Solving Logarithmic Equations - Accuplacer Aaf

If $log_7(5x - 9) = log_7(2x + 3)$ , what is the value of $x$ ?
Answer: $x = 4$ By the one-to-one property of logarithms, if $log_b(M) = log_b(N)$ , then $M = N$ . Therefore, we can set the arguments equal to each other: $5x - 9 = 2x + 3$ . To solve for $x$ , subtract $2x$ from both sides to get $3x - 9 = 3$ . Then, add 9 to both sides to get $3x = 12$ . Finally, divide by 3 to find $x = 4$ . We must check that the arguments are positive for this value of x. For $x=4$ , $5(4)-9=11 > 0$ and $2(4)+3=11 > 0$ , so the solution is valid. Distractor B results from a sign error when moving terms, such as $5x+2x=3+9$ . The other distractors represent other common algebraic errors.
A student solving $\log_3(2x + 1) = \log_3(x - 2) + 1$ arrives at $x = 7$ . Which statement best describes this solution?
Answer: The solution is correct and complete Starting with $\log_3(2x + 1) = \log_3(x - 2) + 1$ , we rewrite the right side: $\log_3(2x + 1) = \log_3(x - 2) + \log_3(3) = \log_3[3(x - 2)]$ . Since the logs have the same base, $2x + 1 = 3(x - 2) = 3x - 6$ . Solving: $2x + 1 = 3x - 6$ , so $x = 7$ . Checking domain: $2x + 1 = 15 > 0$ and $x - 2 = 5 > 0$ , both valid. Verification: $\log_3(15) = \log_3(5) + \log_3(3) = \log_3(5) + 1$ , which is correct. Choice B incorrectly assumes domain violation. Choice C suggests an error that doesn't exist. Choice D incorrectly identifies a property error.
What is the solution to the equation $log_6(x + 5) + log_6(x) = 2$ ?
Answer: $x = 4$ Using the product rule for logarithms, combine the terms on the left: $log_6(x(x + 5)) = 2$ . Convert to exponential form: $x(x + 5) = 6^2$ . Simplify and solve the quadratic equation: $x^2 + 5x = 36$ , or $x^2 + 5x - 36 = 0$ . Factoring gives $(x + 9)(x - 4) = 0$ . The potential solutions are $x = -9$ and $x = 4$ . Checking these in the original equation, $x = -9$ is extraneous because $log_6(-9)$ is undefined. The solution $x = 4$ is valid. Distractor A is the extraneous solution. Distractor C incorrectly includes the extraneous solution. Distractor D results from incorrectly applying the product rule as $log_6(x+5+x)=2$ .
What is the solution set for the equation $log_3(4x - 5) = log_3(2x - 11)$ ?
Answer: No solution By the one-to-one property of logarithms, if $log_b(M) = log_b(N)$ , then $M = N$ . Setting the arguments equal gives $4x - 5 = 2x - 11$ . Subtracting $2x$ from both sides gives $2x - 5 = -11$ . Adding 5 to both sides gives $2x = -6$ . Dividing by 2 yields $x = -3$ . However, this solution must be checked in the original equation. For $x = -3$ , the argument $4x - 5$ becomes $4(-3) - 5 = -17$ . Since the logarithm of a negative number is undefined, $x = -3$ is an extraneous solution. There are no other algebraic solutions, so the equation has no solution. Distractor B is the extraneous algebraic solution. Distractors A and C result from sign errors during the algebraic solution process.
What is the solution to the equation $log_2(3x - 1) = 5$ ?
Answer: $x = 11$ To solve the equation, convert it from logarithmic form to exponential form. The equation $log_b(y) = c$ is equivalent to $b^c = y$ . Thus, $log_2(3x - 1) = 5$ becomes $2^5 = 3x - 1$ . Since $2^5 = 32$ , the equation is $32 = 3x - 1$ . Adding 1 to both sides gives $33 = 3x$ . Dividing by 3 gives $x = 11$ . Distractor A comes from incorrectly converting the equation to $2(3x-1)=5$ . Distractor B comes from mistakenly swapping the base and the result, calculating $5^2 = 3x-1$ . Distractor D is a simple calculation error.
What is the value of $x$ if $log_3(x + 2) = log_2(16)$ ?
