Logarithm Properties - Algebra 3

Solve for $x$ : $\log_2(x-1)+\log_2(x+1)=3.$
Answer: $x=3$ Combine with the product property: $\log_2((x-1)(x+1))=3$ , so $x^2-1=8$ and $x=\pm3$ . The original equation requires $x-1>0$ and $x+1>0$ , so $x>1$ ; therefore only $x=3$ is valid. The options with $x=-3$ or both values fail the domain check, and a real solution does exist.
Given $\log_5 2 = a$ and $\log_5 3 = b$ , which expression equals $\log_2 45$ ?
Answer: $\frac{1+2b}{a}$ By change of base, $\log_2 45 = \frac{\log_5 45}{\log_5 2}$ . Since $45=5\cdot 3^2$ , $\log_5 45=1+2b$ . Dividing by $a$ gives $\frac{1+2b}{a}$ . The option $\frac{1+2b}{1+a}$ uses $1+a$ as $\log_2 5$ ; $\frac{2b}{a}$ forgets the factor $\log_5 5=1$ ; $\frac{2b}{1-a}$ comes from a sign error in the denominator.
Which value is equivalent to $\log_4 27 \cdot \log_9 8$ ?
Answer: $\frac{9}{4}$ When you see a product of logarithms with different bases, your first thought should be: can I rewrite the bases and arguments as powers of the same numbers? Here, $27=3^3$ , $8=2^3$ , $4=2^2$ , and $9=3^2$ . Using the power rule $\log_{a^m} b^n = \frac{n}{m}\log_a b$ , we get $\log_4 27 = \log_{2^2}3^3 = \frac{3}{2}\log_2 3$ , and $\log_9 8 = \log_{3^2}2^3 = \frac{3}{2}\log_3 2$ . Multiplying gives $\frac{9}{4}(\log_2 3)(\log_3 2)$ . Since $\log_a b \cdot \log_b a = 1$ , the product simplifies to $\frac{9}{4}$ . That is the correct value. The choice $\frac{3}{2}$ is a trap for forgetting the reciprocal identity — you might correctly compute each factor as $\frac{3}{2}\log_2 3$ and $\frac{3}{2}\log_3 2$ , then mistakenly multiply only the coefficients but leave the log product as is, or think the log product cancels to zero. The choice $2$ might come from misapplying the change of base formula or incorrectly simplifying $\log_4 27$ to $\frac{3}{2}$ and $\log_9 8$ to $\frac{3}{2}$ then adding instead of multiplying. The choice $3$ could arise from treating $\log_4 27$ as $3\log_4 3$ and $\log_9 8$ as $3\log_9 2$ , then incorrectly assuming $\log_4 3 \cdot \log_9 2 = 1$ without changing bases. Your takeaway: whenever you see $\log_a b \cdot \log_b c$ , consider using $\log_a b = \frac{1}{\log_b a}$ to simplify the product to $\log_a c$ . Here, the key was recognizing that $27$ and $8$ are powers of $3$ and $2$ , allowing you to factor out the exponents cleanly. Practice rewriting logs with matching base-argument pairs to spot these cancellations quickly.
For $x>0$ , which expression is equivalent to $\frac{1}{2}\log_3(9x^4)-\log_3(3x)$ ?
Answer: $\log_3 x$ The power property gives $\frac{1}{2}\log_3(9x^4)=\log_3((9x^4)^{1/2})=\log_3(3x^2)$ . Then the quotient property gives $\log_3(3x^2)-\log_3(3x)=\log_3(x)$ . The value 0 results from incorrectly canceling $3x^2$ and $3x$ ; 1 comes from treating the quotient as 3; $\log_3(x^2)$ comes from subtracting exponents instead of dividing arguments.
Given $\log_{12} 2 = a$ and $\log_{12} 3 = b$ , which expression is equivalent to $\log_6 8$ ?
