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, simplify the right side:

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

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

$$ = -9 + i $$

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:

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

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

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

##[LLM SSE On Failure]

Answer:

🆕 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, simplify the right side:

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

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

$$ = -9 + i $$

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:

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

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

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

##[LLM SSE On Failure]