Solving Exponential Equations with Logarithms - Algebra 2

Use logarithms to solve the equation $2.5\cdot e^{t/4}=40$ for $t$ . Give the exact solution and an approximate value using a calculator.
Answer: $t=4\ln(16)\approx 11.090$ 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 $ab^{ct}$ = d: (1) Isolate the exponential by dividing both sides by a: $b^{ct}$ = d/a, (2) Take log base b of both sides: $\log_b$ ( $b^{ct}$ ) = $\log_b$ (d/a), (3) Use inverse property on left: ct = $\log_b$ (d/a), (4) Solve for variable: t = $\log_b$ (d/a)/c. This systematic approach works for any exponential equation in this form! To solve 2.5· $e^{t/4}$ =40, divide by 2.5: $e^{t/4}$ =16, take natural log: ln( $e^{t/4}$ )=ln(16), simplify to t/4=ln(16), and multiply by 4: t=4 ln(16). Choice A correctly isolates, takes the natural logarithm, uses the inverse property, and approximates to 11.090 using technology. Choice B divides by 4 instead of multiplying, resulting in t=ln(16)/4 which is one-fourth the value—always multiply by the reciprocal of the exponent's coefficient! 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!
Use logarithms to solve $7\cdot 2^{5t}=120$ for $t$ . Express the solution as a logarithm (base 2) and evaluate using a calculator.
Answer: $t=\dfrac{\log_2\left(\frac{120}{7}\right)}{5}\approx 0.820$ 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 7· $2^{5t}$ = 120 follows our systematic approach: (1) Isolate the exponential by dividing both sides by 7: $2^{5t}$ = 120/7, (2) Take log base 2 of both sides: log₂( $2^{5t}$ ) = log₂(120/7), (3) Use inverse property on left: 5t = log₂(120/7), (4) Solve for variable: t = log₂(120/7)/5. This systematic approach works for any exponential equation in this form! Starting with 7· $2^{5t}$ = 120, I divide by 7 to get $2^{5t}$ = 120/7 ≈ 17.143, then take log₂ of both sides: log₂( $2^{5t}$ ) = log₂(120/7), which simplifies to 5t = log₂(120/7) using the inverse property, and finally divide by 5: t = log₂(120/7)/5. Using change of base formula: log₂(120/7) = ln(120/7)/ln(2) ≈ 2.842/0.693 ≈ 4.100, so t ≈ 4.100/5 ≈ 0.820. Choice A correctly isolates the exponential, takes log base 2, and divides by 5 to get t = log₂(120/7)/5 ≈ 0.820. Choice D incorrectly uses common log (base 10) instead of log base 2, which gives log(120/7)/5 ≈ 1.234/5 ≈ 0.246—this is wrong because we need the same base as our exponential expression. Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(120/7)? Use change of base: log₂(120/7) = ln(120/7)/ln(2) or log(120/7)/log(2)—both give the same answer ≈ 4.100. 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!
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 $t$ ?
Answer: $t = \frac{\ln(800/50)}{0.693} = \frac{\ln(16)}{0.693}$ 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.
Solve and evaluate: $2\cdot e^{0.5t}=20$ . Write $t$ in exact logarithmic form and then approximate using technology.
Answer: $t=2\ln(10)\approx 4.605$ 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 $ab^{ct}$ = d: (1) Isolate the exponential by dividing both sides by a: $b^{ct}$ = d/a, (2) Take log base b of both sides: $\log_b$ ( $b^{ct}$ ) = $\log_b$ (d/a), (3) Use inverse property on left: ct = $\log_b$ (d/a), (4) Solve for variable: t = $\log_b$ (d/a)/c. This systematic approach works for any exponential equation in this form! To solve 2· $e^{0.5t}$ =20, divide by 2: $e^{0.5t}$ =10, take natural log: ln( $e^{0.5t}$ )=ln(10), simplify to 0.5t=ln(10), and multiply by 2: t=2 ln(10). Choice A correctly isolates, takes the natural logarithm, applies the inverse property since ln( $e^y$ )=y, and approximates to 4.605. Choice B forgets to multiply by 2, resulting in t=ln(10) which is half the correct value—be sure to solve for t by dividing by the coefficient 0.5, which is the same as multiplying by 2! 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!
Use logarithms to solve the exponential equation $3\cdot 10^{2t}=75$ for $t$ . Give the exact solution (as a logarithm) and an approximate value using technology.
