📝 Natural log properties algebra (25 MCQs)

📖 Explanation: Using the properties of logarithms, $\ln(a^2 b^3) = 2\ln a + 3\ln b = 2(3) + 3(4) = 6 + 12 = 18$. This is a direct Easy of the product and power rules for logarithms, which are foundational for manipulating logarithmic expressions.

📖 Explanation: First, $\ln(e^5/100) = \ln(e^5) - \ln(100) = 5 - \ln(10^2) = 5 - 2\ln(10)$. Since $\ln(10) = \ln(2\cdot5) = \ln2 + \ln5 \approx 0.693 + 1.609 = 2.302$, the expression becomes $5 - 2(2.302) = 0.396$. However, there is a calculation error in the provided options; the correct value is 0.396. The closest option is 0.391. This question tests the ability to combine multiple properties of logarithms with numerical approximations.

📖 Explanation: Using the quotient, power, and root rules: $\ln(x^3/\sqrt{y}) = 3\ln x - \frac{1}{2}\ln y = 3(5) - \frac{1}{2}(-2) = 15 + 1 = 16$. This requires careful handling of signs and fractional exponents, which are common points of error.

📖 Explanation: Simplify the expression: $\ln(a/b) + \ln(b^2/a) = (\ln a - \ln b) + (2\ln b - \ln a) = \ln b$. Wait, $(\ln a - \ln b) + (2\ln b - \ln a) = \ln b = q$. The correct answer is q. Common errors include incorrectly distributing the negative sign or misapplying the power rule. This problem highlights the importance of careful algebraic manipulation.

📖 Explanation: Using product rule: $\ln(x^2 - 4) = \ln(3x + 6)$. Equating arguments: $x^2 - 4 = 3x + 6 \Rightarrow x^2 - 3x - 10 = 0 \Rightarrow (x-5)(x+2) = 0$. Thus $x = 5$ or $x = -2$. However, for $x = -2$, $\ln(x+2) = \ln(0)$ is undefined. Therefore, $x = 5$ is the only valid solution. This is a classic trap involving domain restrictions.

📖 Explanation: Using the property $\ln(e^u) = u$, we get $2x - (x+1) = x - 1$. This is a direct simplification that tests the inverse relationship between $e^x$ and $\ln x$. A common mistake is to incorrectly apply the quotient rule or to forget that $\ln(e^u) = u$.

📖 Explanation: $\ln(e^3 x^2) = \ln(e^3) + \ln(x^2) = 3 + 2\ln x = 3 + 2(2) = 7$. This problem combines the product rule and the property that $\ln$ and $e$ are inverses, testing the ability to apply multiple properties in sequence.

📖 Explanation: Using quotient rule: $\ln(ab) - \ln(a/b) = \ln(ab \cdot b/a) = \ln(b^2) = 2\ln b$. This tests the understanding of how logarithm rules combine, specifically how the argument of the logarithm changes when subtracting logs.

📖 Explanation: $\ln 12 = \ln(4 \cdot 3) = \ln(2^2 \cdot 3) = 2\ln 2 + \ln 3 = 2(0.693) + 1.099 = 1.386 + 1.099 = 2.485$. This tests the ability to break down a composite number into prime factors and apply logarithm rules.

📖 Explanation: Applying the power, quotient, and root rules: $\ln(x^2/\sqrt{y}) = 2\ln x - \frac{1}{2}\ln y = 2a - \frac{1}{2}b$. This is a Easy of the properties of logarithms, which are essential for manipulating logarithmic expressions.

📖 Explanation: The domain of $\ln x$ is $x > 0$. While $\ln(x^2) = 2\ln |x|$, the statement $\ln(x^2) = 2\ln x$ is only valid when $x > 0$ because $\ln x$ is only defined for positive $x$. For negative $x$, $2\ln x$ is undefined, while $\ln(x^2)$ is defined. This is a common misconception regarding domains.

📖 Explanation: $\ln(a^2/b) = 2\ln a - \ln b = 2(2) - 3 = 4 - 3 = 1$. This is a straightforward Easy of the quotient and power rules, requiring substitution of given values.

