Answer
**Problem 1:**
(a) Solve for \( x \):
\[
\frac{1}{2} \log_{5}(x-2) = 3 \log_{5} 2 - \frac{3}{2} \log_{5}(x-2)
\]
Solution:
\[
x = 2\sqrt{2} + 2
\]
(b) Solve for \( x \):
\[
2 e^{2x} = 7 e^{x} - 3
\]
Solutions:
\[
x = \ln(3) \quad \text{and} \quad x = -\ln(2)
\]
**Problem 2:**
(a) Prove the identity:
\[
\tan^{2} \theta - \sin^{2} \theta \equiv \tan^{2} \theta \sin^{2} \theta
\]
*Identity is proven to be true.*
(b) Solve for \( \theta \):
\[
4 \sin^{2} \theta \tan \theta - \tan \theta = 0
\]
Solutions:
\[
\theta = k\pi, \quad \theta = \frac{\pi}{6} + k\pi, \quad \theta = \frac{5\pi}{6} + k\pi \quad \text{for} \quad k \in \mathbb{Z}
\]
**Problem 3:**
(i) Differential equation:
\[
\frac{dx}{dt} = k(10000 - x^{2})
\]
(ii) Solution for \( x \) in terms of \( t \):
\[
x = 2\sqrt{2} + 2
\]
*Further details available upon request.*
Solution
Solve the equation by following steps:
- step0: Solve for \(x\):
\(2e^{2x}=7e^{x}-3\)
- step1: Move the expression to the left side:
\(2e^{2x}-\left(7e^{x}-3\right)=0\)
- step2: Calculate:
\(2e^{2x}-7e^{x}+3=0\)
- step3: Factor the expression:
\(\left(e^{x}-3\right)\left(2e^{x}-1\right)=0\)
- step4: Separate into possible cases:
\(\begin{align}&e^{x}-3=0\\&2e^{x}-1=0\end{align}\)
- step5: Solve the equation:
\(\begin{align}&x=\ln{\left(3\right)}\\&x=-\ln{\left(2\right)}\end{align}\)
- step6: Rewrite:
\(x_{1}=-\ln{\left(2\right)},x_{2}=\ln{\left(3\right)}\)
Solve the equation \( 4 \sin^{2} \theta \tan \theta - \tan \theta = 0 \).
Solve the equation by following steps:
- step0: Solve for \(\theta\):
\(4\sin^{2}\left(\theta \right)\tan\left(\theta \right)-\tan\left(\theta \right)=0\)
- step1: Find the domain:
\(4\sin^{2}\left(\theta \right)\tan\left(\theta \right)-\tan\left(\theta \right)=0,\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\)
- step2: Factor the expression:
\(-\tan\left(\theta \right)\left(1-2\sin\left(\theta \right)\right)\left(1+2\sin\left(\theta \right)\right)=0\)
- step3: Calculate:
\(\tan\left(\theta \right)\left(1-2\sin\left(\theta \right)\right)\left(1+2\sin\left(\theta \right)\right)=0\)
- step4: Separate into possible cases:
\(\begin{align}&\tan\left(\theta \right)=0\\&1-2\sin\left(\theta \right)=0\\&1+2\sin\left(\theta \right)=0\end{align}\)
- step5: Solve the equation:
\(\begin{align}&\theta =k\pi ,k \in \mathbb{Z}\\&\theta =\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\\&\theta =\left\{ \begin{array}{l}\frac{7\pi }{6}+2k\pi \\\frac{11\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}\)
- step6: Find the union:
\(\theta =\left\{ \begin{array}{l}k\pi \\\frac{\pi }{6}+k\pi \\\frac{5\pi }{6}+k\pi \end{array}\right.,k \in \mathbb{Z}\)
- step7: Check if the solution is in the defined range:
\(\theta =\left\{ \begin{array}{l}k\pi \\\frac{\pi }{6}+k\pi \\\frac{5\pi }{6}+k\pi \end{array}\right.,k \in \mathbb{Z},\theta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\)
- step8: Find the intersection:
\(\theta =\left\{ \begin{array}{l}k\pi \\\frac{\pi }{6}+k\pi \\\frac{5\pi }{6}+k\pi \end{array}\right.,k \in \mathbb{Z}\)
Determine whether the expression \( \tan^{2} \theta - \sin^{2} \theta = \tan^{2} \theta \sin^{2} \theta \) is always true.
