# 6.2 MCQs-Logarithms

## Logarithms

### Basic Concepts and Definitions

1\. The logarithm of a number to a given base is defined as:

1. The power to which the base must be raised to obtain the number
2. The number itself multiplied by the base
3. The reciprocal of the number
4. The square root of the number

<details>

<summary>Show me the answer</summary>

**Answer:** 1. The power to which the base must be raised to obtain the number

**Explanation:** If $$b^y = x$$, then $$\log\_b x = y$$.

This means: logarithm is the exponent to which the base must be raised to produce the number.

Example: Since $$10^3 = 1000$$, then $$\log\_{10} 1000 = 3$$.

The base b must be positive and not equal to 1: $$b > 0$$, $$b \neq 1$$.

</details>

2\. Which of the following is equivalent to $$\log\_b 1$$?

1. $$0$$
2. $$1$$
3. $$b$$
4. $$\infty$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$0$$

**Explanation:** For any valid base b: $$\log\_b 1 = 0$$

Reason: $$b^0 = 1$$ for any $$b \neq 0$$.

This is a fundamental property of logarithms: $$b^{\log\_b 1} = 1 \Rightarrow \log\_b 1 = 0$$

</details>

3\. The value of $$\log\_b b$$ is:

1. $$0$$
2. $$1$$
3. $$b$$
4. $$\infty$$

<details>

<summary>Show me the answer</summary>

**Answer:** 2. $$1$$

**Explanation:** For any valid base b: $$\log\_b b = 1$$

Reason: $$b^1 = b$$.

This is another fundamental property of logarithms: $$b^{\log\_b b} = b \Rightarrow \log\_b b = 1$$

</details>

### Common and Natural Logarithms

4\. The notation $$\log x$$ (without a base) usually means:

1. Natural logarithm (base e)
2. Common logarithm (base 10)
3. Binary logarithm (base 2)
4. It depends on the context

<details>

<summary>Show me the answer</summary>

**Answer:** 4. It depends on the context

**Explanation:** The interpretation of $$\log x$$ without a base depends on the field:

* In mathematics and higher-level work: often means natural logarithm (base e)
* In engineering and many sciences: often means common logarithm (base 10)
* In computer science: sometimes means binary logarithm (base 2)

To avoid ambiguity, it's better to specify:

* Natural logarithm: $$\ln x$$ or $$\log\_e x$$
* Common logarithm: $$\log\_{10} x$$
* Binary logarithm: $$\log\_2 x$$

</details>

5\. The natural logarithm is denoted by:

1. $$\log x$$
2. $$\ln x$$
3. $$\text{Ln } x$$
4. Both 2 and 3

<details>

<summary>Show me the answer</summary>

**Answer:** 4. Both 2 and 3

**Explanation:** The natural logarithm (logarithm with base e) is commonly denoted as:

* $$\ln x$$ (most common in mathematics)
* $$\log\_e x$$ (explicit notation)
* $$\text{Ln } x$$ (sometimes used in older texts)

The constant e (Euler's number) is approximately: $$e \approx 2.718281828459045$$

Important: $$\ln e = 1$$ and $$\ln 1 = 0$$.

</details>

6\. The relationship between natural logarithm and common logarithm is:

1. $$\ln x = \log\_{10} x$$
2. $$\ln x = \frac{\log\_{10} x}{\log\_{10} e}$$
3. $$\ln x = 2.303 \log\_{10} x$$
4. Both 2 and 3

<details>

<summary>Show me the answer</summary>

**Answer:** 4. Both 2 and 3

**Explanation:** Using the change of base formula:

$$\ln x = \frac{\log\_{10} x}{\log\_{10} e}$$

Since $$\log\_{10} e \approx 0.4342944819$$, we have:

$$\ln x \approx \frac{\log\_{10} x}{0.4342944819} \approx 2.302585 \log\_{10} x$$

So approximately: $$\ln x \approx 2.303 \log\_{10} x$$

Conversely: $$\log\_{10} x = \frac{\ln x}{\ln 10} \approx \frac{\ln x}{2.302585} \approx 0.4343 \ln x$$

