Algebra 2 Quiz: Solving Exponential Equations With Logarithms

20 questions · exam conditions

0:00

Solving Exponential Equations With LogarithmsQuestion 1 of 20

Use a calculator to solve the exponential equation $6\cdot 2^{0.25t}=90$ for $t$ . Give the exact solution as a logarithm and an approximate value.

t=\dfrac{\log_{2}(15)}{0.25}\approx 15.628

t=0.25\log_{2}(15)\approx 0.977

t=\dfrac{\log_{2}(90)}{0.25}\approx 26.196

t=\dfrac{\log(15)}{0.25}\approx 4.705

← Back to quizzes

Algebra 2 Quiz

Algebra 2 Quiz: Solving Exponential Equations With Logarithms

Practice Solving Exponential Equations With Logarithms in Algebra 2 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 Exponential Equations With Logarithms, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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.

All questions

Question 1

Use a calculator to solve the exponential equation $6\cdot 2^{0.25t}=90$ for $t$ . Give the exact solution as a logarithm and an approximate value.

$t=\dfrac{\log_{2}(15)}{0.25}\approx 15.628$ (correct answer)
$t=0.25\log_{2}(15)\approx 0.977$
$t=\dfrac{\log_{2}(90)}{0.25}\approx 26.196$
$t=\dfrac{\log(15)}{0.25}\approx 4.705$

Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. To solve 6·2^(0.25t) = 90, first isolate the exponential by dividing by 6: 2^(0.25t) = 90/6 = 15. Taking log base 2 of both sides: log₂(2^(0.25t)) = log₂(15), using the inverse property: 0.25t = log₂(15), and dividing by 0.25 (which is the same as multiplying by 4): t = log₂(15)/0.25 = 4·log₂(15). Using change of base: log₂(15) = ln(15)/ln(2) ≈ 2.708/0.693 ≈ 3.907, so t ≈ 4·3.907 ≈ 15.628. Choice A correctly isolates the exponential and divides by the coefficient 0.25 to get t = log₂(15)/0.25 ≈ 15.628. Choice B incorrectly multiplies by 0.25 instead of dividing, while Choice D uses log base 10 instead of log base 2. Remember that dividing by 0.25 is the same as multiplying by 4—this is why the answer is relatively large. Always double-check: if t ≈ 15.628, then 0.25t ≈ 3.907, and 2^3.907 ≈ 15, so 6·15 = 90 ✓

Question 2

Solve for $t$ and express the solution as a logarithm, then approximate using a calculator: $5\cdot 2^{3t}=70$ .

$t=\log_{2}(14)\approx 3.807$
$t=\dfrac{\log_{2}(14)}{3}\approx 1.269$ (correct answer)
$t=\dfrac{\log_{2}(70)}{3}\approx 2.043$
$t=\dfrac{\log(14)}{3}\approx 0.383$

Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for 5·2^(3t) = 70 follows these steps: (1) Isolate the exponential by dividing both sides by 5: 2^(3t) = 14, (2) Take log base 2 of both sides: log₂(2^(3t)) = log₂(14), (3) Use inverse property on left: 3t = log₂(14), (4) Solve for variable: t = log₂(14)/3. This systematic approach gives us the exact answer! To get a decimal approximation, use change of base: log₂(14) = ln(14)/ln(2) ≈ 3.807, so t ≈ 3.807/3 ≈ 1.269. Choice B correctly isolates the exponential, takes log base 2, and divides by the coefficient 3 to get t = log₂(14)/3 ≈ 1.269. Choice A forgets to divide by 3, giving just log₂(14), while Choice C incorrectly uses 70 instead of 14 after isolation. Calculator strategy for non-standard bases: since calculators don't have a log₂ button, use change of base formula log₂(14) = ln(14)/ln(2) or log(14)/log(2). Always isolate the exponential expression before taking logarithms—dividing by 5 first makes the problem much cleaner!

Question 3

An exponential equation has the form $6 \cdot 2^{x-3} = 192$ . Two students solve this differently. Student A simplifies to $2^{x-3} = 32$ and uses $\log_2$ . Student B takes $\ln$ of the original equation immediately. Both methods are valid, but which statement correctly describes the relationship between their approaches?

Student A gets $x-3 = 5$ directly; Student B gets $(x-3)\ln(2) = \ln(32)$ and must divide by $\ln(2)$
Student A gets $x-3 = \log_2(32)$ which equals 5; Student B gets $\ln(6) + (x-3)\ln(2) = \ln(192)$ (correct answer)
Both students get the same equation $x-3 = 5$ but through different logarithmic calculations
Student A's method only works when the right side is a power of 2; Student B's method works for any positive value

Explanation: Student A:

2^{x-3} = 32

, then

\log_2(2^{x-3}) = \log_2(32)

, giving

x-3 = \log_2(32) = \log_2(2^5) = 5

. Student B:

\ln(6 \cdot 2^{x-3}) = \ln(192)

, which becomes

\ln(6) + \ln(2^{x-3}) = \ln(192)

, so

\ln(6) + (x-3)\ln(2) = \ln(192)

. Student B must then solve for

(x-3)

by rearranging:

(x-3)\ln(2) = \ln(192) - \ln(6) = \ln(\frac{192}{6}) = \ln(32)

. Choice A incorrectly describes Student B's equation. Choice C is wrong because Student B gets a more complex equation initially. Choice D incorrectly suggests Student A's method is limited.

Question 4

An investment account grows according to $A = 1200e^{0.045t}$ where $A$ is the account value in dollars and $t$ is time in years. To find when the account reaches $2400, a student sets up the equation $ $1200e^{0.045t} = 2400$ $and simplifies to$ $e^{0.045t} = 2$ $. What is the next correct step to solve for$ $t$ $?

Take the natural logarithm of both sides to get $0.045t = \ln(2)$ (correct answer)
Take the common logarithm of both sides to get $0.045t = \log(2)$
Apply the inverse exponential to get $0.045t = e^{-1}(2)$
Convert to base 10 to get $10^{0.045t \ln(e)} = 2$

Explanation: When solving

e^{0.045t} = 2

, the natural logarithm is the inverse function of the exponential with base

e

. Taking

\ln

of both sides gives

\ln(e^{0.045t}) = \ln(2)

, which simplifies to

0.045t = \ln(2)

using the property that

\ln(e^x) = x

. Choice B incorrectly uses common logarithm instead of natural logarithm. Choice C uses non-standard notation that doesn't represent a valid mathematical operation. Choice D unnecessarily complicates the problem with an incorrect base conversion.

Question 5

A compound interest problem leads to the equation $1000 \cdot e^{0.08t} = 2500$ . A student correctly simplifies this to $e^{0.08t} = 2.5$ and takes the natural logarithm to get $0.08t = \ln(2.5)$ . Using technology, $\ln(2.5) \approx 0.916$ . What is the approximate value of $t$ ?

$t \approx 0.073$
$t \approx 11.45$ (correct answer)
$t \approx 12.50$
$t \approx 87.3$

Explanation: From

0.08t = \ln(2.5) \approx 0.916

, we solve for

t

by dividing both sides by 0.08:

t = \frac{0.916}{0.08} = \frac{0.916}{0.08} = 11.45

. Choice A incorrectly multiplies 0.916 by 0.08 instead of dividing. Choice C appears to come from

\frac{1}{0.08} = 12.5

, ignoring the logarithm value. Choice D might result from calculating

\frac{0.916}{0.125}

(using 0.125 instead of 0.08). The calculation is straightforward division, but the decimal placement and order of operations create multiple opportunities for error.

Question 6

A radioactive substance decays according to the model $N(t) = N_0 e^{-0.693t}$ , where $t$ is in years. If $N_0 = 800$ grams and the substance decays to 50 grams, which equation correctly represents the solution for $t$ ?