Answer: $x = 79$ First, evaluate the right side of the equation. $log_2(16)$ asks for the power to which 2 must be raised to get 16. Since $2^4 = 16$ , $log_2(16) = 4$ . The equation simplifies to $log_3(x + 2) = 4$ . Now, convert this logarithmic equation to its exponential form: $x + 2 = 3^4$ . Since $3^4 = 81$ , we have $x + 2 = 81$ . Subtracting 2 from both sides gives $x = 79$ . Distractor B results from ignoring the bases and incorrectly setting $x+2=16$ . Distractor A results from finding the value of the right side is 4, but incorrectly setting $x+2=4$ .
What is the solution to the equation $log(x + 15) + log(x) = 2$ ?
Answer: $x = 5$ The term $log$ implies a base of 10. Using the product rule, the equation becomes $log(x(x + 15)) = 2$ . Converting to exponential form gives $x(x + 15) = 10^2$ , which simplifies to $x^2 + 15x = 100$ . Rearranging into a standard quadratic form gives $x^2 + 15x - 100 = 0$ . Factoring the quadratic yields $(x + 20)(x - 5) = 0$ . The potential solutions are $x = -20$ and $x = 5$ . We must check these in the original equation. The solution $x = -20$ is extraneous because $log(-20)$ is undefined. The solution $x = 5$ is valid because both $log(5)$ and $log(5+15)=log(20)$ are defined. Distractor A is the extraneous solution. Distractor C incorrectly includes the extraneous solution. Distractor D arises from the common mistake of applying the product rule as $log(x+15+x)=2$ .
If $log_{27}(x) = \frac{2}{3}$ , what is the value of $x$ ?
Answer: $x = 9$ To solve for $x$ , convert the logarithmic equation to its exponential form: $x = 27^{\frac{2}{3}}$ . A rational exponent can be interpreted as a root and a power. The denominator (3) represents the cube root, and the numerator (2) represents the square. So, $x = (\sqrt[3]{27})^2$ . The cube root of 27 is 3, so $x = 3^2 = 9$ . Distractor C comes from incorrectly computing $27 \times \frac{2}{3}$ . Distractor A comes from misinterpreting the exponent as $\sqrt[3]{27} \times 2$ . Distractor D results from miscalculating the exponent, for example as $27^2/3$ .
Which of the following is a solution to the equation $log_3(x) + log_3(x - 8) = 2$ ?
Answer: $x = 9$ First, use the product rule for logarithms, which states that $log_b(M) + log_b(N) = log_b(MN)$ . This transforms the equation to $log_3(x(x - 8)) = 2$ . Convert to exponential form: $x(x - 8) = 3^2$ . Simplify and solve the resulting quadratic equation: $x^2 - 8x = 9$ , which is $x^2 - 8x - 9 = 0$ . Factoring gives $(x - 9)(x + 1) = 0$ , so the potential solutions are $x = 9$ and $x = -1$ . However, logarithms are only defined for positive arguments. The solution $x = -1$ is extraneous because $log_3(-1)$ is undefined. The solution $x = 9$ is valid because both $x$ and $x-8$ are positive. Distractor A is the extraneous solution. Distractor D includes the extraneous solution. Distractor B results from misapplying the product rule as $log_3(x + x - 8) = 2$ .
What is the complete solution set for the equation $log_2(x^2 - 2x) = 3$ ?
Answer: $x = -2$ and $x = 4$ First, convert the equation from logarithmic to exponential form: $x^2 - 2x = 2^3$ . Simplify the right side: $x^2 - 2x = 8$ . Rearrange into a standard quadratic equation: $x^2 - 2x - 8 = 0$ . Factor the quadratic: $(x - 4)(x + 2) = 0$ . The potential solutions are $x = 4$ and $x = -2$ . Both solutions must be checked in the original equation to ensure the argument of the logarithm is positive. For $x = 4$ , $4^2 - 2(4) = 16 - 8 = 8 > 0$ , so it is a valid solution. For $x = -2$ , $(-2)^2 - 2(-2) = 4 + 4 = 8 > 0$ , so it is also a valid solution. Therefore, the complete solution set includes both values. Distractors A and B are incomplete. Distractor C results from an error in converting to exponential form, using $2 \times 3=6$ instead of $2^3=8$ .