Answer: $\frac{3a}{a+b}$ Whenever you see a logarithm in an unfamiliar base, your first move is to change to a base you know. Here all given logs are base $12$ , so rewrite $\log_6 8$ using base $12$ : $\log_6 8=\frac{\log_{12} 8}{\log_{12} 6}$ Now simplify each piece. Since $8=2^3$ , $\log_{12} 8=\log_{12}(2^3)=3\log_{12}2=3a$ And since $6=2\cdot 3$ , $\log_{12} 6=\log_{12}2+\log_{12}3=a+b$ So the expression is $\frac{3a}{a+b}$ The key was recognizing $6=2\cdot 3$ , making the denominator a sum. Now look at the traps. The choice $\frac{3a}{1-b}$ uses $1-b=\log_{12}12-\log_{12}3=\log_{12}4$ , not $\log_{12}6$ . The choice $\frac{3a}{a-b}$ uses $a-b=\log_{12}(2/3)$ , again not $\log_{12}6$ . The choice $\frac{3a}{1-a-b}$ simplifies $1-a-b=\log_{12}(12/(2\cdot3))=\log_{12}2=a$ , so it becomes $3$ , not $\log_6 8$ . Each wrong option comes from mixing up product and quotient log rules. Study tip: after changing base, simplify the numerator and denominator separately. Remember $\log(xy)=\log x+\log y$ and $\log(x/y)=\log x-\log y$ . The denominator must equal the original base's input exactly — here, $6$ .
Which value is equivalent to $\log_2 3 \cdot \log_3 5 \cdot \log_5 8$ ?
Answer: $3$ When you see a product of logarithms with different bases, think about the change-of-base formula: $\log_a b = \frac{\ln b}{\ln a}$ . Write each log in terms of natural logs: $\frac{\ln 3}{\ln 2}\cdot \frac{\ln 5}{\ln 3}\cdot \frac{\ln 8}{\ln 5}$ The $\ln 3$ and $\ln 5$ cancel, leaving $\frac{\ln 8}{\ln 2} = \log_2 8$ . Since $2^3 = 8$ , this equals $3$ . So the correct value is $3$ . The choice $1$ is what you'd get if you mistakenly thought all the logs somehow "cancel out" to leave $1$ . The choice $2$ is a trap if you focus on the bases $2,3,5$ and think there's some averaging, but there is no such operation. The choice $8$ is tempting if you confuse the numeric result with the last argument in the expression; the log product is an exponent-like quantity not equal to the final base's power. In fact, $\log_2 8$ gives $3$ , not $8$ . A strong strategy for logarithm products like this is to rewrite every logarithm using a common base and look for telescoping cancellation. If you see a chain $\log_a b \cdot \log_b c \cdot \log_c d$ , the answer is always $\log_a d$ . This saves time and avoids errors on exam problems that test the change-of-base formula.
Evaluate $\log_4 9 \cdot \log_3 8.$
Answer: $3$ Using change of base, $\log_4 9=\frac{\ln 9}{\ln 4}=\frac{2\ln3}{2\ln2}=\frac{\ln3}{\ln2}$ and $\log_3 8=\frac{\ln8}{\ln3}=\frac{3\ln2}{\ln3}$ . The product is 3. The value 2 comes from simplifying only $\log_4 9$ ; 4 and 6 come from incorrectly combining the bases or exponents.
If $u=\log_3 x$ , which expression is equivalent to $\log_x 27$ ?
Answer: $\frac{3}{u}$ Whenever you see a logarithm with a variable base, your first move should be to rewrite everything in exponential form or use the change-of-base idea. Since $u=\log_3 x$ , you know $x=3^u$ . Also, $27=3^3$ , so the expression becomes $\log_x 27=\log_{3^u}(3^3)=\frac{\ln(3^3)}{\ln(3^u)}=\frac{3\ln 3}{u\ln 3}=\frac{3}{u}.$ So the correct expression is $\frac{3}{u}$ . Now look at the traps. The choice $\frac{u}{3}$ is actually $\log_{27}x$ , not $\log_x 27$ ; it's the reciprocal of the correct answer, often produced when you confuse which quantity is the base. The choice $3u$ comes from noticing $27=3^3$ and multiplying by $\log_3 x$ , but that would represent $\log_3(x^3)$ , not a log with base $x$ . Finally, $u^3$ treats the logarithm like a number to be cubed, which confuses the exponent on the input with the log value itself. A strong strategy for this exam: when you see logs with bases that are powers of one another, rewrite all bases as a common base and compare exponents. Here, $x=3^u$ and $27=3^3$ , so the answer is the ratio of those exponents: $\frac{3}{u}$ .