$t = \frac{\ln(50/800)}{-0.693} = \frac{\ln(1/16)}{-0.693}$
$t = \frac{\ln(800/50)}{0.693} = \frac{\ln(16)}{0.693}$ (correct answer)
$t = \frac{\ln(50)}{-0.693 \ln(800)}$
$t = \frac{-0.693}{\ln(50/800)}$

Explanation: Setting up the equation:

800e^{-0.693t} = 50

. Dividing by 800:

e^{-0.693t} = \frac{50}{800} = \frac{1}{16}

. Taking natural logarithm:

-0.693t = \ln(\frac{1}{16})

. Since

\ln(\frac{1}{16}) = -\ln(16)

, we have

-0.693t = -\ln(16)

. Dividing by

-0.693

t = \frac{\ln(16)}{0.693}

. This can also be written as

t = \frac{\ln(800/50)}{0.693}

. Choice A has the correct setup but doesn't account for the negative sign properly. Choice C incorrectly applies logarithm properties. Choice D inverts the fraction incorrectly.

Question 7

The concentration of a medication in the bloodstream follows $C(t) = 50e^{-0.2t}$ mg/L, where $t$ is hours after administration. A doctor needs to know when the concentration drops to 5 mg/L. After setting up $50e^{-0.2t} = 5$ , which step correctly leads to the logarithmic solution?

$e^{-0.2t} = 0.1$ , so $-0.2t = \ln(0.1)$ and $t = \frac{\ln(0.1)}{-0.2}$
$e^{-0.2t} = 0.1$ , so $-0.2t = \ln(0.1)$ and $t = \frac{-\ln(0.1)}{0.2}$
$e^{-0.2t} = 10$ , so $-0.2t = \ln(10)$ and $t = \frac{\ln(10)}{-0.2}$
$e^{-0.2t} = \frac{1}{10}$ , so $-0.2t = -\ln(10)$ and $t = \frac{\ln(10)}{0.2}$ (correct answer)

Explanation: From

50e^{-0.2t} = 5

, dividing by 50 gives

e^{-0.2t} = \frac{5}{50} = \frac{1}{10} = 0.1

. Taking natural logarithm:

-0.2t = \ln(0.1)

. Since

\ln(0.1) = \ln(\frac{1}{10}) = \ln(1) - \ln(10) = 0 - \ln(10) = -\ln(10)

, we have

-0.2t = -\ln(10)

. Therefore:

t = \frac{-\ln(10)}{-0.2} = \frac{\ln(10)}{0.2}

. Choice A correctly finds

e^{-0.2t} = 0.1

but makes an error in the final division. Choice B has the correct intermediate steps but incorrectly manipulates the negative signs. Choice C incorrectly calculates

\frac{5}{50}

as 10 instead of 0.1.

Question 8

A scientist models the growth of a culture using $P = 200 \cdot 10^{0.5t}$ , where $P$ is the population and $t$ is time in days. She needs the population to reach exactly 20,000. Using technology to evaluate the logarithm, the solution is approximately $t = 4.0$ . Which logarithmic expression did she most likely evaluate?

$\log(100)$
$2\log(10)$
$\frac{\log(100)}{0.5}$ (correct answer)
$\frac{2\log(10)}{0.5}$

Explanation: Setting up:

200 \cdot 10^{0.5t} = 20000

. Dividing by 200:

10^{0.5t} = 100

. Taking common logarithm:

\log(10^{0.5t}) = \log(100)

, which gives

0.5t = \log(100)

. Therefore:

t = \frac{\log(100)}{0.5}

. Since

\log(100) = \log(10^2) = 2

and

\frac{2}{0.5} = 4

, this confirms the answer

t = 4.0

. Choice A gives

\log(100) = 2

, not 4. Choice B gives

2\log(10) = 2 \cdot 1 = 2

, not 4. Choice D gives

\frac{2 \cdot 1}{0.5} = 4

but represents an incorrect setup of the logarithmic equation.

Question 9

The temperature of a cooling object follows Newton's law of cooling: $T(t) = 75 + 125e^{-0.15t}$ , where $T$ is temperature in °F and $t$ is time in minutes. To find when the temperature reaches 100°F, the equation $75 + 125e^{-0.15t} = 100$ must be solved. Which expression gives the exact solution for $t$ ?

$t = \frac{\ln(0.2)}{-0.15}$
$t = \frac{\ln(5)}{0.15}$
$t = \frac{-\ln(0.2)}{0.15}$ (correct answer)
$t = \frac{\ln(25)}{-0.15}$

Explanation: Starting with

75 + 125e^{-0.15t} = 100

. Subtract 75:

125e^{-0.15t} = 25

. Divide by 125:

e^{-0.15t} = \frac{25}{125} = \frac{1}{5} = 0.2

. Take natural logarithm:

-0.15t = \ln(0.2)

. Since

\ln(0.2) < 0

, we solve:

t = \frac{\ln(0.2)}{-0.15} = \frac{-\ln(0.2)}{0.15}

. Choice A has the correct setup but doesn't simplify the negative signs properly. Choice B uses

\ln(5)

instead of

\ln(0.2)

, which would be correct if we wrote

0.2 = \frac{1}{5}

and used

\ln(\frac{1}{5}) = -\ln(5)

, but the final form doesn't match. Choice D incorrectly uses 25 instead of 0.2.

Question 10

The equation $5 \cdot 2^{3x+2} = 320$ is equivalent to $2^{3x+2} = 64$ . A student must solve this using logarithms rather than recognizing powers of 2. Which logarithmic approach will yield the correct solution most efficiently?

Use $\log_2(2^{3x+2}) = \log_2(64)$ to get $3x+2 = 6$ , then solve $x = \frac{4}{3}$ (correct answer)
Use $\ln(2^{3x+2}) = \ln(64)$ to get $(3x+2)\ln(2) = \ln(64)$ , then solve $x = \frac{\ln(64) - 2\ln(2)}{3\ln(2)}$
Use $\log(2^{3x+2}) = \log(64)$ to get $(3x+2)\log(2) = \log(64)$ , then solve $x = \frac{\log(64) - 2\log(2)}{3\log(2)}$
Use $\ln(2^{3x+2}) = \ln(64)$ but convert to $\ln(2^{3x+2}) = \ln(2^6)$ first, giving $(3x+2)\ln(2) = 6\ln(2)$

Explanation: Since we must use logarithms to solve

2^{3x+2} = 64

, the most efficient logarithmic approach uses base-2 logarithms:

\log_2(2^{3x+2}) = \log_2(64)

. This gives

3x+2 = \log_2(64) = \log_2(2^6) = 6

. Solving:

3x = 4

, so

x = \frac{4}{3}

. Choices B and C are mathematically correct but require more complex calculations with change of base formulas. Choice D adds an unnecessary step by converting 64 to

2^6

before applying logarithms, making it less efficient than choice A.

Question 11

Solve for $x$ and evaluate using technology: $8\cdot 2^{x/2}=70$ . Express the exact solution as a logarithm.

$x=\dfrac{\log_2\left(\dfrac{35}{4}\right)}{2}\approx 1.565$
$x=2\log_2\left(\dfrac{70}{8}\right)\approx 3.129$
$x=2\log_2\left(\dfrac{35}{4}\right)\approx 6.258$ (correct answer)
$x=2\log\left(\dfrac{35}{4}\right)\approx 1.792$

2^t

) = log₂(10), and using the inverse property log₂(

2^t

) = t, we get t = log₂(10). This is the exact solution! To get a decimal approximation, use your calculator with change of base: log₂(10) = ln(10)/ln(2) ≈ 3.322. To solve 8·2^{x/2}=70, divide by 8: 2^{x/2}=70/8=35/4, take log base 2: log₂(

2^{x/2}

)=log₂(35/4), simplify to x/2=log₂(35/4), and multiply by 2: x=2 log₂(35/4). Choice A correctly isolates the exponential, takes log base 2, applies the inverse property, and approximates to 6.258 using change of base. Choice B divides by 2 instead of multiplying, giving x=log₂(35/4)/2 which is half the correct value—remember to multiply by 2 to undo the division in the exponent! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(10)? Use change of base: log₂(10) = ln(10)/ln(2) or log(10)/log(2)—both give the same answer ≈ 3.322. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button! Alternatively, leave answers in exact log form if calculator evaluation isn't required. Isolation before logarithms: ALWAYS isolate the exponential expression b^(ct) before taking logarithms. If you have 5·2^t = 40, first divide by 5 to get 2^t = 8, THEN take log. Taking log₂ of both sides of 5·2^t = 40 directly leads to log₂(

5·2^t

), which is more complex (requires log properties). Simple isolation first makes the logarithm application clean: take log of both sides when you have b^(something) = number, with the exponential alone on one side!

Question 12

Solve the exponential equation $3\cdot 10^{2t}=75$ for $t$ . Give the exact solution (as a logarithm) and then approximate $t$ using technology (calculator).

$t=\dfrac{\log(25)}{2}\approx 0.699$ (correct answer)
$t=\log(25)\approx 1.398$
$t=\dfrac{\log(75)}{2}\approx 0.938$
$t=\dfrac{\log(25)}{3}\approx 0.466$

Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 3·10^(2t) = 75 where the exponent contains the variable, logarithms are the tool that unlocks the solution: first isolate the exponential by dividing both sides by 3 to get 10^(2t) = 25, then take log base 10 of both sides to get log(10^(2t)) = log(25), and using the inverse property log(10^(2t)) = 2t, we get 2t = log(25), so t = log(25)/2. To get a decimal approximation, use your calculator: log(25) ≈ 1.398, so t ≈ 1.398/2 ≈ 0.699. Choice A correctly isolates the exponential, takes the logarithm, and divides by 2 to get t = log(25)/2 ≈ 0.699. Choice B incorrectly forgets to divide by 2, giving t = log(25) instead of t = log(25)/2—remember that when the exponent is 2t, you must divide by 2 after taking the logarithm! Calculator strategy for base 10 logarithms: since we're working with 10^(2t), we use the common logarithm (log button on calculator), which is base 10. The exact answer t = log(25)/2 can be evaluated directly as log(25) ÷ 2 ≈ 1.398 ÷ 2 ≈ 0.699.

Question 13

Solve $4e^{0.5t}=50$ for $t$ . Express the solution using a logarithm (exact form) and then approximate using technology.

$t=\dfrac{\ln(50)}{0.5}\approx 7.824$
$t=2\ln\left(\dfrac{25}{2}\right)\approx 5.051$ (correct answer)
$t=\dfrac{\ln\left(\dfrac{25}{2}\right)}{2}\approx 1.263$
$t=\ln\left(\dfrac{25}{2}\right)\approx 2.526$

Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 4e^(0.5t) = 50 where the exponent contains the variable, logarithms are the tool that unlocks the solution: first isolate the exponential by dividing both sides by 4 to get e^(0.5t) = 12.5, then take ln of both sides to get ln(e^(0.5t)) = ln(12.5), and using the inverse property ln(e^(0.5t)) = 0.5t, we get 0.5t = ln(12.5) = ln(25/2), so t = ln(25/2)/0.5 = 2ln(25/2). Starting with 4e^(0.5t) = 50, divide by 4 to get e^(0.5t) = 50/4 = 25/2, then take ln of both sides: 0.5t = ln(25/2), and divide by 0.5 (which is the same as multiplying by 2) to get t = 2ln(25/2). Using a calculator: ln(25/2) ≈ 2.526, so t ≈ 2(2.526) ≈ 5.051. Choice B correctly isolates the exponential, takes the natural logarithm, and multiplies by 2 (since dividing by 0.5 equals multiplying by 2) to get t = 2ln(25/2) ≈ 5.051. Choice C incorrectly divides by 2 instead of multiplying by 2—remember that solving 0.5t = ln(25/2) requires dividing by 0.5, which means multiplying by 2! Isolation before logarithms: ALWAYS isolate the exponential expression e^(0.5t) before taking logarithms. If you have 4e^(0.5t) = 50, first divide by 4 to get e^(0.5t) = 12.5, THEN take ln. This makes the logarithm application clean and straightforward!

