📝 General logarithms base b (29 MCQs)

📖 Explanation: The base $b$ logarithm is defined as the ratio of the natural logarithm of the argument to the natural logarithm of the base. This definition is derived from the inverse relationship between $b^x$ and $\log_b x$, making option B the correct and standard formula.

📖 Explanation: Using the chain rule, the derivative of $\log_b(u)$ is \frac{u'}{u \ln b}. Here, $u = 3x^2+1$, so u' = 6x. Substituting gives $\frac{6x}{(3x^2+1)\ln 5}$. Option B misses the $\ln 5$ factor, and option C incorrectly places it in the numerator.

📖 Explanation: The student is correct. The product rule for logarithms states $\log_b(xy) = \log_b x + \log_b y$. Since $8 = 4 \times 2$, $\log_2 8 = \log_2 4 + \log_2 2 = 2 + 1 = 3$. The misconception might be that the rule doesn't apply to all bases, but it does. Option D, while providing a correct calculation, supports the statement, making A the best answer.

📖 Explanation: The equation $\log_x 27 = 3$ means $x^3 = 27$. Taking the cube root of both sides gives $x = 3$. Option C, $\sqrt[3]{27}$, equals 3 as well, but it's a calculation step, not the final simplified value. Options B and D are common errors from misapplying the exponent rules.

📖 Explanation: We solve by converting to exponential form: $(\sqrt{2})^y = 8$. Since $\sqrt{2} = 2^{1/2}$ and $8 = 2^3$, we have $2^{y/2} = 2^3$, so $y/2=3$ and $y=6$. Option B is the argument, option C is a common mistake in solving the exponent, and option D is often chosen if the base is misinterpreted.

📖 Explanation: The property $\log_{1/b} x = -\log_b x$ shows that the two functions are opposites. Reflection of a graph across the x-axis transforms $y = f(x)$ into $y = -f(x)$. Therefore, the graphs are reflections across the x-axis. Option C is the reflection of a function and its inverse, which does not apply here.

📖 Explanation: $\log_b \sqrt{6} = \log_b (6^{1/2}) = \frac{1}{2} \log_b 6 = \frac{1}{2} (\log_b 2 + \log_b 3) = \frac{1}{2} (0.3010 + 0.4771) = 0.38905$. Option A is the sum without dividing by 2. Option C is $\log_b(3/2)$, and D is $\log_b 4$.

📖 Explanation: For $x > 1$, a logarithm with a smaller base yields a larger value. This is because you need a higher exponent to reach $x$ from a smaller base. For example, $\log_2 8 = 3$, while $\log_4 8 = 1.5$. Thus, $\log_2 x$ gives the largest value. This concept is often confused with the idea that a larger base means a larger number.

📖 Explanation: This uses the 'change of base' or inverse property: $\log_a b = \frac{1}{\log_b a}$. If $\log_b a = 3$, then $\log_a b = \frac{1}{3}$. Option A is a common error, thinking the relationship is reciprocal, and option D is squaring the value instead of taking the reciprocal.

📖 Explanation: The argument of a logarithm must be positive. So $x^2 - 4x + 3 > 0$. Factoring gives $(x-1)(x-3) > 0$. This inequality holds when $x > 3$ or $x < 1$. Option A misses the $x < 1$ part, and C is the union of intervals where the inequality is not satisfied. Option D is only one part of the solution.

📖 Explanation: The equation is false. The correct product rule is $\log_b(xy) = \log_b x + \log_b y$, which applies to the product of two arguments, not the sum. $\log_2 (x+3)$ is not equal to $\log_2 x + \log_2 3$. Option A is a common mistake of confusing the rule for products with sums. Option B is incorrect because the rule doesn't apply, period.

📖 Explanation: The point $(16, 2)$ means $\log_b 16 = 2$. This converts to $b^2 = 16$. Since $b > 0$ and $b \neq 1$, $b = 4$. Option B is a common error, thinking $b$ is the argument. Option C is the exponent value, and D is another common miscalculation of the square root.

📖 Explanation: Since the bases are the same and the log function is one-to-one, we can equate the arguments: $x+1 = 2x-3$. Solving gives $x = 4$. Option B is the solution to $x+1 = -(2x-3)$, which is incorrect. Option C is a common miscalculation. Option D would be considered if the solution didn't satisfy the domain; $x=4$ gives arguments 5 and 5, which are positive, so it is valid.

📖 Explanation: The population triples in 4 hours, so $3N_0 = N_0 b^4$. Solving for $b$: $b^4 = 3$ implies $b = 3^{1/4}$. Option A incorrectly multiplies the exponent. Option C misinterprets the relationship between time and growth factor. Option D is a simple addition of the two numbers.

📖 Explanation: $\log_{10} 1000 = 3$ because $10^3 = 1000$. $\log_{10} 0.01 = \log_{10} (10^{-2}) = -2$. The difference is $3 - (-2) = 5$. Option B is only the first term. Option C is only the second term. Option D is a common error in handling the negative sign in subtraction.

📖 Explanation: Options B and C are standard properties of logarithms. Option A is a common error; the logarithm of a sum is not the sum of the logarithms. Therefore, both B and C are correct, making D the right choice. This question tests the ability to identify and distinguish between valid and invalid logarithmic properties.

