Math 3 Quiz: Solving With Logarithms

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Solving With LogarithmsQuestion 1 of 15

A radioactive substance decays according to $N(t) = N_0 e^{-0.0231t}$ , where $t$ is in years. If 75% of the original amount has decayed, how many years have passed?

Approximately 52 years

Approximately 45 years

Approximately 30 years

Approximately 60 years

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Math 3 Quiz

Math 3 Quiz: Solving With Logarithms

Practice Solving With Logarithms in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving With Logarithms, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A radioactive substance decays according to $N(t) = N_0 e^{-0.0231t}$ , where $t$ is in years. If 75% of the original amount has decayed, how many years have passed?

Approximately 52 years
Approximately 45 years
Approximately 30 years
Approximately 60 years (correct answer)

Explanation: When you encounter exponential decay problems, you're working with situations where a quantity decreases at a rate proportional to its current amount. The key insight here is understanding what "75% has decayed" means for your calculation. If 75% of the substance has decayed, then 25% (or 0.25) of the original amount remains. This means

N(t) = 0.25N_0

. Substituting into the decay equation:

0.25N_0 = N_0 e^{-0.0231t}

Dividing both sides by

N_0

0.25 = e^{-0.0231t}

Taking the natural logarithm of both sides:

\ln(0.25) = -0.0231t

-1.386 = -0.0231t

t = \frac{1.386}{0.0231} \approx 60 \text{ years}

This confirms answer D is correct. Answer A (52 years) would correspond to about 70% decay, not 75%. Answer B (45 years) represents roughly 65% decay. Answer C (30 years) would only result in about 50% decay—a common trap if you confused "75% decayed" with the half-life. The most common error on these problems is misinterpreting the percentage. Remember: if X% has decayed, then (100-X)% remains. Always set up your equation using the remaining percentage, not the decayed percentage. Also, when solving exponential equations, natural logarithms are your tool for "undoing" the exponential function.

Question 2

A carbon-14 dating formula is $t = \frac{-\ln(R)}{0.000121}$ , where $R$ is the ratio of current carbon-14 to original carbon-14, and $t$ is years since death. If an artifact shows 15% of its original carbon-14, approximately how old is it?

12,600 years
15,700 years (correct answer)
18,900 years
21,200 years

Explanation: Given

R = 0.15

, substitute into the formula:

t = \frac{-\ln(0.15)}{0.000121}

. Calculate

\ln(0.15) \approx -1.897

, so

t = \frac{-(-1.897)}{0.000121} = \frac{1.897}{0.000121} \approx 15,700

years. Choice A incorrectly uses

\ln(0.2)

. Choice C incorrectly uses

\ln(0.1)

. Choice D incorrectly uses

\ln(0.05)

Question 3

A bacterial culture grows according to the model $P(t) = 250e^{0.4t}$ , where $t$ is time in hours. If the culture needs to reach 8000 bacteria before a certain treatment can be applied, and the treatment must be administered within a 2-hour window, what is the latest time (in hours) the treatment window can begin?

6.75 hours (correct answer)
7.25 hours
8.75 hours
9.25 hours

Explanation: First, solve for when the population reaches 8000:

8000 = 250e^{0.4t}

. Dividing by 250:

32 = e^{0.4t}

. Taking natural log:

\ln(32) = 0.4t

, so

t = \frac{\ln(32)}{0.4} \approx 8.75

hours. Since the treatment must be administered within a 2-hour window, the latest the window can begin is

8.75 - 2 = 6.75

hours. Choice B incorrectly adds 0.5 hours. Choice C gives the time when 8000 is reached, not when the window begins. Choice D adds 2 hours instead of subtracting.

Question 4

A sound intensity level in decibels is given by $L = 10\log\left(\frac{I}{I_0}\right)$ , where $I$ is the intensity and $I_0$ is a reference intensity. If one sound measures 60 dB and another measures 80 dB, what is the ratio of their intensities $\frac{I_2}{I_1}$ ?

$\frac{4}{3}$
$20$
$100$ (correct answer)
$\frac{80}{60}$

Explanation: For the first sound:

60 = 10\log\left(\frac{I_1}{I_0}\right)

, so

6 = \log\left(\frac{I_1}{I_0}\right)

, giving

\frac{I_1}{I_0} = 10^6

. For the second sound:

80 = 10\log\left(\frac{I_2}{I_0}\right)

, so

8 = \log\left(\frac{I_2}{I_0}\right)

, giving

\frac{I_2}{I_0} = 10^8

. Therefore:

\frac{I_2}{I_1} = \frac{10^8}{10^6} = 10^2 = 100

. Choice A incorrectly assumes a linear relationship. Choice B uses the difference

80-60=20

. Choice D uses the simple ratio of decibel values.