Answer: $t=\dfrac{\log(25)}{2}\approx 0.699$ 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 term, takes the common logarithm, and divides by 2 to get t = log(25)/2 ≈ 0.699. Choice B incorrectly uses log(75) without first isolating the exponential by dividing by 3, while Choice C forgets to divide by 2 after applying the logarithm. Isolation before logarithms: ALWAYS isolate the exponential expression $10^{2t}$ before taking logarithms. If you have 3· $10^{2t}$ = 75, first divide by 3 to get $10^{2t}$ = 25, THEN take log. Taking log of both sides of 3· $10^{2t}$ = 75 directly leads to log(3· $10^{2t}$ ), which is more complex (requires log properties). Simple isolation first makes the logarithm application clean!
Solve and evaluate using technology: $7\cdot 2^{t/4}=60$ . Give $t$ in exact logarithmic form and as a decimal approximation.
Answer: $t=4\log_2\!\left(\dfrac{60}{7}\right)\approx 12.40$ 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 $2^t$ = 10 where the exponent contains the variable, logarithms are the tool that unlocks the solution: taking log base 2 of both sides gives log₂( $2^t$ ) = 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. For 7· $2^{t/4}$ =60, isolate: $2^{t/4}$ =60/7, log base 2: t/4 = log₂(60/7), t = 4 log₂(60/7) ≈ 12.40. Choice A correctly multiplies by 4 after taking the log and approximates using change of base. Choice B divides by 4 instead of multiplying, which reverses the operation—keep track of whether the coefficient is in the numerator or denominator! 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!
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 $t$ ?
Answer: $t \approx 11.45$ 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.
Solve $6\cdot 10^{0.2t}=48$ for $t$ . Express your solution as a logarithm and then approximate using technology.
Answer: $t=5\log(8)\approx 4.515$ 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 $ab^{ct}$ = d: (1) Isolate the exponential by dividing both sides by a: $b^{ct}$ = d/a, (2) Take log base b of both sides: $\log_b$ ( $b^{ct}$ ) = $\log_b$ (d/a), (3) Use inverse property on left: ct = $\log_b$ (d/a), (4) Solve for variable: t = $\log_b$ (d/a)/c. This systematic approach works for any exponential equation in this form! To solve 6· $10^{0.2t}$ =48, divide by 6: $10^{0.2t}$ =8, take common log: log( $10^{0.2t}$ )=log(8), simplify to 0.2t=log(8), and divide by 0.2 (or multiply by 5): t=5 log(8). Choice A correctly isolates the exponential, takes the common logarithm, applies the inverse property, and approximates to 4.515. Choice B divides by 5 instead of multiplying, giving t=log(8)/5 which is too small—remember that dividing by 0.2 is multiplying by 5 to solve 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( $5·2^t$ ), 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!
Use a calculator to solve the exponential equation $6\cdot 10^{4t}=95$ . Give $t$ in exact logarithmic form and as an approximation.
Answer: $t=\dfrac{\log\!\left(\dfrac{95}{6}\right)}{4}\approx 0.300$ 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 $2^t$ = 10 where the exponent contains the variable, logarithms are the tool that unlocks the solution: taking log base 2 of both sides gives log₂( $2^t$ ) = 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. For 6· $10^{4t}$ =95, isolate: $10^{4t}$ =95/6, take log base 10: 4t = log(95/6), t = log(95/6)/4 ≈ 0.300. Choice A correctly uses base-10 log, divides by 4, and approximates. Choice B multiplies by 4 instead, which is the opposite—ensure you divide 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( $5·2^t$ ), 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!
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?
Answer: Use $\log_2(2^{3x+2}) = \log_2(64)$ to get $3x+2 = 6$ , then solve $x = \frac{4}{3}$ 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.
Solve for $t$ : $12\cdot 2^{t/4}=100$ . Give the exact solution as a logarithm and then approximate using technology.
