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# In-Class Activity 3.2

### Question 1: Rewrite as 'simple' natural logarithms

#### a) Rewrite $\ln(x^8)$:

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<summary> Example: </summary>

a) Rewrite $\ln(x^{6})$:

| Explanation | Equation |
|--------------------------------------------------|-------------------------------------|
| Use the power rule of logarithms: $\ln(a^b) = b\ln(a)$ | $\ln(x^{6}) = 6\ln(x)$ |

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#### b) Rewrite $\ln\left(\frac{1}{x^3}\right)$:

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<summary> Example: </summary>

Rewrite $\ln\left(\frac{1}{x^7}\right)$

1. Use the fact that $\frac{1}{x^7} = x^{-7}$:
$$ \ln\left(\frac{1}{x^7}\right) = \ln(x^{-7}) $$

2. Apply the power rule: $\ln(A^B) = B\ln(A)$:
$$ \ln(x^{-7}) = -7\ln(x) $$

### Final Answer:
$$ \ln\left(\frac{1}{x^7}\right) = -7\ln(x) $$

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#### c) Rewrite $\ln(2x^5)$:

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<summary> Example: </summary>

### Rewrite $\ln(3x^4)$ in a simple form

| Explanation | Expression |
|--------------------------------------------------|---------------------------------------|
| Start with the given expression | $\ln(3x^4)$ |
| Apply the product rule: $\ln(AB) = \ln(A) + \ln(B)$ | $\ln(3) + \ln(x^4)$ |
| Apply the power rule: $\ln(A^B) = B\ln(A)$ to $\ln(x^4)$ | $\ln(3) + 4\ln(x)$ |

### Final Expression:
$$ \ln(3x^4) = \ln(3) + 4\ln(x) $$

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#### d) Rewrite $\ln\left(\frac{3}{4x^5}\right)$:

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<summary> Example: </summary>

### Rewrite $\ln\left(\frac{4}{6x^7}\right)$ in a simple form

| Explanation | Expression |
|--------------------------------------------------|---------------------------------------|
| Start with the given expression | $\ln\left(\frac{4}{6x^7}\right)$ |
| Apply the quotient rule: $\ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B)$ | $\ln(4) - \ln(6x^7)$ |
| Use the product rule: $\ln(AB) = \ln(A) + \ln(B)$ to split $\ln(6x^7)$ | $\ln(4) - [\ln(6) + \ln(x^7)]$ |
| Apply the power rule: $\ln(A^B) = B\ln(A)$ to $\ln(x^7)$ | $\ln(4) - [\ln(6) + 7\ln(x)]$ |
| Simplify the expression | $\ln(4) - \ln(6) - 7\ln(x)$ |

### Final Expression:
$$ \ln\left(\frac{4}{6x^7}\right) = \ln(4) - \ln(6) - 7\ln(x) $$

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#### e) Rewrite $\log_2(4x^5)$:

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<summary> Example: </summary>

Rewrite $\log_3(5x^6)$ as simple natural logarithms

| Explanation | Expression |
|--------------------------------------------------|---------------------------------------|
| Start with the given expression | $\log_3(5x^6)$ |
| Use the change of base formula: $\log_b(A) = \frac{\ln(A)}{\ln(b)}$ to convert the logarithm to natural logarithms. | $\frac{\ln(5x^6)}{\ln(3)}$ |
| Apply the product rule: $\ln(AB) = \ln(A) + \ln(B)$ | $\frac{\ln(5) + \ln(x^6)}{\ln(3)}$ |
| Apply the power rule: $\ln(A^B) = B\ln(A)$ to $\ln(x^6)$ | $\frac{\ln(5) + 6\ln(x)}{\ln(3)}$ |

### Final Expression:
$$ \log_3(5x^6) = \frac{\ln(5) + 6\ln(x)}{\ln(3)} $$

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#### f) Rewrite $\log_{10}\left(\frac{100}{2x^3}\right)$:

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<summary> Example: </summary>

Rewrite $\log_{10}\left(\frac{1000}{3x^4}\right)$ as simple natural logarithms

| Explanation | Expression |
|--------------------------------------------------|---------------------------------------|
| Start with the given expression | $\log_{10}\left(\frac{1000}{3x^4}\right)$ |
| Use the change of base formula: $\log_b(A) = \frac{\ln(A)}{\ln(b)}$ to convert the logarithm to natural logarithms. | $\frac{\ln\left(\frac{1000}{3x^4}\right)}{\ln(10)}$ |
| Apply the quotient rule: $\ln\left(\frac{A}{B}\right) = \ln(A) - \ln(B)$ | $\frac{\ln(1000) - \ln(3x^4)}{\ln(10)}$ |
| Apply the product rule: $\ln(AB) = \ln(A) + \ln(B)$ to $\ln(3x^4)$ | $\frac{\ln(1000) - [\ln(3) + \ln(x^4)]}{\ln(10)}$ |
| Apply the power rule: $\ln(A^B) = B\ln(A)$ to $\ln(x^4)$ | $\frac{\ln(1000) - \ln(3) - 4\ln(x)}{\ln(10)}$ |

### Final Expression:
$$ \log_{10}\left(\frac{1000}{3x^4}\right) = \frac{\ln(1000) - \ln(3) - 4\ln(x)}{\ln(10)} $$

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### Question 2: Find $f'(x)$

#### a) For $f(x) = 5e^x + 3x + 1$:

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<summary> Example: </summary>

### Find $f'(x)$ for $f(x) = 3e^x + 4x + 2$

| Explanation | Expression |
|--------------------------------------------------|---------------------------------------|
| Start with the given function | $f(x) = 3e^x + 4x + 2$ |
| Differentiate $3e^x$. The derivative of $e^x$ is $e^x$, so $3e^x$ becomes $3e^x$. | $f'(x) = 3e^x + \dots$ |
| Differentiate $4x$. The derivative of $x$ is $1$, so $4x$ becomes $4$. | $f'(x) = 3e^x + 4 + \dots$ |
| The derivative of the constant $2$ is $0$. | $f'(x) = 3e^x + 4$ |

### Final Expression:
$$ f'(x) = 3e^x + 4 $$

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#### b) For $f(x) = 6\ln(x) - x^3 + 2$:

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<summary> Example: </summary>

### Find $f'(x)$ for $f(x) = 7\ln(x) - x^5 + 4$

| Explanation | Expression |
|--------------------------------------------------|---------------------------------------|
| Start with the given function | $f(x) = 7\ln(x) - x^5 + 4$ |
| Differentiate $7\ln(x)$. The derivative of $\ln(x)$ is $\frac{1}{x}$, so $7\ln(x)$ becomes $7 \cdot \frac{1}{x} = \frac{7}{x}$. | $f'(x) = \frac{7}{x} - \dots$ |
| Differentiate $-x^5$. Use the power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$, so $-x^5$ becomes $-5x^4$. | $f'(x) = \frac{7}{x} - 5x^4 + \dots$ |
| The derivative of the constant $4$ is $0$. | $f'(x) = \frac{7}{x} - 5x^4$ |

### Final Expression:
$$ f'(x) = \frac{7}{x} - 5x^4 $$

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#### c) For $f(x) = \ln(x^8)$:

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<summary> Example: </summary>

### Find $f'(x)$ for $f(x) = \ln(x^9)$

| Explanation | Expression |
|--------------------------------------------------|---------------------------------------|
| Start with the given function | $f(x) = \ln(x^9)$ |
| Apply the power rule: $\ln(A^B) = B\ln(A)$. Rewrite $\ln(x^9)$ as $9\ln(x)$. | $f(x) = 9\ln(x)$ |
| Differentiate $9\ln(x)$. The derivative of $\ln(x)$ is $\frac{1}{x}$, so $9\ln(x)$ becomes $9 \cdot \frac{1}{x} = \frac{9}{x}$. | $f'(x) = \frac{9}{x}$ |

