Definition of Logarithms - College Algebra

1. If $\log_b(x) = y$, then which statement must be true about the relationship between $b^{y+1}$ and $x$?
   Answer: $b^{y+1} = bx$ Given $\log_b(x) = y$, by the definition of logarithm, this means $b^y = x$. To find $b^{y+1}$, we use the property $b^{y+1} = b^y \cdot b^1 = b^y \cdot b = x \cdot b = bx$. Choice A incorrectly adds $b$ to $x$. Choice C incorrectly raises $x$ to the power $b$. Choice D incorrectly adds $b$ and $x$.
2. A student claims that since $2^3 = 8$, then $\log_2(8) + \log_2(8) = \log_2(16)$. Which statement best describes this claim?
   Answer: The claim is incorrect because $\log_2(8) + \log_2(8) = \log_2(64)$, not $\log_2(16)$ When you encounter logarithm addition problems, the key is remembering the fundamental logarithm properties and how they relate to exponents. Let's work through this step by step. First, we need to find $\log_2(8)$. Since $2^3 = 8$, we have $\log_2(8) = 3$. Now, $\log_2(8) + \log_2(8) = 3 + 3 = 6$. The crucial insight is applying the logarithm addition rule: $\log_b(x) + \log_b(y) = \log_b(xy)$. Therefore, $\log_2(8) + \log_2(8) = \log_2(8 \times 8) = \log_2(64)$. Since $2^6 = 64$, we have $\log_2(64) = 6$. The student's claim that this equals $\log_2(16)$ is wrong because $\log_2(16) = 4$ (since $2^4 = 16$). Looking at the answer choices: Choice A correctly calculates $\log_2(8) = 3$ and $3 + 3 = 6$, but incorrectly states that $\log_2(16) = 6$ when it actually equals 4. Choice B is too vague and doesn't identify the specific error. Choice C makes the fundamental mistake of thinking that if $8 + 8 = 16$, then $\log_2(8) + \log_2(8) = \log_2(16)$, which confuses addition of numbers with addition of logarithms. Choice D correctly identifies that $\log_2(8) + \log_2(8) = \log_2(64)$, not $\log_2(16)$. Remember: when adding logarithms with the same base, you multiply the arguments, not add them. Always apply $\log_b(x) + \log_b(y) = \log_b(xy)$.
3. If $4^{x-1} = 16$ and $\log_2(y) = x$, then what is the value of $y$?
   Answer: $8$ First solve $4^{x-1} = 16$. Since $4 = 2^2$ and $16 = 2^4$, we have $(2^2)^{x-1} = 2^4$, so $2^{2(x-1)} = 2^4$, giving us $2(x-1) = 4$, thus $x-1 = 2$ and $x = 3$. Then from $\log_2(y) = x = 3$, by definition of logarithm, $y = 2^3 = 8$. Choice A results from using $x = 2$ instead of $x = 3$. Choice C results from confusing $y$ with the original equation's result. Choice D results from calculating $2^{3+2} = 2^5 = 32$, adding an extra power.
4. Given that $\log_5(p) = r$ and $\log_5(q) = s$, which equation must be satisfied by $\log_5(p^2q^{-1})$?
   Answer: $\log_5(p^2q^{-1}) = 2r - s$ Using logarithm properties: $\log_5(p^2q^{-1}) = \log_5(p^2) + \log_5(q^{-1}) = \log_5(p^2) - \log_5(q) = 2\log_5(p) - \log_5(q) = 2r - s$. Choice A incorrectly multiplies $2r$ by $s^{-1}$ instead of subtracting $s$. Choice B treats the logarithm of a quotient as if it were a quotient of logarithms. Choice D incorrectly squares $r$ and treats the expression as a fraction rather than using logarithm properties.
5. Given that $\log_3(m) = 2$ and $\log_3(n) = -1$, what is the value of $\log_3\left(\frac{m^2}{n}\right)$?
   Answer: $5$ Using the definition of logarithms and properties: $\log_3\left(\frac{m^2}{n}\right) = \log_3(m^2) - \log_3(n) = 2\log_3(m) - \log_3(n) = 2(2) - (-1) = 4 + 1 = 5$. Choice A results from calculating $2(2) - 1 = 3$, forgetting that $\log_3(n) = -1$. Choice C results from calculating $2(2) + 2(1) + 1 = 7$, incorrectly applying properties. Choice D results from calculating $2(2) + 2(1) = 6$, incorrectly treating the subtraction as addition.
6. If $a = \log_2(8)$ and $b = \log_4(8)$, then which expression correctly represents $\frac{a}{b}$?
   Answer: $\frac{3}{2}$ First find $a$ and $b$. Since $2^3 = 8$, we have $a = \log_2(8) = 3$. For $b = \log_4(8)$, since $4 = 2^2$ and $8 = 2^3$, we can write $4^b = 8$ as $(2^2)^b = 2^3$, giving us $2^{2b} = 2^3$, so $2b = 3$ and $b = \frac{3}{2}$. Therefore $\frac{a}{b} = \frac{3}{3/2} = 3 \times \frac{2}{3} = \frac{6}{3} = 2$. However, this contradicts our answer choice. Let me recalculate: $\frac{a}{b} = \frac{3}{3/2} = 3 \div \frac{3}{2} = 3 \times \frac{2}{3} = \frac{6}{3} = 2$. Wait, this gives us 2, but the correct answer is listed as C (3/2). Let me check: Actually, $\frac{a}{b} = \frac{3}{3/2} = \frac{3 \times 2}{3} = 2$. So the correct answer should be D, not C.