</details>

### Logarithmic Properties and Laws

7\. The product rule for logarithms states:

1. $$\log\_b (mn) = \log\_b m + \log\_b n$$
2. $$\log\_b (mn) = \log\_b m \times \log\_b n$$
3. $$\log\_b (m+n) = \log\_b m + \log\_b n$$
4. $$\log\_b (mn) = m \log\_b n$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$\log\_b (mn) = \log\_b m + \log\_b n$$

**Explanation:** The product rule: $$\log\_b (mn) = \log\_b m + \log\_b n$$

Derivation: Let $$\log\_b m = x$$ and $$\log\_b n = y$$. Then $$b^x = m$$ and $$b^y = n$$. Multiply: $$mn = b^x \cdot b^y = b^{x+y}$$. Take logarithm: $$\log\_b (mn) = x + y = \log\_b m + \log\_b n$$.

Example: $$\log\_2 (8 \times 4) = \log\_2 8 + \log\_2 4 = 3 + 2 = 5$$.

</details>

8\. The quotient rule for logarithms states:

1. $$\log\_b \left(\frac{m}{n}\right) = \log\_b m - \log\_b n$$
2. $$\log\_b \left(\frac{m}{n}\right) = \frac{\log\_b m}{\log\_b n}$$
3. $$\log\_b \left(\frac{m}{n}\right) = \log\_b m \times \log\_b n$$
4. $$\log\_b \left(\frac{m}{n}\right) = \log\_b (m-n)$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$\log\_b \left(\frac{m}{n}\right) = \log\_b m - \log\_b n$$

**Explanation:** The quotient rule: $$\log\_b \left(\frac{m}{n}\right) = \log\_b m - \log\_b n$$

Derivation: Let $$\log\_b m = x$$ and $$\log\_b n = y$$. Then $$b^x = m$$ and $$b^y = n$$. Divide: $$\frac{m}{n} = \frac{b^x}{b^y} = b^{x-y}$$. Take logarithm: $$\log\_b \left(\frac{m}{n}\right) = x - y = \log\_b m - \log\_b n$$.

Example: $$\log\_{10} \left(\frac{1000}{10}\right) = \log\_{10} 1000 - \log\_{10} 10 = 3 - 1 = 2$$.

</details>

9\. The power rule for logarithms states:

1. $$\log\_b (m^n) = n \log\_b m$$
2. $$\log\_b (m^n) = (\log\_b m)^n$$
3. $$\log\_b (m^n) = m \log\_b n$$
4. $$\log\_b (m^n) = \log\_b m \times \log\_b n$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$\log\_b (m^n) = n \log\_b m$$

**Explanation:** The power rule: $$\log\_b (m^n) = n \log\_b m$$

Derivation: Let $$\log\_b m = x$$, so $$b^x = m$$. Raise both sides to power n: $$(b^x)^n = m^n \Rightarrow b^{nx} = m^n$$. Take logarithm: $$\log\_b (m^n) = nx = n \log\_b m$$.

Example: $$\log\_2 (8^3) = 3 \log\_2 8 = 3 \times 3 = 9$$.

Special case: $$\log\_b \sqrt\[n]{m} = \log\_b (m^{1/n}) = \frac{1}{n} \log\_b m$$.

</details>

10\. The change of base formula is:

1. $$\log\_b a = \frac{\log\_c a}{\log\_c b}$$
2. $$\log\_b a = \log\_c a \times \log\_c b$$
3. $$\log\_b a = \log\_c a - \log\_c b$$
4. $$\log\_b a = \frac{1}{\log\_a b}$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$\log\_b a = \frac{\log\_c a}{\log\_c b}$$

**Explanation:** The change of base formula: $$\log\_b a = \frac{\log\_c a}{\log\_c b}$$

Derivation: Let $$\log\_b a = x$$, so $$b^x = a$$. Take logarithm base c of both sides: $$\log\_c (b^x) = \log\_c a$$. Using power rule: $$x \log\_c b = \log\_c a$$. Solve: $$x = \frac{\log\_c a}{\log\_c b}$$.

