Below is our AP Calculus AB unit test on logarithms. These questions cover basic logarithmic properties, such as the sum and difference of logs, the logarithmic exponent rule, and logarithmic base changes. These log properties will become extremely useful when solving complicated derivatives and integrals.

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Question 1 |

**Which of the following is a simplification for: $\log(3x)+\log(x)$**

$\log(4x)$ | |

$\log(3x^2 )$ | |

$\log(2x)$ | |

$\log(3)$ |

Question 1 Explanation:

The correct answer is (B). The addition rule for logarithms is:

$\log(a)+\log(b)=\log(ab)$

In this case, we see that:

$\log(3x)+\log(x)=$ $\log(3x⋅x)=\log(3x^2)$

$\log(a)+\log(b)=\log(ab)$

In this case, we see that:

$\log(3x)+\log(x)=$ $\log(3x⋅x)=\log(3x^2)$

Question 2 |

**Which of the following is a simplification for: $\log(7x)+\log(3xy)-\log(3x)$**

$\log(x^3 y)$ | |

$\log(7x^2 y-3x)$ | |

$\log\left(\frac{63x^2}{y^2}\right)$ | |

$\log(7xy)$ |

Question 2 Explanation:

The correct answer is (D). The first two logarithms can be combined using the addition rule. So:

$\log(7x)+\log(3xy)=$ $\log(21x^2 y)$

Since the third logarithm is being subtracted from the other two, we must divide $21x^2 y$ by the argument of $\log(3x)$. This results in:

$\log\left(\frac{21x^2 y} {3x}\right) = \log(7xy)$

$\log(7x)+\log(3xy)=$ $\log(21x^2 y)$

Since the third logarithm is being subtracted from the other two, we must divide $21x^2 y$ by the argument of $\log(3x)$. This results in:

$\log\left(\frac{21x^2 y} {3x}\right) = \log(7xy)$

Question 3 |

**Which of the following is equivalent to:**$\log_2 (2^{\log_2(2^7)} )$

$2$ | |

$7$ | |

$7^2$ | |

$2^7$ |

Question 3 Explanation:

The correct answer is (B). To solve this problem quickly, start with the innermost logarithm. Since $\log_2 2^7 = 7$, the upper expression can be rewritten as $\log_2 (2^7 )$. This logarithm can be further simplified as $\log_2 (2^7 ) = 7$.

Question 4 |

**Which of the following is equal to:**$\log_2 (3) \cdot \log_3 (4) \cdot \log_4 (5) \cdot$ $\ldots \cdot \log_{98} (99) \cdot \log_{99} (100)$

$\log_{99} (100)$ | |

$\log_{100} (2)$ | |

$\dfrac{\log(2)}{2}$ | |

$\dfrac{2}{\log(2)}$ |

Question 4 Explanation:

The correct answer is (D). The log change of base formula is: $\log_a (b) = \frac{\log(b)}{\log(a)}$

We can apply this formula to a few elements in this series and look for a pattern.

$\frac{\log(3)}{\log(2)} \cdot \frac{\log(4)}{\log(3)} \cdot \frac{\log(5)}{\log(4)} \cdot$ $... \cdot \frac{\log(99)}{\log(98)} \cdot \frac{\log(100)}{\log(99)}$ $= \frac{\log(100)}{\log(2)} = \frac{2}{\log(2)}$

We can apply this formula to a few elements in this series and look for a pattern.

$\frac{\log(3)}{\log(2)} \cdot \frac{\log(4)}{\log(3)} \cdot \frac{\log(5)}{\log(4)} \cdot$ $... \cdot \frac{\log(99)}{\log(98)} \cdot \frac{\log(100)}{\log(99)}$ $= \frac{\log(100)}{\log(2)} = \frac{2}{\log(2)}$

Question 5 |

**Which of the following is equal to:**$\log(z^2) + \log(y^2) + \log(x^2)$,

**where**$\log(x) + \log(y) + \log(z) = 8$

$8$ | |

$16$ | |

$64$ | |

$128$ |

Question 5 Explanation:

The correct answer is (B). The logarithmic exponent rule states that: $\log(x^n) = n × \log(x)$

We can rewrite the sum of the three logarithms in the given expression as: $2\log(z) + 2\log(y) + 2\log(x)$

Factoring out $2$ gives: $2(\log(x) + \log(y) + \log(z))$

Since the problem stated that $\log(x) + \log(y) + \log(z) = 8$, substituting this value into the expression gives: $2(8) = 16$

We can rewrite the sum of the three logarithms in the given expression as: $2\log(z) + 2\log(y) + 2\log(x)$

Factoring out $2$ gives: $2(\log(x) + \log(y) + \log(z))$

Since the problem stated that $\log(x) + \log(y) + \log(z) = 8$, substituting this value into the expression gives: $2(8) = 16$

Question 6 |

**Which of the following is a simplification of:**$\ln(x^2) + \ln(e^4)$

$2 \ln(x) + 4$ | |

$2\ln(x) + e^4$ | |

$\ln(xe^4)$ | |

$\ln(8x)$ |

Question 6 Explanation:

The correct answer is (A). The first logarithm can be simplified to $2 \ln(x)$ using the logarithmic exponent rule. The second logarithm can be simplified to $4$, since the natural logarithm is base $e$. The sum of these two logarithms is $2 \ln(x) + 4$.

Question 7 |

**Which of the following is equivalent to:**$\ln(x) + \ln(x) − \ln(x^2)$

$\ln(x^4)$ | |

$\ln(x^2 + 2)$ | |

$0$ | |

$1$ |

Question 7 Explanation:

The correct answer is (C). Using the sum and difference rules of logarithms, the expression can be rewritten as:

$\ln(\frac{x^2}{x^2}) = \ln(1) = 0$

$\ln(\frac{x^2}{x^2}) = \ln(1) = 0$

Question 8 |

**Which of the following values of $x$ satisfies the equation:**$\log_x(7) = 3$

$3^7$ | |

$7^3$ | |

$3^\frac{1}{7}$ | |

$7^\frac{1}{3}$ |

Question 8 Explanation:

The correct answer is (D). From the definition of a logarithm, we know that $x^3 = 7$. Therefore, the value of $x$ is the cube root of $7$, or $7^\frac{1}{3}$.

Question 9 |

**Which of the following values of $x$ satisfies the equation:**$\log_{(x + 4)}(4) = 2$

$−2$ | |

$0$ | |

$2$ | |

$4$ |

Question 9 Explanation:

The correct answer is (A). From the definition of logarithm: $(x + 4)^2 = 4$

Expanding and simplifying gives: $x^2 + 8x − 12 = 0$

Using the quadratic formula, the possible solutions for $x$ are:

$\dfrac{–8±4}{2}$

This means $x = −6$ or $x = −2$. Since log bases cannot be negative, $x = −6$ cannot be a valid answer.

Thus, the only possible value for $x$ that satisfies the equation is: $x = −2$

Expanding and simplifying gives: $x^2 + 8x − 12 = 0$

Using the quadratic formula, the possible solutions for $x$ are:

$\dfrac{–8±4}{2}$

This means $x = −6$ or $x = −2$. Since log bases cannot be negative, $x = −6$ cannot be a valid answer.

Thus, the only possible value for $x$ that satisfies the equation is: $x = −2$

Question 10 |

**Which of the following values of $x$ satisfies the equation:**$\log_8(x^2 + 16x + 128) = 2$

$-8$ | |

$-4$ | |

$-2$ | |

$0$ |

Question 10 Explanation:

The correct answer is (A). From the definition of logarithm: $x^2 + 16x + 128 = 64$

Subtracting $64$ from both sides gives $x^2 + 16x + 64 = 0$. This quadratic can be factored as $(x + 8)^2 = 0$.

The only possible solution for $x$ that satisfies this equation is: $x = −8$

Subtracting $64$ from both sides gives $x^2 + 16x + 64 = 0$. This quadratic can be factored as $(x + 8)^2 = 0$.

The only possible solution for $x$ that satisfies this equation is: $x = −8$

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