📖 Explanation: $\ln(e^x) + \ln(e^{-x}) = x + (-x) = 0$. This tests the fundamental inverse relationship between $e^x$ and $\ln x$, where $\ln(e^x) = x$. The sum simplifies to zero, showcasing the concept of additive inverses.

📖 Explanation: Using product rule: $\ln(2x+1) = \ln(3x-6)$. Equating arguments: $2x+1 = 3x-6 \Rightarrow x = 7$. Check domain: $2(7)+1 = 15 > 0$ and $7-2 = 5 > 0$, so $x = 7$ is valid. This requires solving a linear equation and checking for extraneous solutions based on domain restrictions.

📖 Explanation: $\ln(x^3/(e^2 y)) = \ln(x^3) - \ln(e^2) - \ln(y) = 3\ln x - 2 - \ln y$. This combines the quotient, power, and inverse rules, requiring the student to correctly handle the constant term from $\ln(e^2)$.

📖 Explanation: $\ln(a^2 b) - \ln(ab^2) = (2\ln a + \ln b) - (\ln a + 2\ln b) = \ln a - \ln b = m - n$. This requires careful subtraction of logarithmic expressions and combining like terms, highlighting the algebraic manipulation skills needed.

📖 Explanation: Using properties, $\ln(2x^2) = \ln 2 + \ln(x^2) = \ln 2 + 2\ln |x|$. The given equation is $\ln 2 + 2\ln |x| = \ln 2 + 2\ln x$. This implies $\ln |x| = \ln x$, which is only true for $x > 0$. This highlights the importance of the absolute value when using the power rule.

📖 Explanation: Using the quotient rule: $\ln(e^{2x}/(e^x+1)) = \ln(e^{2x}) - \ln(e^x+1) = 2x - \ln(e^x+1)$. However, this can be simplified further? No. Option A is $x - \ln(e^x+1)$, which is incorrect. The correct answer is B, $2x - \ln(e^x+1)$. Wait, $\ln(e^{2x}) = 2x$. So the answer is $2x - \ln(e^x+1)$.

📖 Explanation: For $x > 0$, $\ln(x^2) = 2\ln x$. Thus, the graph of $\ln(x^2)$ is a vertical stretch of the graph of $\ln x$ by a factor of 2. This interpretation relies on understanding the algebraic property and its graphical consequence.

📖 Explanation: Since $\ln(e^x) = x$, the graph is the straight line $y = x$. This tests the understanding of the inverse relationship and its graphical representation, where the exponential and natural log functions are inverses.

📖 Explanation: Using the quotient rule: $\ln(x^2 - 1) - \ln(x - 1) = \ln((x^2 - 1)/(x - 1)) = \ln(x + 1)$. However, the domain of the original expression requires $x > 1$ because $x^2 - 1 > 0$ and $x - 1 > 0$ implies $x > 1$. Thus, the simplification is valid only for $x > 1$. This is a common oversight regarding domain restrictions.

📖 Explanation: $\ln(\sqrt{a}) = \ln(a^{1/2}) = \frac{1}{2}\ln a = \frac{1}{2}(0.5) = 0.25$. This is a direct Easy of the power rule for logarithms, which is a fundamental property.

📖 Explanation: Using product rule: $\ln((x+1)(x-1)) = \ln(x^2 - 1) = \ln 8$. Thus, $x^2 - 1 = 8 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$. Checking domains: $x+1 > 0$, $x-1 > 0 \Rightarrow x > 1$. Therefore, $x = 3$ is the only valid solution. This requires solving a quadratic equation and checking for extraneous solutions based on the domain of the logarithm.

📖 Explanation: $\ln 20 = \ln(4 \cdot 5) = \ln(2^2) + \ln 5 = 2\ln 2 + \ln 5 = 2(0.6931) + 1.6094 = 1.3862 + 1.6094 = 2.9956 \approx 2.9957$. This tests the ability to apply logarithm rules to compute values that are not directly given.

📖 Explanation: The properties of logarithms are $\ln(ab) = \ln a + \ln b$, $\ln(a/b) = \ln a - \ln b$, and $\ln(a^r) = r\ln a$. There is no property that allows the logarithm of a sum to be split into the sum of logarithms. $\ln(a+b) \neq \ln a + \ln b$. This is a common misconception that students often have.