Question 5

A learning curve model shows that typing speed follows $S(t) = 85(1 - e^{-0.15t})$ words per minute, where $t$ is weeks of practice. To reach 95% of the maximum speed, which equation must be solved?

$0.95 = 1 - e^{-0.15t}$
$80.75 = 85(1 - e^{-0.15t})$
$\ln(0.05) = -0.15t$
$0.05 = e^{-0.15t}$ (correct answer)

Explanation: When you encounter exponential growth or decay models, you're often asked to find when the function reaches a certain percentage of its maximum value. The key insight is identifying what "maximum" means and translating the percentage correctly. In this typing speed model

S(t) = 85(1 - e^{-0.15t})

, the maximum speed occurs as

t \to \infty

. Since

e^{-0.15t} \to 0

t

increases, the maximum speed is

85(1 - 0) = 85

words per minute. To reach 95% of this maximum, you need

S(t) = 0.95 \times 85 = 80.75

words per minute. Setting up the equation:

80.75 = 85(1 - e^{-0.15t})

. Dividing both sides by 85 gives

0.95 = 1 - e^{-0.15t}

. Rearranging:

e^{-0.15t} = 1 - 0.95 = 0.05

. This matches answer choice D. Answer A stops at

0.95 = 1 - e^{-0.15t}

but doesn't complete the isolation of the exponential term. Answer B gives the correct setup but hasn't simplified to isolate the exponential expression. Answer C jumps ahead to taking the natural logarithm (

\ln(0.05) = -0.15t

) but the question asks which equation "must be solved," not the final logarithmic form. Remember: when working with exponential models, carefully identify the maximum value, calculate the target percentage, then systematically isolate the exponential term. The question often asks for an intermediate step in your solution process, not necessarily the final form.

Question 6

An investment account follows the compound interest formula $A = P(1.08)^t$ , where $A$ is the final amount, $P$ is the principal, and $t$ is time in years. If an investor wants to triple their initial investment, which logarithmic equation correctly represents the time needed using natural logarithms?

$t = \frac{\ln(3)}{\ln(1.08)}$ (correct answer)
$t = \frac{\ln(1.08)}{\ln(3)}$
$t = \ln(3) \cdot \ln(1.08)$
$t = \ln(3) + \ln(1.08)$

Explanation: To triple the investment:

3P = P(1.08)^t

, which simplifies to

3 = (1.08)^t

. Taking natural log of both sides:

\ln(3) = t\ln(1.08)

, so

t = \frac{\ln(3)}{\ln(1.08)}

. Choice B inverts the fraction. Choice C incorrectly multiplies the logarithms. Choice D incorrectly adds the logarithms.

Question 7

A radioactive substance decays according to $N(t) = N_0 e^{-0.693t/5.7}$ , where $t$ is in years and $5.7$ years is the half-life. If a sample starts with 1000 grams and must be reduced to less than 50 grams before disposal, what is the minimum number of complete years that must pass?

24 years
25 years
26 years (correct answer)
27 years

Explanation: Set up the inequality:

50 > 1000e^{-0.693t/5.7}

. Dividing by 1000:

0.05 > e^{-0.693t/5.7}

. Taking natural log:

\ln(0.05) > -0.693t/5.7

. Since

\ln(0.05) < 0

, when we divide by the negative coefficient, the inequality flips:

t > \frac{-5.7\ln(0.05)}{0.693} = \frac{5.7\ln(20)}{0.693} \approx 24.6

years. Since we need complete years and the amount must be less than 50 grams, we need at least 25 complete years. Checking: after 25 years,

N(25) \approx 52.8

grams (still too much), but after 26 years,

N(26) \approx 48.9

grams (acceptable). Choice A gives the approximate solution without considering it must be complete years. Choice B gives 25 years but this doesn't quite reach the threshold. Choice D is one year too many.

Question 8

A medication concentration in the bloodstream follows $C(t) = 12e^{-0.25t}$ mg/L, where $t$ is hours after administration. For therapeutic effectiveness, the concentration must remain above 3 mg/L. After how many hours will the medication drop below the therapeutic level?

