Question

\\\log_2(x + 1) = \log_3(27)\\
\\(1 + i)(x - yi) = i(14 + 7i) - (2 + 13i)\\
\\3x + (3x - y)i = 4 - 6i\\
\\\text{roots } -6x^2 + 36x - 59\\
\\\text{solve for } t\text{: } 2t - s = p\\
\\\text{derivative of } f(x) = \ln(x)\text{, at } x = 17\\
\\\cos(x) - \sin(x) = 0\\

Explanation:

🆕 New Concept Discovered: Solving Multi-Topic Math Problems
A collection of algebra, complex numbers, calculus, and trigonometry.

Step 1: Solve the logarithmic equation

First, we simplify the right side of the equation:

$$ \log_3(27) = \log_3(3^3) = 3 $$

Now, substitute this back into the equation:

$$ \log_2(x + 1) = 3 $$

Convert the logarithmic equation into its exponential form:

$$ x + 1 = 2^3 $$

$$ x + 1 = 8 $$

$$ x = 7 $$

Step 2: Solve the complex number equation for \(x\) and \(y\)

We are given:

$$ (1 + i)(x - yi) = i(14 + 7i) - (2 + 13i) $$

First, expand the left side:

$$ (1 + i)(x - yi) = x - yi + xi - yi^2 $$

Since \(i^2 = -1\):

$$ x - yi + xi + y = (x + y) + (x - y)i $$

Next, expand and simplify the right side:

$$ i(14 + 7i) - (2 + 13i) = 14i + 7i^2 - 2 - 13i $$

$$ = 14i - 7 - 2 - 13i $$

$$ = -9 + i $$

Now, equate the real and imaginary parts from both sides:

1. Real parts:

$$ x + y = -9 $$

1. Imaginary parts:

$$ x - y = 1 $$

Add the two equations together:

$$ 2x = -8 \implies x = -4 $$

Substitute \(x = -4\) back into the second equation:

$$ -4 - y = 1 \implies y = -5 $$

Step 3: Solve the third equation for \(x\) and \(y\)

We are given:

$$ 3x + (3x - y)i = 4 - 6i $$

Equate the real and imaginary parts:

1. Real parts:

$$ 3x = 4 \implies x = \frac{4}{3} $$

1. Imaginary parts:

$$ 3x - y = -6 $$

Substitute \(3x = 4\) into the imaginary equation:

$$ 4 - y = -6 \implies y = 10 $$

Step 4: Find the roots of the quadratic expression

We want to find the roots of:

$$ -6x^2 + 36x - 59 = 0 $$

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -6\), \(b = 36\), and \(c = -59\):

$$ x = \frac{-36 \pm \sqrt{36^2 - 4(-6)(-59)}}{2(-6)} $$

$$ x = \frac{-36 \pm \sqrt{1296 - 1416}}{-12} $$

$$ x = \frac{-36 \pm \sqrt{-120}}{-12} $$

Simplify the radical using complex numbers (\(\sqrt{-1} = i\)):

$$ \sqrt{-120} = i\sqrt{120} = 2i\sqrt{30} $$

$$ x = \frac{-36 \pm 2i\sqrt{30}}{-12} $$

$$ x = 3 \pm \frac{\sqrt{30}}{6}i $$

Step 5: Solve the literal equation for \(t\)

We want to isolate \(t\) in the equation:

$$ 2t - s = p $$

Add \(s\) to both sides:

$$ 2t = p + s $$

Divide by 2:

$$ t = \frac{p + s}{2} $$

Step 6: Find the derivative of \(f(x) = \ln(x)\) at \(x = 17\)

The derivative of \(f(x) = \ln(x)\) is:

$$ f'(x) = \frac{1}{x} $$

Evaluating this derivative at \(x = 17\):

$$ f'(17) = \frac{1}{17} $$

Step 7: Solve the trigonometric equation

We want to solve:

$$ \cos(x) - \sin(x) = 0 $$

Add \(\sin(x)\) to both sides:

$$ \cos(x) = \sin(x) $$

Divide both sides by \(\cos(x)\) (assuming \(\cos(x)
eq 0\)):

$$ 1 = \tan(x) $$

The general solution for \(x\) is:

$$ x = \frac{\pi}{4} + k\pi \quad \text{for } k \in \mathbb{Z} $$

Answer:

1. Logarithmic Equation:

$$ x = 7 $$

1. First Complex Equation:

$$ x = -4, \quad y = -5 $$

1. Second Complex Equation:

$$ x = \frac{4}{3}, \quad y = 10 $$

1. Quadratic Roots:

$$ x = 3 \pm \frac{\sqrt{30}}{6}i $$

1. Literal Equation:

$$ t = \frac{p + s}{2} $$

1. Derivative Value:

$$ f'(17) = \frac{1}{17} $$

1. Trigonometric Equation:

$$ x = \frac{\pi}{4} + k\pi \quad (k \in \mathbb{Z}) $$