What is the value of $x$ in the equation $log_x(81) = 4$ ?
Answer: $x = 3$ Convert the logarithmic equation to its exponential form: $x^4 = 81$ . To solve for $x$ , take the fourth root of both sides: $x = \pm \sqrt[4]{81}$ . Since $3^4 = 81$ , the algebraic solutions are $x = 3$ and $x = -3$ . However, the base of a logarithm must be positive and not equal to 1. Therefore, $x = -3$ is not a valid solution. The only valid solution is $x = 3$ . Distractor A is invalid because the base cannot be negative. Distractor C incorrectly includes the invalid negative base. Distractor D results from an incorrect calculation, $81/4$ .
Solve for $x$ : $3log_4(x - 2) + 5 = 11$
Answer: $x = 18$ First, isolate the logarithm term. Subtract 5 from both sides: $3log_4(x - 2) = 6$ . Divide by 3: $log_4(x - 2) = 2$ . Now, convert the equation to exponential form: $x - 2 = 4^2$ . Simplify: $x - 2 = 16$ . Add 2 to both sides to find $x = 18$ . The solution is valid as $18-2=16$ is positive. Distractor A comes from an error in converting to exponential form, calculating $x-2 = 4 \times 2 = 8$ . Distractor D comes from an order of operations error where 3 is subtracted before division. Distractor C comes from incorrectly dividing the argument of the log by 3.
Solve for $x$ : $log_4(log_2(x)) = 1$
Answer: $x = 16$ This is a nested logarithm problem that must be solved from the outside in. First, convert the outer logarithm to exponential form. The equation $log_4(A) = 1$ where $A = log_2(x)$ becomes $A = 4^1$ , so $log_2(x) = 4$ . Now, solve this second logarithmic equation by converting it to exponential form: $x = 2^4$ . Calculating the power gives $x = 16$ . Distractor A results from incorrectly solving $log_2(x) = 1$ , perhaps by confusing the bases. Distractor C results from an error in the second conversion step, calculating $x = 2 \times 4$ instead of $x = 2^4$ . Distractor B comes from stopping after the first step, where $log_2(x) = 4$ , and confusing the result with the variable.
What is the solution to the equation $log_3(x + 24) = 2 + log_3(x)$ ?
Answer: $x = 3$ First, isolate the logarithm terms on one side: $log_3(x + 24) - log_3(x) = 2$ . Use the quotient rule for logarithms: $log_3(\frac{x+24}{x}) = 2$ . Convert to exponential form: $\frac{x+24}{x} = 3^2 = 9$ . Solve for $x$ : $x + 24 = 9x$ . Subtract $x$ from both sides: $24 = 8x$ . Divide by 8 to get $x = 3$ . Checking the solution shows it is valid. Distractor D results from misinterpreting $2 + log_3(x)$ as $log_3(2x)$ . Distractor B results from an error in exponential conversion, using $3 \times 2 = 6$ instead of $3^2 = 9$ .
What is the solution to the equation $log_4(x+1) + log_4(x-5) = log_4(2x-9)$ ?
Answer: $x = 3 + \sqrt{5}$ First, use the product rule on the left side: $log_4((x+1)(x-5)) = log_4(2x-9)$ . By the one-to-one property, $(x+1)(x-5) = 2x-9$ . Expand the left side: $x^2 - 4x - 5 = 2x - 9$ . Move all terms to one side to form a quadratic equation: $x^2 - 6x + 4 = 0$ . This does not factor easily, so use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ : $x = \frac{6 \pm \sqrt{36-16}}{2} = \frac{6 \pm \sqrt{20}}{2} = \frac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5}$ . We must check both potential solutions. The domain requires $x+1>0$ , $x-5>0$ , and $2x-9>0$ , which simplifies to $x>5$ . The solution $x = 3 + \sqrt{5} \approx 3+2.24 = 5.24$ is greater than 5 and is valid. The solution $x = 3 - \sqrt{5} \approx 3-2.24 = 0.76$ is not greater than 5 and is extraneous. Distractor A is the extraneous solution. Distractor C includes the extraneous solution. Distractor D might be chosen if one misapplies log rules to get an impossible statement like $-4=-9$ .
What is the solution to the equation $2 + ln(x) = 5$ ?
