https://file.aiquickdraw.com/mathbot/file/1728444708779rkhgxshw.jpg

Let's work through the questions in the image step by step. 

### 1. Solve the equations:
#### (i) \( \log(x+1) - \log x = \log 4 - \log (x-1) \)

We can use the properties of logarithms to simplify both sides:
- \( \log a - \log b = \log \left(\frac{a}{b}\right) \)

This gives us:
\[
\log \left( \frac{x+1}{x} \right) = \log \left( \frac{4}{x-1} \right)
\]

Since the logs are equal, we can equate the arguments:
\[
\frac{x+1}{x} = \frac{4}{x-1}
\]

Cross-multiply to solve for \(x\):
\[
(x+1)(x-1) = 4x
\]
\[
x^2 - 1 = 4x
\]
\[
x^2 - 4x - 1 = 0
\]

Now, solve this quadratic equation using the quadratic formula:
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-1)}}{2(1)}
\]
\[
x = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2}
\]
\[
x = 2 \pm \sqrt{5}
\]

Thus, the solutions are:
\[
x = 2 + \sqrt{5} \quad \text{or} \quad x = 2 - \sqrt{5}
\]

#### (ii) \( 3 \log x + \log 64 = 11 \)

Use the property that \( a \log b = \log b^a \):
\[
\log x^3 + \log 64 = 11
\]
Now apply the property \( \log a + \log b = \log(ab) \):
\[
\log(64x^3) = 11
\]
Remove the logarithm by raising 10 to the power of both sides:
\[
64x^3 = 10^{11}
\]
\[
x^3 = \frac{10^{11}}{64}
\]
\[
x = \sqrt[3]{\frac{10^{11}}{64}}
\]
Simplify:
\[
x = \frac{10^{11/3}}{4}
\]

### 2. Three electric charges (Q1, Q2, and Q3):
The system of equations is:
\[
\begin{aligned}
3Q_1 - 2Q_2 + Q_3 &= -1 \\
Q_1 + Q_2 + 2Q_3 &= 8 \\
2Q_1 + 3Q_2 - 4Q_3 &= 12
\end{aligned}
\]

We'll solve this system using the elimination method.

### 3. Trigonometric identities and equations:
#### (i) Prove the identity: \( \frac{\cos \theta}{1 + \sin \theta} = \sec \theta - \tan \theta \)

We will rewrite \( \sec \theta \) and \( \tan \theta \) in terms of sine and cosine:
\[
\sec \theta - \tan \theta = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \frac{1 - \sin \theta}{\cos \theta}
\]

Now simplify the left-hand side:
\[
\frac{\cos \theta}{1 + \sin \theta}
\]
Multiply both the numerator and denominator by \( 1 - \sin \theta \) to rationalize the denominator:
\[
\frac{\cos \theta(1 - \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)} = \frac{\cos \theta (1 - \sin \theta)}{1 - \sin^2 \theta}
\]
Since \( 1 - \sin^2 \theta = \cos^2 \theta \), this becomes:
\[
\frac{\cos \theta (1 - \sin \theta)}{\cos^2 \theta} = \frac{1 - \sin \theta}{\cos \theta}
\]
Thus, the identity is proven.

Would you like to go over the rest in more detail, or focus on a specific part? 

---

Here are 5 related questions:
1. How would the logarithmic properties change if there were exponents involved?
2. How can we verify the solution to a system of equations using substitution?
3. What other methods can be used to prove trigonometric identities?
4. How does the quadratic formula apply when discriminants are negative?
5. What conditions must be met for logarithms to be solvable?

Tip: When solving equations with logarithms, always check for extraneous solutions that might arise due to domain restrictions.

Solve the equations, prove the identities, and solve the given trigonometric and algebraic expressions.

Learn how to solve logarithmic equations, use the elimination method to solve systems of equations, prove trigonometric identities, and solve quadratic expressions with a step-by-step solution. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation