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Quizzes > High School Quizzes > Mathematics

Ace the Log Quiz Practice Test

Sharpen skills with interactive questions and tips

Difficulty: Moderate
Grade: Grade 9
Study OutcomesCheat Sheet
Colorful paper art promoting The Log Quiz Challenge, a high school-level math quiz on logarithms.

This logarithms quiz helps you practice key rules for high school math and solve basic to midlevel problems. Work through 20 fast questions to check gaps before an exam, build confidence, and see which topics to review next, like properties of logs, simplifying expressions, and solving equations.

What is the definition of log₝(b)?
The exponent to which b must be raised to yield a
The product of a and b
The exponent to which a must be raised to yield b
The difference between a and b
Logarithm log₝(b) is defined as the exponent c such that aᶜ = b. It answers the question: 'to what power should the base a be raised to obtain b?'
Which of the following is equivalent to log_b(MN)?
log_bM + log_bN
log_bM * log_bN
log_bM - log_bN
log_b(M/N)
The product rule for logarithms states that log_b(MN) equals log_bM + log_bN. This is fundamental for simplifying logarithmic expressions.
What is the value of log₝₀(100)?
100
10
2
5
Since 10 raised to the power of 2 equals 100, log₝₀(100) equals 2. This tests understanding of converting from exponential form back to logarithm.
Convert the exponential equation 2ˣ = 16 to its logarithmic form.
x = log₂(16)
x = log₝₆(2)
16 = log₂(x)
16 = logₓ(2)
Converting 2ˣ = 16 to logarithmic form involves identifying x as the exponent to which 2 must be raised to obtain 16, yielding x = log₂(16). This uses the definition of a logarithm.
What is the logarithm property for the quotient log_b(M/N)?
log_bM * log_bN
log_bM - log_bN
log_bM + log_bN
log_b(M+N)
The quotient rule for logarithms states that log_b(M/N) = log_bM - log_bN. This rule is essential for breaking down logarithms of fractions.
Solve for x: log₃(x) = 4.
x = 64
x = 81
x = 7
x = 12
The equation log₃(x) = 4 means that 3 raised to the power of 4 equals x, hence x = 81. This tests understanding of converting a logarithmic equation into exponential form.
Using the power rule, simplify 3 * log...(x).
log₃(x❵)
log...(x³)
3 + log...(x)
log...(3x)
The power rule of logarithms allows us to rewrite k * log_b(x) as log_b(xᵝ), so 3 * log...(x) simplifies to log...(x³). This is a common manipulation of logarithmic expressions.
If log₂(x) + log₂(x-2) = 3, what is the value of x?
x = -4
x = 4
x = 6
x = 2
Combining the logarithms gives log₂[x(x-2)] = 3, so x(x-2) = 2³ = 8. Solving the quadratic and considering domain restrictions eliminates extraneous solutions, leaving x = 4.
What is the change of base formula for log_b(a)?
log_b(a) = log_c(a) / log_c(b)
log_b(a) = log(a) - log(b)
log_b(a) = log_c(b) / log_c(a)
log_b(a) = log(a) * log(b)
The change of base formula expresses log_b(a) as the ratio log_c(a) divided by log_c(b), where c can be any positive number. This method is especially useful on calculators.
Solve for x: log(x-1) + log(x+1) = log(15).
x = 4
x = 2
x = 5
x = 8
Combining the two logarithms using the product rule yields log[(x-1)(x+1)] = log(15), so (x-1)(x+1) = 15. Solving x² - 1 = 15 gives x² = 16, hence x = 4 (rejecting the negative solution due to domain constraints).
Express the exponential equation e^(2x) = 7 in logarithmic form.
2x = log₝₀(7)
x = ln(7)
x = log'(7)
2x = ln(7)
Taking the natural logarithm of both sides of e^(2x) = 7 gives 2x = ln(7), converting the equation to logarithmic form. Recognizing that ln is the natural logarithm (base e) is essential.
Which property does the equation log_b(M) - log_b(N) = log_b(M/N) illustrate?
Quotient Rule
Product Rule
Change of Base Rule
Power Rule
This equation demonstrates the quotient rule for logarithms, which states that subtracting the logarithms of two numbers with the same base is equivalent to taking the logarithm of their quotient.
What is the domain of the function f(x) = log₃(x-5)?
x ≠ 5
x > 5
x < 5
x ≥ 5
For the logarithm to be defined, the argument must be positive. Setting x - 5 > 0 gives x > 5, which is the domain of the function.
Solve for x: log₄(2x+8) = 2.
x = 0
x = 8
x = 4
x = 2
Converting log₄(2x+8) = 2 into exponential form gives 2x + 8 = 4² = 16. Solving 2x + 8 = 16 results in 2x = 8 and consequently x = 4.
Determine x if log₂(x) - log₂(3) = 5.
x = 96
x = 64
x = 32
x = 48
Using the quotient rule, log₂(x) - log₂(3) becomes log₂(x/3). Setting log₂(x/3) equal to 5 implies x/3 = 2❵ = 32, hence x = 96.
Solve for x: log₂(x² - 5x + 6) = 1.
x = 1 only
x = 1 or x = 4
x = 4 only
x = -1 or x = 4
Converting log₂(x² - 5x + 6) = 1 to exponential form gives x² - 5x + 6 = 2. Simplifying to x² - 5x + 4 = 0 and factoring yields (x - 1)(x - 4) = 0, so x = 1 or x = 4. Both solutions are valid as they keep the logarithm's argument positive.
Find the solution for x in the equation log₃(2x+1) + log₃(x-2) = 1.
x = 2.5
x = 4
x = 3
x = -1
Combining the logarithms using the product rule gives log₃[(2x+1)(x-2)] = 1, which implies (2x+1)(x-2) = 3. Solving the resulting quadratic produces two potential solutions; however, only x = 2.5 satisfies the domain conditions for both logarithms.
Solve: log...(x) + log...(x-4) = log...(21).
x = 7
x = 3
x = -3
x = 21
Combining the logarithms results in log...[x(x-4)] = log...(21), so setting x(x-4) equal to 21 leads to the quadratic equation x² - 4x - 21 = 0. This factors to give x = 7 or a negative solution, but only x = 7 is acceptable based on the domain restrictions.
Determine the value of y in the equation ln(y) - ln(y-2) = ln(3).
y = 3
y = 6
y = 2
y = 1
Using the logarithm quotient rule, ln(y) - ln(y-2) simplifies to ln[y/(y-2)] = ln(3), implying y/(y-2) = 3. Solving for y yields y = 3, which also satisfies the necessary domain conditions.
If log₝(x) = 2 and log₝(y) = 3, what is log₝(x² * y)?
8
6
7
5
Using the power and product rules, log₝(x² * y) can be written as 2·log₝(x) + log₝(y) which is 2·2 + 3 = 7. This problem reinforces how to combine and simplify logarithmic expressions.
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Study Outcomes