Answer: $x = e^3$ First, isolate the natural logarithm term. Subtract 2 from both sides of the equation: $ln(x) = 5 - 2$ , which simplifies to $ln(x) = 3$ . The notation $ln(x)$ is shorthand for $log_e(x)$ . To solve for $x$ , convert the equation to its exponential form: $x = e^3$ . Distractor C results from confusing the natural log (base $e$ ) with the common log (base 10). Distractor B results from an order of operations error, evaluating $ln(2+x)=5$ instead. Distractor D represents a misunderstanding of how to convert from logarithmic to exponential form.
If $2\log_4(x) = \log_4(x + 12)$ , what is the value of $x$ ?
Answer: $x = 4$ Using the logarithm property $n\log_a(x) = \log_a(x^n)$ , we rewrite the left side: $\log_4(x^2) = \log_4(x + 12)$ . Since the logarithms have the same base, their arguments must be equal: $x^2 = x + 12$ . Rearranging: $x^2 - x - 12 = 0$ . Factoring: $(x - 4)(x + 3) = 0$ , giving $x = 4$ or $x = -3$ . Checking domain restrictions: for $\log_4(x)$ to be defined, we need $x > 0$ . Therefore, $x = -3$ is extraneous, leaving $x = 4$ as the only valid solution. Verification: $2\log_4(4) = 2(1) = 2$ and $\log_4(4 + 12) = \log_4(16) = \log_4(4^2) = 2$ . Choice B gives $2\log_4(6) \neq \log_4(18)$ . Choice C gives $2\log_4(12) \neq \log_4(24)$ . Choice D gives $2\log_4(3) \neq \log_4(15)$ .
What is the solution to the equation $2log_5(x) = log_5(3x + 10)$ ?
Answer: $x = 5$ Use the power rule for logarithms, $n log_b(M) = log_b(M^n)$ , to rewrite the left side: $log_5(x^2) = log_5(3x + 10)$ . By the one-to-one property, $x^2 = 3x + 10$ . Rearrange into a quadratic equation: $x^2 - 3x - 10 = 0$ . Factoring gives $(x - 5)(x + 2) = 0$ . The potential solutions are $x = 5$ and $x = -2$ . We must check for extraneous solutions. For $x = 5$ , the arguments are positive. For $x = -2$ , the term $log_5(-2)$ is undefined. Thus, $x = -2$ is an extraneous solution. The only valid solution is $x = 5$ . Distractor B is the extraneous solution. Distractor D incorrectly includes the extraneous solution. Distractor A results from misapplying the power rule as $log_5(2x) = log_5(3x+10)$ .
What is the solution to the equation $log_4(x + 12) - log_4(x) = 2$ ?
Answer: $x = \frac{4}{5}$ Using the quotient rule for logarithms, $log_b(M) - log_b(N) = log_b(M/N)$ , the equation becomes $log_4(\frac{x+12}{x}) = 2$ . Converting to exponential form gives $\frac{x+12}{x} = 4^2 = 16$ . To solve for $x$ , multiply both sides by $x$ : $x + 12 = 16x$ . Subtract $x$ from both sides: $12 = 15x$ . Divide by 15: $x = \frac{12}{15} = \frac{4}{5}$ . This solution is valid as it makes the arguments of both logarithms positive. Distractor C results from incorrectly flipping the fraction when solving $12=15x$ . Distractor D results from misinterpreting the conversion to exponential form as $\frac{x+12}{x} = 4 \times 2 = 8$ . Distractor A is a sign error.
To solve $\log_5(x + 4) - \log_5(x - 1) = 1$ , a student first combines the logarithms and then converts to exponential form. What equation does the student obtain after these two steps?
Answer: $\frac{x + 4}{x - 1} = 5$ Using the logarithm property $\log_a(m) - \log_a(n) = \log_a(\frac{m}{n})$ , the equation becomes $\log_5\left(\frac{x+4}{x-1}\right) = 1$ . Converting to exponential form using the definition $\log_a(y) = b$ means $a^b = y$ : $5^1 = \frac{x+4}{x-1}$ , which gives $\frac{x+4}{x-1} = 5$ . Choice A shows the equation after cross-multiplication (a third step). Choice C incorrectly treats the logarithm operation as simple subtraction. Choice D results from incorrectly substituting before applying properties.

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