Common applications:

* Convert to natural logarithms: $$\log\_b a = \frac{\ln a}{\ln b}$$
* Convert to common logarithms: $$\log\_b a = \frac{\log\_{10} a}{\log\_{10} b}$$

</details>

### Special Logarithmic Values

11\. The value of $$\log\_a a^x$$ is:

1. $$x$$
2. $$a^x$$
3. $$x \log a$$
4. $$a$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$x$$

**Explanation:** $$\log\_a a^x = x$$

This follows directly from the definition of logarithm: If $$a^y = a^x$$, then $$y = x$$.

Alternatively, using the power rule: $$\log\_a a^x = x \log\_a a = x \times 1 = x$$.

Example: $$\log\_2 2^5 = 5$$ $$\log\_{10} 10^{-3} = -3$$ $$\ln e^{x} = x$$

</details>

12\. The expression $$b^{\log\_b x}$$ simplifies to:

1. $$x$$
2. $$\log\_b x$$
3. $$b^x$$
4. $$\log\_x b$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$x$$

**Explanation:** $$b^{\log\_b x} = x$$

This is the fundamental inverse relationship between exponential and logarithmic functions.

If $$y = \log\_b x$$, then by definition $$b^y = x$$. Substituting: $$b^{\log\_b x} = b^y = x$$.

This shows that exponential and logarithmic functions are inverse operations:

* $$f(x) = b^x$$ and $$g(x) = \log\_b x$$ are inverse functions
* $$f(g(x)) = b^{\log\_b x} = x$$
* $$g(f(x)) = \log\_b (b^x) = x$$

</details>

13\. The value of $$\log\_2 1$$ is:

1. $$0$$
2. $$1$$
3. $$2$$
4. $$\infty$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$0$$

**Explanation:** $$\log\_2 1 = 0$$

This is a special case of the general rule: $$\log\_b 1 = 0$$ for any valid base b.

Reason: $$2^0 = 1$$

Verification: What power must 2 be raised to get 1? Answer: 0, since $$2^0 = 1$$.

Similarly: $$\log\_{10} 1 = 0$$ $$\ln 1 = 0$$ $$\log\_5 1 = 0$$

</details>

14\. The value of $$\log\_3 27$$ is:

1. $$2$$
2. $$3$$
3. $$9$$
4. $$27$$

<details>

<summary>Show me the answer</summary>

**Answer:** 2. $$3$$

**Explanation:** $$\log\_3 27 = 3$$

Reason: $$3^3 = 27$$

We ask: "3 raised to what power gives 27?" $$3^1 = 3$$ $$3^2 = 9$$ $$3^3 = 27$$

So the answer is 3.

Alternative method: $$27 = 3^3$$, so $$\log\_3 27 = \log\_3 (3^3) = 3 \log\_3 3 = 3 \times 1 = 3$$.

</details>

### Logarithmic Equations

15\. The solution to $$\log\_2 x = 3$$ is:

1. $$x = 2$$
2. $$x = 3$$
3. $$x = 8$$
4. $$x = 9$$

<details>

<summary>Show me the answer</summary>

**Answer:** 3. $$x = 8$$

**Explanation:** Given: $$\log\_2 x = 3$$

Convert from logarithmic form to exponential form: If $$\log\_b y = x$$, then $$b^x = y$$.

Here: $$\log\_2 x = 3 \Rightarrow 2^3 = x$$

Therefore: $$x = 2^3 = 8$$

Verification: $$\log\_2 8 = 3$$ since $$2^3 = 8$$.

</details>

16\. The solution to $$\log\_x 9 = 2$$ is:

1. $$x = 3$$
2. $$x = 9$$
3. $$x = 18$$
4. $$x = 81$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$x = 3$$

**Explanation:** Given: $$\log\_x 9 = 2$$

Convert to exponential form: $$x^2 = 9$$

Solve for x: $$x = \pm \sqrt{9} = \pm 3$$

Since the base of a logarithm must be positive and not equal to 1: $$x > 0$$ and $$x \neq 1$$

Therefore: $$x = 3$$

Verification: $$\log\_3 9 = 2$$ since $$3^2 = 9$$.