7. Which equation represents the inverse relationship correctly if $f(x) = \log_5(x+2)$?
   Answer: $f^{-1}(x) = 5^x - 2$ To find the inverse of $f(x) = \log_5(x+2)$, let $y = \log_5(x+2)$. By the definition of logarithm, this means $5^y = x+2$, so $x = 5^y - 2$. Therefore $f^{-1}(x) = 5^x - 2$. Choice B incorrectly subtracts 2 in the exponent. Choice C incorrectly adds 2 to the exponent. Choice D incorrectly divides $5^x$ by 2 instead of subtracting 2 from the entire expression.
8. If $\log_a(b) = c$ where $a > 0$, $a \neq 1$, $b > 0$, then which statement about $\log_b(a)$ is correct?
   Answer: $\log_b(a) = \frac{1}{c}$ only when $c \neq 0$ When you encounter logarithm problems involving the relationship between $\log_a(b)$ and $\log_b(a)$, you're working with the reciprocal property of logarithms. This is a fundamental relationship that's worth memorizing. Starting with $\log_a(b) = c$, we can convert this to exponential form: $a^c = b$. Now we want to find $\log_b(a)$. Let's call this value $x$, so $\log_b(a) = x$, which means $b^x = a$. Substituting our first equation into the second: $(a^c)^x = a$, which simplifies to $a^{cx} = a^1$. Therefore, $cx = 1$, and solving for $x$ gives us $x = \frac{1}{c}$. This means $\log_b(a) = \frac{1}{c}$, but only when $c \neq 0$ (since division by zero is undefined). Answer A is correct because it includes the crucial restriction that $c \neq 0$. Answer B fails to acknowledge this restriction—when $c = 0$, the reciprocal relationship breaks down because $\log_a(b) = 0$ means $b = 1$, and $\log_1(a)$ is undefined since 1 cannot be a logarithm base. Answer C incorrectly suggests the relationship is $\log_b(a) = -c$, which would be the negative reciprocal. Answer D uses the notation $c^{-1}$ (which does equal $\frac{1}{c}$) but incorrectly restricts it to $c > 1$ when it actually works for any $c \neq 0$. Remember: $\log_a(b)$ and $\log_b(a)$ are reciprocals, but always check that neither equals zero before applying this relationship.
9. For what value of $k$ does the equation $\log_k(27) = \frac{3}{2}$ hold true?
   Answer: $k = 9$ When you encounter a logarithmic equation like $\log_k(27) = \frac{3}{2}$, you're dealing with the fundamental definition of logarithms. This equation is asking: "What base $k$ raised to the power $\frac{3}{2}$ equals 27?" To solve this, convert the logarithmic form to exponential form. The equation $\log_k(27) = \frac{3}{2}$ means $k^{3/2} = 27$. Now you need to solve for $k$. Since $k^{3/2} = 27$, raise both sides to the power $\frac{2}{3}$ to isolate $k$: $(k^{3/2})^{2/3} = 27^{2/3}$ $k = 27^{2/3}$ To evaluate $27^{2/3}$, recognize that $27 = 3^3$, so: $27^{2/3} = (3^3)^{2/3} = 3^{3 \cdot 2/3} = 3^2 = 9$ Therefore, $k = 9$, which is answer choice D. Let's check why the other answers are wrong: - A) $k = 3$: This gives $3^{3/2} = 3\sqrt{3} \approx 5.2$, not 27 - B) $k = 3\sqrt{3}$: This equals $3^{3/2}$, which when raised to $\frac{3}{2}$ gives $3^{9/4} \approx 15.6$, not 27 - C) $k = 18$: This gives $18^{3/2} = 18\sqrt{18} \approx 76.4$, not 27 Study tip: Always convert logarithmic equations to exponential form first—it makes the algebra much clearer. Remember that $a^{m/n} = \sqrt[n]{a^m}$, and look for perfect powers to simplify your work.
10. If $\log_3(x-1) = 2$ and $\log_3(y+1) = 2$, what can be concluded about the relationship between $x$ and $y$?
    Answer: $x + y = 18$ because both expressions equal 9 when converted From $\log_3(x-1) = 2$, by definition, $x-1 = 3^2 = 9$, so $x = 10$. From $\log_3(y+1) = 2$, we have $y+1 = 3^2 = 9$, so $y = 8$. Therefore $x + y = 10 + 8 = 18$. Choice A is incorrect because while both logarithmic expressions equal 2, the arguments are different ($x-1$ vs $y+1$). Choice B incorrectly assumes a direct relationship between the variables. Choice D correctly identifies that $x - y = 10 - 8 = 2$, but states it as $-2$.

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