Which expression is equivalent to $2\log_3 x - \log_3(x+1) + \log_3(x-1)$ for $x>1$ ?
Answer: $\log_3\left(\frac{x^2(x-1)}{x+1}\right)$ When you see logarithms being added and subtracted with coefficients, your first thought should be the product, quotient, and power properties. Combine the terms into a single logarithm, being careful that subtraction means the quotient and that a coefficient becomes an exponent. Start with the coefficient: $2\log_3 x = \log_3(x^2)$ . Then handle the difference and sum of the remaining logs: $-\log_3(x+1)+\log_3(x-1) = \log_3\left(\frac{x-1}{x+1}\right)$ . Now multiply the two results inside one log: $\log_3(x^2) + \log_3\left(\frac{x-1}{x+1}\right) = \log_3\left(\frac{x^2(x-1)}{x+1}\right)$ . That is the correct expression. The choice $\log_3\left(\frac{x^2(x+1)}{x-1}\right)$ flips the quotient, turning $(x-1)/(x+1)$ into $(x+1)/(x-1)$ — a sign-order error. The choice $\log_3\left(\frac{x^2}{x^2-1}\right)$ shows the common mistake of writing $x^2/(x^2-1)$ instead of $x^2(x-1)/(x+1)$ ; these are not equivalent because the $x-1$ belongs in the numerator, not as part of a factored denominator. Finally, $\log_3\left(\frac{2x(x-1)}{x+1}\right)$ confuses $2\log_3 x$ with $\log_3(2x)$ — but the power property says the coefficient becomes an exponent, so $2\log_3 x = \log_3(x^2)$ , not $\log_3(2x)$ . A reliable study tip: before combining logs, rewrite every coefficient as an exponent, and always test with a simple value like $x=2$ to verify equivalence. If the two expressions give the same number, you've likely combined them correctly.
If $\log_2 3 = a$ and $\log_3 5 = b$ , which expression equals $\log_{10} 5$ ?
Answer: $\frac{ab}{1+ab}$ Use change of base to base 2: $\log_{10}5=\frac{\log_2 5}{\log_2 10}$ . Since $\log_2 5=ab$ and $\log_2 10=\log_2(2\cdot5)=1+ab$ , the result is $\frac{ab}{1+ab}$ . The other options use $a$ or $b$ alone in place of $ab$ , or incorrectly add $a$ and $b$ in the denominator.
If $\log_a b = p$ and $\log_b c = q$ , which expression equals $\log_a(b^2 c^3)$ ?
Answer: $p(2+3q)$ First use change of base: $\log_a c = \log_a b \cdot \log_b c = pq$ . Then $\log_a(b^2c^3)=2\log_a b+3\log_a c=2p+3pq=p(2+3q)$ . The expression $2p+3q$ treats $q$ as if it were $\log_a c$ ; the options with $q^2$ mishandle the exponent on $q$ .
Suppose $\log_b 2 = p$ , $\log_b 3 = q$ , and $\log_b 5 = r$ . Which expression is equivalent to $\log_b\left(\frac{5\sqrt{5}}{12}\right)$ ?
Answer: $\frac{3}{2}r - 2p - q$ Whenever you see a logarithm of a fraction involving products and roots, your first move is to expand using the properties: $\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N$ and $\log_b(M^k)=k\log_b M$ . Here, rewrite the expression carefully. Notice that $5\sqrt{5} = 5^{3/2}$ because $\sqrt{5}=5^{1/2}$ , and $12 = 3 \cdot 4 = 3 \cdot 2^2$ . So the log becomes: $\log_b\left(\frac{5^{3/2}}{3\cdot 2^2}\right) = \frac{3}{2}\log_b 5 - \log_b 3 - 2\log_b 2 = \frac{3}{2}r - q - 2p.$ That matches the choice $\frac{3}{2}r - 2p - q$ , which is correct. Now, why are the others off? The choice $\frac{3}{2}r - p - q$ loses the factor of 2 in front of $p$ — it treats $12$ as $3\cdot 2$ instead of $3\cdot 2^2$ . The choice $\frac{5}{2}r - 2p - q$ comes from thinking $5\sqrt{5} = 5^{5/2}$ , but that would be $5^2\cdot 5^{1/2}$ , not $5^1\cdot 5^{1/2}$ . Finally, $\frac{3}{2}r - 2p - \frac{1}{2}q$ incorrectly halves the $q$ term — the denominator has $3$ , not $\sqrt{3}$ . Your takeaway: before expanding a logarithm, rewrite any radicals as fractional exponents and factor out all integers completely (like $12 = 2^2 \cdot 3$ ). Then apply the power rule to each term, paying special attention to coefficients. This avoids the most common traps on this exam.