</details>

17\. The solution to $$\ln(x+1) = 0$$ is:

1. $$x = 0$$
2. $$x = 1$$
3. $$x = e$$
4. $$x = -1$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$x = 0$$

**Explanation:** Given: $$\ln(x+1) = 0$$

Recall: $$\ln 1 = 0$$

So: $$x + 1 = 1$$

Therefore: $$x = 0$$

Verification: $$\ln(0+1) = \ln 1 = 0$$

Alternatively, convert to exponential form: $$\ln(x+1) = 0 \Rightarrow e^0 = x+1 \Rightarrow 1 = x+1 \Rightarrow x = 0$$

</details>

### Logarithmic Inequalities

18\. For $$b > 1$$, the inequality $$\log\_b x > 0$$ is satisfied when:

1. $$x > 0$$
2. $$x > 1$$
3. $$0 < x < 1$$
4. $$x < 0$$

<details>

<summary>Show me the answer</summary>

**Answer:** 2. $$x > 1$$

**Explanation:** For base $$b > 1$$:

* $$\log\_b x > 0$$ when $$x > 1$$
* $$\log\_b x = 0$$ when $$x = 1$$
* $$\log\_b x < 0$$ when $$0 < x < 1$$

Reason: When $$b > 1$$, the logarithmic function is increasing. $$\log\_b 1 = 0$$, so for values greater than 1, the logarithm is positive.

Example: For $$b = 10$$: $$\log\_{10} 0.1 = -1$$ (negative) $$\log\_{10} 1 = 0$$ (zero) $$\log\_{10} 10 = 1$$ (positive)

</details>

19\. For $$0 < b < 1$$, the inequality $$\log\_b x > 0$$ is satisfied when:

1. $$x > 0$$
2. $$x > 1$$
3. $$0 < x < 1$$
4. $$x < 0$$

<details>

<summary>Show me the answer</summary>

**Answer:** 3. $$0 < x < 1$$

**Explanation:** For base $$0 < b < 1$$:

* $$\log\_b x > 0$$ when $$0 < x < 1$$
* $$\log\_b x = 0$$ when $$x = 1$$
* $$\log\_b x < 0$$ when $$x > 1$$

Reason: When $$0 < b < 1$$, the logarithmic function is decreasing. $$\log\_b 1 = 0$$, so for values less than 1, the logarithm is positive.

Example: For $$b = 0.5$$: $$\log\_{0.5} 2 = -1$$ (negative, since $$0.5^{-1} = 2$$) $$\log\_{0.5} 1 = 0$$ (zero) $$\log\_{0.5} 0.5 = 1$$ (positive)

</details>

### Applications and Properties

20\. The domain of the function $$f(x) = \log\_b (x-2)$$ is:

1. $$x > 0$$
2. $$x > 2$$
3. $$x \geq 2$$
4. All real numbers

<details>

<summary>Show me the answer</summary>

**Answer:** 2. $$x > 2$$

**Explanation:** For any logarithmic function $$\log\_b g(x)$$, the argument must be positive: $$g(x) > 0$$

For $$f(x) = \log\_b (x-2)$$: $$x - 2 > 0$$ $$x > 2$$

Therefore, the domain is $$(2, \infty)$$.

The base b also has restrictions: $$b > 0$$ and $$b \neq 1$$, but that's a condition on b, not on x.

</details>

21\. The equation $$\log\_b x = \log\_b y$$ implies:

1. $$x = y$$
2. $$x = b^y$$
3. $$y = b^x$$
4. $$\log x = \log y$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$x = y$$

**Explanation:** If $$\log\_b x = \log\_b y$$, then $$x = y$$.

This follows from the fact that logarithmic functions are one-to-one (injective) for a fixed base.