Question 14

Solve and evaluate: $12\cdot 10^{t/3}=90$ . Give $t$ in exact logarithmic form and as a decimal (use a calculator).

$t=3\log(7.5)\approx 2.625$ (correct answer)
$t=\dfrac{\log(90)}{3}\approx 0.651$
$t=3\log(90/12)\approx 2.875$
$t=\log(7.5)\approx 0.875$

Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for 12·10^(t/3) = 90: (1) Isolate the exponential by dividing both sides by 12: 10^(t/3) = 7.5, (2) Take log (base 10) of both sides: log(10^(t/3)) = log(7.5), (3) Use inverse property on left: t/3 = log(7.5), (4) Solve for variable: t = 3·log(7.5). This systematic approach works for any exponential equation in this form! Starting with 12·10^(t/3) = 90, divide by 12 to get 10^(t/3) = 7.5, then take log of both sides: log(10^(t/3)) = log(7.5), which simplifies to t/3 = log(7.5), so t = 3·log(7.5). To evaluate with a calculator: log(7.5) ≈ 0.875, so t ≈ 3(0.875) ≈ 2.625. Choice B correctly isolates the exponential, takes common logarithm, and multiplies by 3 to get t = 3·log(7.5) ≈ 2.625. Choice A incorrectly leaves t = log(7.5) without multiplying by 3, while Choice C uses log(90) without first isolating the exponential. When the variable appears in a fraction like t/3 in the exponent, remember to multiply both sides by the denominator (3) after applying the logarithm. The pattern is: if you have 10^(t/n) = k, then t/n = log(k), so t = n·log(k). This multiplication step is crucial for getting the correct answer!

Question 15

Find $x$ in $15e^{2x}=400$ . Give the exact solution as a logarithm and an approximate value using technology.

$x=\dfrac{\ln\left(\frac{400}{15}\right)}{2}\approx 1.642$ (correct answer)
$x=\ln\left(\frac{400}{15}\right)\approx 3.284$
$x=\dfrac{\ln\left(\frac{15}{400}\right)}{2}\approx -1.642$
$x=\dfrac{\log\left(\frac{400}{15}\right)}{2}\approx 0.713$

Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 15e^(2x) = 400 where e is the base and the coefficient of x is 2, natural logarithms are the perfect tool: first isolate the exponential by dividing both sides by 15 to get e^(2x) = 400/15 ≈ 26.667, then take ln of both sides to get ln(e^(2x)) = ln(400/15), and using the inverse property ln(e^(2x)) = 2x, we get 2x = ln(400/15), so x = ln(400/15)/2. To solve 15e^(2x) = 400 step by step: divide by 15 to get e^(2x) = 26.667, take natural log of both sides: ln(e^(2x)) = ln(26.667), apply the inverse property to get 2x = ln(26.667) ≈ 3.283, and divide by 2: x = ln(26.667)/2 = ln(400/15)/2 ≈ 3.283/2 ≈ 1.642. Choice A correctly isolates the exponential, takes the natural logarithm, and divides by 2 to get x = ln(400/15)/2 ≈ 1.642. Choice B forgets to divide by 2, giving x = ln(400/15) ≈ 3.284, which would mean e^(2x) = e^6.568, far larger than 26.667. Isolation before logarithms: ALWAYS isolate the exponential expression e^(2x) before taking logarithms. If you have 15e^(2x) = 400, first divide by 15 to get e^(2x) = 26.667, THEN take ln. When working with base e, always use natural log (ln) for the cleanest solution—the inverse property ln(

e^x

) = x makes the algebra straightforward!

Question 16

Use logarithms to solve $5\cdot 2^{3t}=60$ for $t$ . Express your answer as a logarithm (exact form) and then approximate using a calculator.