Which expression is NOT equivalent to $\log_2\left(\frac{16x^2}{y}\right)$ for $x,y>0$ ?
Answer: $4+\log_2 x^2 - \log_2(2y)$ Whenever you see a logarithmic expression with products, quotients, and powers, your first move should be to expand it using the log laws: $\log_b(MN)=\log_b M+\log_b N$ , $\log_b(M/N)=\log_b M-\log_b N$ , and $\log_b(M^k)=k\log_b M$ . Starting from $\log_2\left(\frac{16x^2}{y}\right)$ , split the quotient: $\log_2(16x^2)-\log_2 y$ , then the product and power: $\log_2 16+2\log_2 x-\log_2 y=4+2\log_2 x-\log_2 y$ . That is the target form. The choice $4+2\log_2 x-\log_2 y$ is exactly this expansion, so it is equivalent. The choice $\log_2(16x^2)-\log_2 y$ is just the first quotient-splitting step, also equivalent. The choice $2\log_2(4x)-\log_2 y$ works because $2\log_2(4x)=2(\log_2 4+\log_2 x)=2(2+\log_2 x)=4+2\log_2 x$ . All three simplify to the original. The non-equivalent expression is $4+\log_2 x^2-\log_2(2y)$ . It looks close, but $\log_2(2y)=\log_2 2+\log_2 y=1+\log_2 y$ , so it becomes $4+2\log_2 x-(1+\log_2 y)=3+2\log_2 x-\log_2 y$ , missing 1. That extra factor of 2 inside the logarithm becomes $+1$ when pulled out, and subtracting it changes the constant term. Study tip: when comparing logarithmic forms, expand every choice completely into the same base terms. Watch for coefficients hidden inside logs, like $\log_2(2y)$ , which secretly contribute a constant.
Which expression is equivalent to $\ln\left(\frac{e^2\sqrt{x}}{(x+1)^3}\right)$ for $x>0$ ?
Answer: $2+\frac{1}{2}\ln x - 3\ln(x+1)$ Whenever you see a logarithm of a fraction containing products, powers, and roots, apply the three core log rules: $\ln(ab)=\ln a+\ln b$ , $\ln(a/b)=\ln a-\ln b$ , and $\ln(a^r)=r\ln a$ . Here, break the expression apart before simplifying. Start with the numerator: $\ln(e^2\sqrt{x})=\ln(e^2)+\ln(\sqrt{x})$ . Since $\ln(e^2)=2$ and $\sqrt{x}=x^{1/2}$ , this becomes $2+\frac12\ln x$ . Then subtract the log of the denominator: $\ln((x+1)^3)=3\ln(x+1)$ . Because the denominator is divided, you subtract it, giving $2+\frac12\ln x-3\ln(x+1).$ This matches the expression $2+\frac{1}{2}\ln x - 3\ln(x+1)$ exactly. Now look at the wrong choices. The expression $2+\frac{1}{2}\ln x + 3\ln(x+1)$ uses a plus sign before the denominator term, but division inside the logarithm becomes subtraction, not addition. The expression $\frac{1}{2}\ln x - 3\ln(x+1) - 2$ uses $-2$ instead of $+2$ ; the $e^2$ is in the numerator, so its log is positive $2$ . Finally, $2+\ln x - 3\ln(x+1)$ treats $\sqrt{x}$ as $x$ , forgetting that the square root means the exponent is $\frac12$ , not $1$ . A useful habit: rewrite roots as fractional exponents first, then expand the logarithm term by term. Check every sign carefully — the most common trap on this exam is losing the subtraction for a denominator or misreading a root as a plain power.