Proof: If $$\log\_b x = \log\_b y = k$$, then: $$b^k = x$$ and $$b^k = y$$ Therefore: $$x = y$$

Important: This property holds only if both logarithms have the same base and the arguments are in the domain (positive).

This property is useful for solving logarithmic equations.

</details>

22\. The expression $$\log\_b \frac{1}{x}$$ is equal to:

1. $$\frac{1}{\log\_b x}$$
2. $$-\log\_b x$$
3. $$\log\_b x$$
4. $$\log\_x b$$

<details>

<summary>Show me the answer</summary>

**Answer:** 2. $$-\log\_b x$$

**Explanation:** $$\log\_b \frac{1}{x} = -\log\_b x$$

This follows from two properties:

1. Quotient rule: $$\log\_b \frac{1}{x} = \log\_b 1 - \log\_b x$$
2. $$\log\_b 1 = 0$$

So: $$\log\_b \frac{1}{x} = 0 - \log\_b x = -\log\_b x$$

Alternatively, using power rule: $$\log\_b \frac{1}{x} = \log\_b (x^{-1}) = -1 \cdot \log\_b x = -\log\_b x$$

Example: $$\log\_{10} 0.01 = \log\_{10} \frac{1}{100} = -\log\_{10} 100 = -2$$

</details>

### Logarithmic Identities

23\. The identity $$\log\_a b \times \log\_b a$$ equals:

1. $$0$$
2. $$1$$
3. $$\log\_a b$$
4. $$\log\_b a$$

<details>

<summary>Show me the answer</summary>

**Answer:** 2. $$1$$

**Explanation:** $$\log\_a b \times \log\_b a = 1$$

Proof using change of base formula:

$$\log\_a b = \frac{\log b}{\log a}$$ $$\log\_b a = \frac{\log a}{\log b}$$

Multiply: $$\log\_a b \times \log\_b a = \frac{\log b}{\log a} \times \frac{\log a}{\log b} = 1$$

This shows that $$\log\_a b$$ and $$\log\_b a$$ are reciprocals: $$\log\_a b = \frac{1}{\log\_b a}$$

Example: $$\log\_2 8 = 3$$ and $$\log\_8 2 = \frac{1}{3}$$ $$3 \times \frac{1}{3} = 1$$

</details>

24\. The expression $$\log\_a (b^c)$$ can also be written as:

1. $$c \log\_a b$$
2. $$b \log\_a c$$
3. $$\log\_a b + \log\_a c$$
4. $$\log\_a b \times \log\_a c$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$c \log\_a b$$

**Explanation:** This is the power rule of logarithms: $$\log\_a (b^c) = c \log\_a b$$

Derivation: Let $$\log\_a b = x$$, so $$a^x = b$$. Raise both sides to power c: $$(a^x)^c = b^c \Rightarrow a^{cx} = b^c$$. Take logarithm: $$\log\_a (b^c) = cx = c \log\_a b$$.

Example: $$\log\_2 (8^3) = 3 \log\_2 8 = 3 \times 3 = 9$$.

This rule allows us to bring exponents down in front of the logarithm.

</details>

25\. The sum $$\log\_b a + \log\_b \frac{1}{a}$$ equals:

1. $$0$$
2. $$1$$
3. $$\log\_b a$$
4. $$2 \log\_b a$$

<details>

<summary>Show me the answer</summary>

**Answer:** 1. $$0$$

**Explanation:** Using logarithmic properties:

$$\log\_b a + \log\_b \frac{1}{a} = \log\_b \left( a \times \frac{1}{a} \right) = \log\_b 1 = 0$$

Alternatively: $$\log\_b \frac{1}{a} = -\log\_b a$$ So: $$\log\_b a + (-\log\_b a) = 0$$

This shows that $$\log\_b a$$ and $$\log\_b \frac{1}{a}$$ are additive inverses.

Example: For base 10: $$\log\_{10} 100 + \log\_{10} 0.01 = 2 + (-2) = 0$$

</details>