$t=\log_2(12)\approx 3.585$
$t=\dfrac{\log_2(60)}{3}\approx 1.969$
$t=\dfrac{\log(12)}{3}\approx 0.360$
$t=\dfrac{\log_2(12)}{3}\approx 1.195$ (correct answer)

Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for 5·2^(3t) = 60: (1) Isolate the exponential by dividing both sides by 5: 2^(3t) = 12, (2) Take log base 2 of both sides: log₂(2^(3t)) = log₂(12), (3) Use inverse property on left: 3t = log₂(12), (4) Solve for variable: t = log₂(12)/3. This systematic approach works for any exponential equation in this form! Starting with 5·2^(3t) = 60, divide by 5 to get 2^(3t) = 12, then take log₂ of both sides to get 3t = log₂(12), and finally divide by 3 to get t = log₂(12)/3. To approximate with a calculator, use change of base: log₂(12) = ln(12)/ln(2) ≈ 3.585, so t ≈ 3.585/3 ≈ 1.195. Choice A correctly isolates the exponential, takes log base 2, and divides by 3 to get t = log₂(12)/3 ≈ 1.195. Choice B incorrectly uses 60 instead of 12 after isolation—remember to divide 60 by 5 first to get 2^(3t) = 12, not 2^(3t) = 60! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(12)? Use change of base: log₂(12) = ln(12)/ln(2) or log(12)/log(2)—both give the same answer ≈ 3.585. The formula is log_b(x) = ln(x)/ln(b) for any base b.

Question 17

A population of bacteria grows according to the model $P(t) = 500 \cdot 2^{0.3t}$ , where $t$ is time in hours. If the population reaches 8000 bacteria, which expression gives the exact time when this occurs?

$\frac{\log_2(16)}{0.3}$
$\frac{\log(16)}{0.3 \log(2)}$
$\frac{\ln(16)}{0.3 \ln(2)}$ (correct answer)
$\frac{\log_2(8000)}{0.3}$

Explanation: Setting up the equation:

500 \cdot 2^{0.3t} = 8000

. Dividing both sides by 500:

2^{0.3t} = 16

. Taking the natural logarithm of both sides:

\ln(2^{0.3t}) = \ln(16)

. Using the logarithm property:

0.3t \cdot \ln(2) = \ln(16)

. Solving for

t

t = \frac{\ln(16)}{0.3 \ln(2)}

. Choice A uses the wrong logarithm base conversion. Choice B uses common log instead of natural log but has the correct structure. Choice D fails to simplify by dividing out the initial population.

Question 18

Use a calculator to solve the exponential equation $4\cdot 2^{t/5}=30$ . Give the exact solution as a logarithm and an approximate decimal value.

$t=5\log_2\left(\dfrac{15}{2}\right)\approx 14.535$ (correct answer)
$t=\dfrac{\log_2\left(\dfrac{15}{2}\right)}{5}\approx 0.581$
$t=5\log_2\left(\dfrac{30}{4}\right)\approx 5.814$
$t=5\log\left(\dfrac{15}{2}\right)\approx 4.375$

2^t

) = log₂(10), and using the inverse property log₂(

2^t

) = t, we get t = log₂(10). This is the exact solution! To get a decimal approximation, use your calculator with change of base: log₂(10) = ln(10)/ln(2) ≈ 3.322. To solve 4·2^{t/5}=30, divide by 4: 2^{t/5}=30/4=15/2, take log base 2: log₂(

2^{t/5}

)=log₂(15/2), simplify to t/5=log₂(15/2), and multiply by 5: t=5 log₂(15/2). Choice A correctly isolates, takes log base 2, applies the inverse property, and approximates to 14.535 using change of base on a calculator. Choice B divides by 5 instead of multiplying, resulting in t=log₂(15/2)/5 which is one-fifth the size—remember that to undo division in the exponent, you multiply when solving for t! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(10)? Use change of base: log₂(10) = ln(10)/ln(2) or log(10)/log(2)—both give the same answer ≈ 3.322. The formula is log_b(x) = ln(x)/ln(b) for any base b. This lets you evaluate any logarithm using just the ln button! Alternatively, leave answers in exact log form if calculator evaluation isn't required. Isolation before logarithms: ALWAYS isolate the exponential expression b^(ct) before taking logarithms. If you have 5·2^t = 40, first divide by 5 to get 2^t = 8, THEN take log. Taking log₂ of both sides of 5·2^t = 40 directly leads to log₂(

5·2^t

Question 19

Solve the equation $8\cdot e^{t/3}=90$ for $t$ . Give the exact answer as a logarithm and then approximate using technology.