Solve for $x$ : $\log_2 x + \log_4 x + \log_8 x = 11.$
Answer: $64$ Convert to base 2: $\log_4 x=\frac{1}{2}\log_2 x$ and $\log_8 x=\frac{1}{3}\log_2 x$ . Thus $\left(1+\frac12+\frac13\right)\log_2 x=\frac{11}{6}\log_2 x=11$ , so $\log_2 x=6$ and $x=64$ . The other values result from using only one or two of the base conversions.
A student writes $\log(x^2-1)-\log(x-1)=\log(x+1).$ Which statement best describes the validity of this simplification?
Answer: It is correct for all $x>1$ , but the two sides are not equivalent on all of $\mathbb{R}$ . On the domain of the left side, $x^2-1>0$ and $x-1>0$ , so $x>1$ . There, $\frac{x^2-1}{x-1}=x+1$ , so the identity is valid on that domain. But $\log(x+1)$ is defined for $x>-1$ , so the two sides are not equivalent for every real $x$ . The other statements misidentify the domain or falsely deny that the quotient property applies.
If $\log_a 3 = p$ , what is $\log_3(a^4)$ ?
Answer: $\frac{4}{p}$ From $\log_a 3=p$ , the reciprocal change-of-base property gives $\log_3 a=\frac{1}{p}$ . Therefore $\log_3(a^4)=4\log_3 a=\frac{4}{p}$ . The choice $4p$ treats $p$ as $\log_3 a$ ; $\frac{p}{4}$ inverts the exponent; $\frac{1}{4p}$ inverts both the base and the exponent.
Which expression is equivalent to $\log_5(25x^3) - 2\log_5(5x)$ for $x>0$ ?
Answer: $\log_5 x$ When you see a logarithm expression with coefficients and products, your first move should be to expand using the product and power rules: $\log_b(MN)=\log_b M+\log_b N$ and $\log_b(M^a)=a\log_b M$ . Here, $\log_5(25x^3)=\log_5 25+\log_5(x^3)=2+3\log_5 x$ . Also, $2\log_5(5x)=2(\log_5 5+\log_5 x)=2(1+\log_5 x)=2+2\log_5 x$ . Subtracting gives $(2+3\log_5 x)-(2+2\log_5 x)=\log_5 x$ , so the equivalent expression is $\log_5 x$ . The choice $\log_5(5x)$ is a trap from moving the coefficient 2 inside incorrectly: $2\log_5(5x)$ means $\log_5((5x)^2)=\log_5(25x^2)$ , not $\log_5(5x^2)$ . The choice $2\log_5 x$ comes from distributing the 2 only to part of the log: $2\log_5(5x)$ must become $2+2\log_5 x$ , not $2+\log_5 x$ . The choice $1$ treats $\log_5(5x)$ as if it were just $\log_5 5=1$ , forgetting the $x$ inside; actually $\log_5(5x)=1+\log_5 x$ . A good check is to plug in a non-special value like $x=2$ . Avoid testing $x=5$ , because $\log_5 5=1$ would make the constant $1$ look falsely correct. Always expand fully, and make sure coefficients apply to the entire argument inside the log.
Simplify $\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdots \log_{15} 16.$
Answer: $4$ Using the change-of-base property, each factor equals $\frac{\ln(k+1)}{\ln k}$ . The product telescopes to $\frac{\ln 16}{\ln 2} = \log_2 16 = 4$ . The value 2 is just the first base, 16 is the final argument, and 5 is not produced by the telescoping product.
Which of the following is NOT equivalent to $\log_b\left(\frac{x^2}{y^3}\right)$ for $x,y>0$ ?
Answer: $\frac{2\log_b x}{3\log_b y}$ The quotient property says $\log_b\left(\frac{x^2}{y^3}\right)=\log_b(x^2)-\log_b(y^3)$ , so the first three options are equivalent forms. Logs do not distribute over a quotient as a quotient of logs, so $\frac{2\log_b x}{3\log_b y}$ is not equivalent.

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