ$, $\theta \approx 52.24^\circ, 232.24^\circ$.

For $\tan \theta = -\sqrt{\frac{5}{3}} \approx -1.291$:
The principal value is $\theta_2 = \arctan\left(-\sqrt{\frac{5}{3}}\right) \approx -52.24^\circ$.
The general solution is $\theta = -52.24^\circ + 180^\circ n$.
In the range $0^\circ \le \theta < 360^\circ$, $\theta \approx 180^\circ - 52.24^\circ = 127.76^\circ$ and $\theta \approx 360^\circ - 52.24^\circ = 307.76^\circ$.

$\boxed{\theta = \arctan\left(\pm\sqrt{\frac{5}{3}}\right) + 180^\circ n \text{ (where } n \text{ is an integer)}}$
$\boxed{\theta \approx 52.24^\circ, 127.76^\circ, 232.24^\circ, 307.76^\circ}$

b) Prove the following trigonometric identities

i) $\sin x \sec x \equiv \tan x$

Recall that $\sec x = \frac{1}{\cos x}$.

$$\text{LHS} = \sin x \cdot \frac{1}{\cos x}$$

$$\text{LHS} = \frac{\sin x}{\cos x}$$

Recall that $\tan x = \frac{\sin x}{\cos x}$.

$$\text{LHS} = \tan x$$

This matches the Right Hand Side (RHS).

$$\text{LHS} \equiv \text{RHS}$$

The identity is proven.

ii) $\frac{1}{1-\sin \theta} + \frac{1}{1+\sin \theta} \equiv 2\sec^2 \theta$

The common denominator is $(1-\sin \theta)(1+\sin \theta)$.

$$\text{LHS} = \frac{1(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)} + \frac{1(1-\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$$ $$\text{LHS} = \frac{(1+\sin \theta) + (1-\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$$

Numerator: $(1+\sin \theta) + (1-\sin \theta) = 1 + \sin \theta + 1 - \sin \theta = 2$.
Denominator: $(1-\sin \theta)(1+\sin \theta)$ is a difference of squares, which is $1^2 - \sin^2 \theta = 1 - \sin^2 \theta$.
Recall the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, so $1 - \sin^2 \theta = \cos^2 \theta$.

$$\text{LHS} = \frac{2}{\cos^2 \theta}$$

Recall that $\sec \theta = \frac{1}{\cos \theta}$, so $\sec^2 \theta = \frac{1}{\cos^2 \theta}$.

$$\text{LHS} = 2 \cdot \frac{1}{\cos^2 \theta}$$ $$\text{LHS} = 2\sec^2 \theta$$

This matches the Right Hand Side (RHS).

$$\text{LHS} \equiv \text{RHS}$$

The identity is proven.