$t=3\ln\left(\dfrac{45}{4}\right)\approx 7.266$ (correct answer)
$t=\dfrac{\ln\left(\dfrac{45}{4}\right)}{3}\approx 0.807$
$t=3\ln\left(\dfrac{90}{8}\right)\approx 2.422$
$t=\ln\left(\dfrac{45}{4}\right)\approx 2.422$

Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. The solving strategy for 8·e^(t/3) = 90: (1) Isolate the exponential by dividing both sides by 8: e^(t/3) = 90/8 = 45/4, (2) Take ln of both sides: ln(e^(t/3)) = ln(45/4), (3) Use inverse property on left: t/3 = ln(45/4), (4) Solve for variable: t = 3ln(45/4). This systematic approach works for any exponential equation in this form! Starting with 8·e^(t/3) = 90, divide by 8 to get e^(t/3) = 45/4, then take ln of both sides to get t/3 = ln(45/4), and multiply by 3 to get t = 3ln(45/4). Using a calculator: ln(45/4) = ln(11.25) ≈ 2.422, so t ≈ 3(2.422) ≈ 7.266. Choice A correctly isolates the exponential, takes the natural logarithm, and multiplies by 3 to get t = 3ln(45/4) ≈ 7.266. Choice B incorrectly divides by 3 instead of multiplying—when the exponent is t/3 and you get t/3 = ln(45/4), you must multiply both sides by 3 to solve for t! Isolation before logarithms: ALWAYS isolate the exponential expression e^(t/3) before taking logarithms. If you have 8e^(t/3) = 90, first divide by 8 to get e^(t/3) = 45/4, THEN take ln. Simple isolation first makes the logarithm application clean!

Question 20

Use logarithms to solve $15\cdot 2^{2t}=500$ for $t$ . Express the solution as a logarithm (exact form) and then approximate using a calculator.

$t=\dfrac{\log_2\left(\dfrac{100}{3}\right)}{2}\approx 2.529$ (correct answer)
$t=\dfrac{\log_2\left(\dfrac{500}{15}\right)}{2}\approx 1.265$
$t=\dfrac{\log\left(\dfrac{100}{3}\right)}{2}\approx 0.761$
$t=2\log_2\left(\dfrac{100}{3}\right)\approx 10.116$

Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 15·2^(2t) = 500 where the exponent contains the variable, logarithms are the tool that unlocks the solution: first isolate the exponential by dividing both sides by 15 to get 2^(2t) = 500/15 = 100/3, then take log base 2 of both sides to get log₂(2^(2t)) = log₂(100/3), and using the inverse property log₂(2^(2t)) = 2t, we get 2t = log₂(100/3), so t = log₂(100/3)/2. Starting with 15·2^(2t) = 500, divide by 15 to get 2^(2t) = 100/3, then take log₂ of both sides: 2t = log₂(100/3), and divide by 2 to get t = log₂(100/3)/2. To approximate with a calculator, use change of base: log₂(100/3) = ln(100/3)/ln(2) ≈ 3.507/0.693 ≈ 5.058, so t ≈ 5.058/2 ≈ 2.529. Choice A correctly isolates the exponential, takes log base 2, and divides by 2 to get t = log₂(100/3)/2 ≈ 2.529. Choice B incorrectly keeps the original fraction 500/15 without simplifying—while 500/15 = 100/3, the simplified form makes the calculation clearer! Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(100/3)? Use change of base: log₂(100/3) = ln(100/3)/ln(2) ≈ 5.058. The formula is log_b(x) = ln(x)/ln(b) for any base b.