### Final Expression:
$$ f'(x) = \frac{9}{x} $$

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#### d) For $f(x) = \ln(x^{10}) + 2\ln(x)$:

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<summary> Example: </summary>

### Find $f'(x)$ for $f(x) = \ln(x^{11}) + 4\ln(x)$

| Explanation | Expression |
|--------------------------------------------------|---------------------------------------|
| Start with the given function | $f(x) = \ln(x^{11}) + 4\ln(x)$ |
| Apply the power rule: $\ln(A^B) = B\ln(A)$. Rewrite $\ln(x^{11})$ as $11\ln(x)$. | $f(x) = 11\ln(x) + 4\ln(x)$ |
| Combine like terms: $11\ln(x) + 4\ln(x) = 15\ln(x)$. | $f(x) = 15\ln(x)$ |
| Differentiate $15\ln(x)$. The derivative of $\ln(x)$ is $\frac{1}{x}$, so $15\ln(x)$ becomes $15 \cdot \frac{1}{x} = \frac{15}{x}$. | $f'(x) = \frac{15}{x}$ |

### Final Expression:
$$ f'(x) = \frac{15}{x} $$

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### Question 3: Find the equation of the tangent line to the graph of $f(x)$ at $x = 1$

Given: $f(x) = 3 + \ln(x)$

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### Find the equation of the tangent line to the graph of $f(x)$ at $x = 2$
Given: $f(x) = 4 + \ln(x)$

| Explanation | Expression |
|--------------------------------------------------|---------------------------------------|
| Start with the given function | $f(x) = 4 + \ln(x)$ |
| Find the derivative $f'(x)$. The derivative of $\ln(x)$ is $\frac{1}{x}$. | $f'(x) = \frac{1}{x}$ |
| Evaluate the slope at $x = 2$. | $f'(2) = \frac{1}{2}$ |
| Find $f(2)$ by substituting $x = 2$ into $f(x)$. | $f(2) = 4 + \ln(2)$ |
| Use point-slope form: $y - f(a) = f'(a)(x - a)$. Substitute $f'(2) = \frac{1}{2}$ and $f(2) = 4 + \ln(2)$. | $y - (4 + \ln(2)) = \frac{1}{2}(x - 2)$ |
| Distribute the $\frac{1}{2}$ to simplify the equation. | $y = \frac{1}{2}x - 1 + 4 + \ln(2)$ |
| Simplify further. | $y = \frac{1}{2}x + 3 + \ln(2)$ |

### Final Equation of the Tangent Line:
$$ y = \frac{1}{2}x + 3 + \ln(2) $$


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### Question 4: Find the equation of the tangent line to the graph of $f(x)$ at $x = 0$

Given: $f(x) = e^x + 1$

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### Question 4: Find the equation of the tangent line to the graph of $f(x)$ at $x = 1$
Given: $f(x) = e^x + 2$

| Explanation | Expression |
|--------------------------------------------------|---------------------------------------|
| Start with the given function | $f(x) = e^x + 2$ |
| Find the derivative $f'(x)$. The derivative of $e^x$ is $e^x$, and the constant 2 has a derivative of 0. | $f'(x) = e^x$ |
| Evaluate the slope at $x = 1$. | $f'(1) = e^1 = e$ |
| Find $f(1)$ by substituting $x = 1$ into $f(x)$. | $f(1) = e^1 + 2 = e + 2$ |
| Use the point-slope form: $y - f(a) = f'(a)(x - a)$. Substitute $f'(1) = e$ and $f(1) = e + 2$. | $y - (e + 2) = e(x - 1)$ |
| Distribute $e$ to simplify the equation. | $y = e(x - 1) + e + 2$ |
| Simplify further. | $y = ex - e + e + 2 = ex + 2$ |

### Final Equation of the Tangent Line:
$$ y = ex + 2 $$


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