  1. Apply logarithmic properties to simplify complex expressions.
  2. Analyze the relationship between exponential functions and logarithms.
  3. Solve equations involving logarithms and exponents.
  4. Interpret logarithmic expressions in practical contexts.
  5. Evaluate the impact of domain restrictions on logarithmic functions.

Log Quiz Practice Test Cheat Sheet

  1. Understanding Logarithms - Think of a logarithm as the "undo" button for exponents: it tells you what power you raise the base to in order to get a certain number. Once you master this, you'll breeze through huge or tiny values without breaking a sweat.
  2. Product Rule - Logs turn multiplication into addition, so log₝(xy) becomes log₝(x) + log₝(y). This trick is a lifesaver when simplifying big products or solving growth problems in one neat step.
  3. Quotient Rule - When you divide inside a log, it splits into subtraction: log₝(x/y) = log₝(x) − log₝(y). Perfect for chopping down messy fractions into simpler bits you can handle.
  4. Power Rule - A log of something raised to an exponent just pulls the exponent out front: log₝(x❿) = n·log₝(x). This rule is your go‑to for exponents inside logs, making expansions quick and painless.
  5. Change of Base Formula - Stuck with a base your calculator doesn't support? Use log₝(b) = logₓ(b)/logₓ(a) (often x is 10 or e) to switch bases in a flash. Ideal for when you only have log or ln buttons!
  6. Logarithm of 1 - No matter the base, log₝(1) always equals 0, because any number to the zero power is 1. It's a simple but crucial fact you'll use in proofs and shortcuts.
  7. Logarithm of the Base - If you take log₝(a), you get 1, since a¹ = a. This is another key anchor point when you're matching logs to their exponential counterparts.
  8. Inverse Relationship - Exponentials and logs are perfect inverses: a^(log₝(x)) brings you back to x, and log₝(a^x) returns x. Think of them as two dance partners always stepping on each other's toes!
  9. Expanding Logarithmic Expressions - Combine the product, quotient, and power rules to break a big, scary log into bite‑sized pieces. Expansion helps when you need to isolate unknowns or simplify before solving.
  10. Condensing Logarithmic Expressions - Reverse the expansion: pack multiple logs into one by using the same rules in reverse. This is super helpful for solving equations where you want a single log term.
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