$t = 4\ln(2)$ hours
$t = \frac{\ln(4)}{0.25}$ hours (correct answer)
$t = \frac{\ln(12)}{0.25}$ hours
$t = 4\ln(4)$ hours

Explanation: Set up the equation:

3 = 12e^{-0.25t}

. Dividing by 12:

\frac{1}{4} = e^{-0.25t}

. Taking natural log:

\ln(\frac{1}{4}) = -0.25t

. Since

\ln(\frac{1}{4}) = -\ln(4)

-\ln(4) = -0.25t

, so

t = \frac{\ln(4)}{0.25}

. Choice A incorrectly uses

\ln(2)

instead of

\ln(4)

. Choice C uses

\ln(12)

from the initial concentration rather than the ratio. Choice D has the correct

\ln(4)

but wrong coefficient.

Question 9

A population model is given by $P(t) = \frac{5000}{1 + 9e^{-0.5t}}$ , where $t$ is time in months. To find when the population reaches 2500, which equation must be solved?

$\ln(1) = -0.5t$
$\ln(\frac{1}{9}) = -0.5t$
$\ln(9) = 0.5t$ (correct answer)
$\ln(\frac{9}{1}) = 0.5t$

Explanation: Set

2500 = \frac{5000}{1 + 9e^{-0.5t}}

. Cross-multiply:

2500(1 + 9e^{-0.5t}) = 5000

. Dividing by 2500:

1 + 9e^{-0.5t} = 2

. Subtracting 1:

9e^{-0.5t} = 1

. Dividing by 9:

e^{-0.5t} = \frac{1}{9}

. Taking natural log:

-0.5t = \ln(\frac{1}{9}) = -\ln(9)

. Multiplying by -1:

0.5t = \ln(9)

, or equivalently

\ln(9) = 0.5t

. Choice A results from incorrectly getting

e^{-0.5t} = 1

. Choice B has the correct logarithm but wrong sign for

t

. Choice D uses unnecessary notation

\frac{9}{1}

Question 10

An investment grows according to $A = 1000(1.06)^t$ , where $A$ is the amount in dollars and $t$ is time in years. At what time will the investment first exceed $2500?

Between 15 and 16 years (correct answer)
Between 14 and 15 years
Between 16 and 17 years
Between 13 and 14 years

Explanation: Set up the inequality:

2500 < 1000(1.06)^t

. Divide by 1000:

2.5 < (1.06)^t

. Take logarithm of both sides:

\ln(2.5) < t\ln(1.06)

. Therefore:

t > \frac{\ln(2.5)}{\ln(1.06)} = \frac{0.916}{0.0583} ≈ 15.73

years. Since we want when it first exceeds $2500, this occurs between 15 and 16 years. Choice B uses $

\log_{10}

$ instead of natural log. Choice C misinterprets 'first exceeds' as 'reaches exactly'. Choice D makes a calculation error in the division.

Question 11

A certain medication's concentration in the bloodstream decreases according to $C(t) = 20e^{-0.15t}$ , where $C(t)$ is concentration in mg/L and $t$ is time in hours. A second dose should be given when the concentration drops to 5 mg/L. After how many hours should the second dose be administered?

Approximately 8.2 hours
Approximately 9.2 hours (correct answer)
Approximately 10.4 hours
Approximately 7.7 hours

Explanation: Set up the equation:

5 = 20e^{-0.15t}

. Divide both sides by 20:

0.25 = e^{-0.15t}

. Take the natural logarithm:

\ln(0.25) = -0.15t

. Therefore:

t = \frac{\ln(0.25)}{-0.15} = \frac{-1.386}{-0.15} ≈ 9.24

hours. Choice A uses an incorrect decay constant. Choice C makes an error in the logarithm calculation. Choice D uses the wrong initial concentration in the setup.

Question 12

The equation $3^{2x-1} = 27^{x-2}$ has a solution that can be found using logarithms. When this equation is solved correctly, what is the value of $x^2 + 1$ ?

26 (correct answer)
17
10
37

Explanation: Rewrite

27 = 3^3

, so

3^{2x-1} = (3^3)^{x-2} = 3^{3(x-2)} = 3^{3x-6}

. Since the bases are equal:

2x - 1 = 3x - 6

. Solving:

-1 + 6 = 3x - 2x

, so

x = 5

. Therefore

x^2 + 1 = 25 + 1 = 26

. Choice B results from the error

x = 4

. Choice C results from the error

x = 3

. Choice D results from the error

x = 6

Question 13

The equation $\log(x^2 - 6x + 8) = \log(2x - 6)$ has potential solutions that must be verified. After solving and checking domain restrictions, how many valid solutions exist?

No valid solutions exist
Exactly one valid solution exists (correct answer)
Exactly two valid solutions exist
Three or more valid solutions exist

Explanation: Since the logarithms are equal:

x^2 - 6x + 8 = 2x - 6

. Rearranging:

x^2 - 8x + 14 = 0

. Using the quadratic formula:

x = \frac{8 \pm \sqrt{64 - 56}}{2} = \frac{8 \pm \sqrt{8}}{2} = \frac{8 \pm 2\sqrt{2}}{2} = 4 \pm \sqrt{2}

. For domain restrictions, we need

x^2 - 6x + 8 > 0

and

2x - 6 > 0

(so

x > 3

). Note that

x^2 - 6x + 8 = (x-2)(x-4)

, which is positive when

x < 2

x > 4

. Since we also need

x > 3

, we need

x > 4

. Check solutions:

4 + \sqrt{2} \approx 5.41 > 4

(valid), but

4 - \sqrt{2} \approx 2.59 < 4

(invalid). Therefore exactly one valid solution exists. Choice A incorrectly concludes no solutions. Choice C doesn't account for domain restrictions. Choice D suggests too many solutions.

Question 14

An exponential equation $4^{x+1} = 16^{2x-3}$ needs to be solved. After expressing both sides with base 4, what equation results?

$4^{x+1} = 4^{2(2x-3)}$
$4^{x+1} = 4^{4x-6}$ (correct answer)
$4^{x+1} = 4^{8x-12}$
$4^{x+1} = 4^{4x-12}$

Explanation: Since

16 = 4^2

, we can rewrite

16^{2x-3}

(4^2)^{2x-3} = 4^{2(2x-3)} = 4^{4x-6}

. Therefore, the equation becomes

4^{x+1} = 4^{4x-6}

. Setting the exponents equal:

x+1 = 4x-6

, which gives

7 = 3x

, so

x = \frac{7}{3}

. Choice A doesn't simplify the exponent completely. Choice C incorrectly multiplies the exponent by 4 instead of 2. Choice D makes an error in the constant term of the exponent.

Question 15

To solve $5^{2x-1} = 3^{x+2}$ for $x$ , logarithms must be used. Which equation correctly represents the first step after taking the natural logarithm of both sides?

$2x\ln(5) - 1 = x\ln(3) + 2$
$\ln(2x-1) \cdot \ln(5) = \ln(x+2) \cdot \ln(3)$
$(2x-1)\ln(5) = (x+2)\ln(3)$ (correct answer)
$(2x-1) + \ln(5) = (x+2) + \ln(3)$

Explanation: When you encounter an exponential equation where the bases are different (like 5 and 3), taking logarithms of both sides is the key to solving it. This transforms the equation from exponential form into linear form by bringing the exponents down as coefficients. Starting with

5^{2x-1} = 3^{x+2}

, when you take the natural logarithm of both sides, you get

\ln(5^{2x-1}) = \ln(3^{x+2})

. The crucial step is applying the logarithm power rule:

\ln(a^b) = b\ln(a)

. This rule allows you to bring the exponents down as multipliers in front of the logarithms. Applying this rule:

\ln(5^{2x-1}) = (2x-1)\ln(5)

and

\ln(3^{x+2}) = (x+2)\ln(3)

. Therefore, the equation becomes

(2x-1)\ln(5) = (x+2)\ln(3)

, which is answer C. Let's examine why the other options are incorrect. Option A,

2x\ln(5) - 1 = x\ln(3) + 2

, incorrectly distributes the logarithm over subtraction and addition within the exponents. Option B,

\ln(2x-1) \cdot \ln(5) = \ln(x+2) \cdot \ln(3)

, confuses the logarithm of a product rule with the power rule—it treats the exponents as if they're being logged separately. Option D,

(2x-1) + \ln(5) = (x+2) + \ln(3)

, uses addition instead of multiplication, which isn't how the power rule works. Remember: when applying

\ln(a^b) = b\ln(a)

, the exponent becomes a coefficient that multiplies the entire logarithm—don't break apart the exponent's components.