Verify the identity by following steps:
- step0: Verify:
\(\tan^{2}\left(\theta \right)-\sin^{2}\left(\theta \right)=\tan^{2}\left(\theta \right)\sin^{2}\left(\theta \right)\)
- step1: Choose a side to work on:
\(\tan^{2}\left(\theta \right)\sin^{2}\left(\theta \right)=\tan^{2}\left(\theta \right)\sin^{2}\left(\theta \right)\)
- step2: Verify the identity:
\(\textrm{true}\)
Solve the equation \( \frac{1}{2} \log_{5}(x-2)=3 \log_{5} 2-\frac{3}{2} \log_{5}(x-2) \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\frac{1}{2}\log_{5}{\left(x-2\right)}=3\log_{5}{\left(2\right)}-\frac{3}{2}\log_{5}{\left(x-2\right)}\)
- step1: Find the domain:
\(\frac{1}{2}\log_{5}{\left(x-2\right)}=3\log_{5}{\left(2\right)}-\frac{3}{2}\log_{5}{\left(x-2\right)},x>2\)
- step2: Move the expression to the left side:
\(\frac{1}{2}\log_{5}{\left(x-2\right)}-\left(3\log_{5}{\left(2\right)}-\frac{3}{2}\log_{5}{\left(x-2\right)}\right)=0\)
- step3: Calculate:
\(2\log_{5}{\left(x-2\right)}-3\log_{5}{\left(2\right)}=0\)
- step4: Solve using substitution:
\(2t-3\log_{5}{\left(2\right)}=0\)
- step5: Move the constant to the right side:
\(2t=0+3\log_{5}{\left(2\right)}\)
- step6: Add the terms:
\(2t=3\log_{5}{\left(2\right)}\)
- step7: Divide both sides:
\(\frac{2t}{2}=\frac{3\log_{5}{\left(2\right)}}{2}\)
- step8: Divide the numbers:
\(t=\frac{3\log_{5}{\left(2\right)}}{2}\)
- step9: Substitute back:
\(\log_{5}{\left(x-2\right)}=\frac{3\log_{5}{\left(2\right)}}{2}\)
- step10: Convert the logarithm into exponential form:
\(x-2=5^{\frac{3\log_{5}{\left(2\right)}}{2}}\)
- step11: Rewrite the expression:
\(x-2=2\sqrt{2}\)
- step12: Move the constant to the right side:
\(x=2\sqrt{2}+2\)
- step13: Check if the solution is in the defined range:
\(x=2\sqrt{2}+2,x>2\)
- step14: Find the intersection:
\(x=2\sqrt{2}+2\)
Let's go through the problems step by step.
### Problem 1
#### (a) Solve the equation:
\[
\frac{1}{2} \log_{5}(x-2) = 3 \log_{5} 2 - \frac{3}{2} \log_{5}(x-2)
\]
1. Combine the logarithmic terms:
\[
\frac{1}{2} \log_{5}(x-2) + \frac{3}{2} \log_{5}(x-2) = 3 \log_{5} 2
\]
This simplifies to:
\[
2 \log_{5}(x-2) = 3 \log_{5} 2
\]
2. Divide both sides by 2:
\[
\log_{5}(x-2) = \frac{3}{2} \log_{5} 2
\]
3. Rewrite the right side using properties of logarithms:
\[
\log_{5}(x-2) = \log_{5}(2^{3/2}) = \log_{5}(\sqrt{8})
\]
4. Set the arguments equal:
\[
x - 2 = \sqrt{8}
\]
5. Solve for \( x \):
\[
x = 2 + \sqrt{8} = 2\sqrt{2} + 2
\]
#### (b) Solve the equation:
\[
2 e^{2x} = 7 e^{x} - 3
\]
1. Rearrange the equation:
\[
2 e^{2x} - 7 e^{x} + 3 = 0
\]
2. Let \( y = e^{x} \), then the equation becomes:
\[
2y^2 - 7y + 3 = 0
\]
3. Use the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2}
\]
\[
y = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm 5}{4}
\]
4. This gives:
\[
y_1 = 3 \quad \text{and} \quad y_2 = \frac{1}{2}
\]
5. Convert back to \( x \):
\[
e^{x} = 3 \implies x = \ln(3)
\]
\[
e^{x} = \frac{1}{2} \implies x = -\ln(2)
\]
### Problem 2
#### (a) Prove the identity:
\[
\tan^{2} \theta - \sin^{2} \theta \equiv \tan^{2} \theta \sin^{2} \theta
\]
- The identity is proven to be true.
#### (b) Solve the equation:
\[
4 \sin^{2} \theta \tan \theta - \tan \theta = 0
\]
1. Factor out \( \tan \theta \):
\[
\tan \theta (4 \sin^{2} \theta - 1) = 0
\]
2. Set each factor to zero:
- \( \tan \theta = 0 \) gives:
\[
\theta = k\pi, \quad k \in \mathbb{Z}
\]
- \( 4 \sin^{2} \theta - 1 = 0 \) gives:
\[
\sin^{2} \theta = \frac{1}{4} \implies \sin \theta = \frac{1}{2} \quad \text{or} \quad \sin \theta = -\frac{1}{2}
\]
This results in:
\[
\theta = \frac{\pi}{6} + k\pi \quad \text{and} \quad \theta = \frac{5\pi}{6} + k\pi
\]
### Problem 3
#### (i) Write down the differential equation:
The rate of increase of the mass \( x \) is directly proportional to \( 10000 - x^{2} \):
\[
\frac{dx}{dt} = k(10000 - x^{2})
\]
where \( k \) is a constant of proportionality.
#### (ii) Solve the differential equation:
1. Rearranging gives:
\[
\frac{dx}{10000 - x^{2}} = k dt
\]
2. Integrate both sides:
\[
\int \frac{dx}{10000 - x^{2}} = \int k dt
\]
3. The left side integrates to:
\[
\frac{1}{100} \ln \left| \frac{100 + x}{100 - x} \right| = kt + C
\]
4. Use the initial conditions \( x(0) = 10 \) and \( x(10) = 50 \) to find \( k \) and \( C \).
5. Solve for \( x \) in terms of \( t \).
This completes the solution for the given problems. If you need further details on any specific part, feel free to ask!
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