📖 Explanation: A horizontal shift to the left by 3 units transforms $f(x)$ into $f(x+3)$. Therefore, the new function is $g(x) = \log_2 (x+3)$. Option A represents a vertical shift up. Option B represents a shift to the right. Option D represents a vertical stretch.

📖 Explanation: The inverse of a logarithmic function $y = \log_b x$ is the exponential function $y = b^x$. Therefore, the inverse of $\log_3 x$ is $3^x$. Option B is the reciprocal of the argument, which is incorrect. Option C is the inverse of $x^3$, not $\log_3 x$. Option D is the reciprocal of the function, not the inverse.

📖 Explanation: We can use the change of base formula: $\log_b a = \frac{\ln a}{\ln b}$. The expression becomes $\frac{\ln 5}{\ln 2} \cdot \frac{\ln 8}{\ln 5} = \frac{\ln 8}{\ln 2} = \log_2 8 = 3$. Option A is the product of the arguments, which is incorrect. Option B is only the second term. Option D is equivalent to 3 but is in the form of a logarithm, making C the simplified and correct answer.

📖 Explanation: The student is correct. $\log_2 (4x^2) = \log_2 4 + \log_2 x^2 = 2 + 2\log_2 x$. Option B is incorrect as it doesn't correctly separate the factors. Option C is a common error in applying the power rule. Option D is false because the expression is indeed simplifiable.

📖 Explanation: The equation $\log_x 64 = \frac{3}{2}$ means $x^{3/2} = 64$. Raising both sides to the $2/3$ power gives $x = 64^{2/3} = (64^{1/3})^2 = 4^2 = 16$. Option B is a common mistake of multiplying 64 by 3/2. Option C is a misEasy of the exponent rule. Option D is often chosen if the problem is misread as $\log_x 64 = 3$.

📖 Explanation: Using the quotient rule: $\log_2 \frac{x^2-1}{x-1} = 2$. $\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1$ (for $x \neq 1$). So $\log_2 (x+1) = 2$, giving $x+1=4$, $x=3$. Option B is correct. Option A is incorrect. Option C, $x=-1$, makes the argument of the original logs negative or zero, so it's extraneous. Option D is a common error in solving the resulting linear equation.

📖 Explanation: The base of a logarithm must be positive and not equal to 1. Therefore, the domain is $x > 0$ and $x \neq 1$. Option B misses the $x \neq 1$ restriction. Option C is too restrictive. Option D is incorrect because $x$ cannot be negative or zero. This is a common point of confusion where students forget the base restrictions.

📖 Explanation: Given $50 = 100 \cdot b^2$, we have $b^2 = 0.5$. Since $b > 0$, $b = \sqrt{0.5}$. Option B would be $b = 0.25$, which is incorrect. Option C is the value for $b$ if the relationship were $P=100\cdot b^{d/2}$. Option D is the value for $b$ if the percentage at 1 meter were 50%.

📖 Explanation: Using the product rule: $\log_4 [(x+2)(x-4)] = 2$. So $(x+2)(x-4) = 4^2 = 16$. This simplifies to $x^2 - 2x - 24 = 0$, giving $(x-6)(x+4)=0$. $x = 6$ or $x = -4$. Checking the domain, $x$ must be > 4, so $x = 6$. Option B is incorrect. Option C includes the extraneous solution $x=2$ from an incorrect factorization. Option D is the extraneous negative solution.

📖 Explanation: $A = \log_2 4 = 2$ because $2^2 = 4$. $B = \log_4 2 = \frac{1}{2}$ because $4^{1/2} = 2$. Therefore, $A > B$. Option C is a common misconception that these are reciprocals in value, but they are reciprocals of the exponents. Option D is incorrect; their values are easily determined.

📖 Explanation: Using the change of base formula, $\log_x 10 = \frac{\log_{10} 10}{\log_{10} x} = \frac{1}{\log_{10} x}$. The equation becomes $y = \frac{1}{y}$ where $y = \log_{10} x$. This gives $y^2 = 1$, so $y = \pm 1$. Since $x > 0$, $y = 1$ gives $x = 10$, and $y = -1$ gives $x = 0.1$. Option A is the correct set of both solutions. Option B misses the $x=0.1$ solution. Option C is incorrect, as $x=1$ makes the log base 1, which is undefined.

📖 Explanation: Equating arguments: $x^2 = 3x-2$, so $x^2 - 3x + 2 = 0$, giving $x = 1, 2$. Checking the domain, for $x = 1$, the arguments are $1$ and $1$, both positive. For $x = 2$, the arguments are $4$ and $4$, both positive. So both are valid. Option A is correct. Option C is a common mistake of assuming the smaller integer is the only solution. Option D is incorrect because both values are valid.

📖 Explanation: The vertical asymptote of a logarithmic function $y = \log_b (x-h)$ occurs when the argument is zero: $x+2=0$, so $x = -2$. Since the base $2 > 1$, the function is increasing. Option A has the incorrect sign for the asymptote. Option C correctly identifies the asymptote but incorrectly states the function is decreasing. Option D has both incorrect.