Answer: $t=4\log_2\left(\dfrac{25}{3}\right)\approx 12.235$ 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· $2^{t/4}$ = 100: (1) Isolate the exponential by dividing both sides by 12: $2^{t/4}$ = 100/12 = 25/3, (2) Take log base 2 of both sides: log₂( $2^{t/4}$ ) = log₂(25/3), (3) Use inverse property on left: t/4 = log₂(25/3), (4) Solve for variable: t = 4log₂(25/3). This systematic approach works for any exponential equation in this form! Starting with 12· $2^{t/4}$ = 100, divide by 12 to get $2^{t/4}$ = 25/3, then take log₂ of both sides to get t/4 = log₂(25/3), and multiply by 4 to get t = 4log₂(25/3). To approximate with a calculator, use change of base: log₂(25/3) = ln(25/3)/ln(2) ≈ 2.120/0.693 ≈ 3.059, so t ≈ 4(3.059) ≈ 12.235. Choice A correctly isolates the exponential, takes log base 2, and multiplies by 4 to get t = 4log₂(25/3) ≈ 12.235. Choice B incorrectly divides by 4 instead of multiplying—when the exponent is t/4 and you get t/4 = log₂(25/3), you must multiply both sides by 4 to solve 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(25/3)? Use change of base: log₂(25/3) = ln(25/3)/ln(2) ≈ 3.059. Always multiply by the coefficient when the variable appears as t/4 in the exponent!
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?
Answer: $\frac{\ln(16)}{0.3 \ln(2)}$ 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.
Solve $2\cdot 10^{3x}=65$ for $x$ . Express the solution as a logarithm and evaluate using technology.
Answer: $x=\dfrac{\log\left(\frac{65}{2}\right)}{3}\approx 0.504$ 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 2· $10^{3x}$ = 65 where the exponent contains 3x, logarithms are the tool that unlocks the solution: first isolate the exponential by dividing both sides by 2 to get $10^{3x}$ = 65/2 = 32.5, then take log base 10 of both sides to get log( $10^{3x}$ ) = log(32.5), and using the inverse property log( $10^{3x}$ ) = 3x, we get 3x = log(32.5), so x = log(32.5)/3. To solve 2· $10^{3x}$ = 65, I first divide by 2: $10^{3x}$ = 32.5, then take log of both sides: log( $10^{3x}$ ) = log(32.5), which simplifies to 3x = log(32.5) using the inverse property, and finally divide by 3: x = log(32.5)/3 = log(65/2)/3 ≈ 1.512/3 ≈ 0.504. Choice A correctly isolates the exponential, takes the common logarithm, and divides by 3 to get x = log(65/2)/3 ≈ 0.504. Choice D uses natural log (ln) instead of common log, giving x = ln(65/2)/3 ≈ 3.482/3 ≈ 1.161, which is wrong because we need log base 10 to match our base 10 exponential—using ln would require additional conversion. Calculator strategy for base 10 logarithms: since we're using base 10, just use the 'log' button on your calculator directly—log(32.5) ≈ 1.512, then divide by 3 to get x ≈ 0.504. Always remember to isolate the exponential expression first, then take the logarithm, and finally solve for the variable by dividing by any coefficient of x in the exponent!
Solve for $t$ : $15\cdot e^{t/4}=120$ . Express $t$ exactly using a logarithm and then approximate using a calculator.
Answer: $t=4\ln(8)\approx 8.318$ 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 15· $e^{t/4}$ = 120: (1) Isolate the exponential by dividing both sides by 15: $e^{t/4}$ = 8, (2) Take ln (natural log) of both sides: ln( $e^{t/4}$ ) = ln(8), (3) Use inverse property on left: t/4 = ln(8), (4) Solve for variable: t = 4·ln(8). This systematic approach works for any exponential equation in this form! Starting with 15· $e^{t/4}$ = 120, divide by 15 to get $e^{t/4}$ = 8, then take ln of both sides: ln( $e^{t/4}$ ) = ln(8), which simplifies to t/4 = ln(8), so t = 4·ln(8). To evaluate with a calculator: ln(8) ≈ 2.079, so t ≈ 4(2.079) ≈ 8.318. Choice A correctly isolates the exponential, takes natural logarithm, and multiplies by 4 to get t = 4·ln(8) ≈ 8.318. Choice B incorrectly divides by 4 instead of multiplying, while Choice C uses ln(120) without first isolating the exponential, and Choice D uses common log instead of natural log. When the variable appears in a fraction like t/4 in the exponent, remember to multiply both sides by the denominator (4) after applying the logarithm. The pattern is: if you have $e^{t/n}$ = k, then t/n = ln(k), so t = n·ln(k). This multiplication step is crucial for getting the correct answer!
Solve the exponential equation $3\cdot 10^{2x}=75$ for $x$ . Give the exact solution (as a logarithm) and an approximate value using technology.
Answer: $x=\dfrac{\log(25)}{2}\approx 0.699$ 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 $ab^{ct}$ = d: (1) Isolate the exponential by dividing both sides by a: $b^{ct}$ = d/a, (2) Take log base b of both sides: $\log_b$ ( $b^{ct}$ ) = $\log_b$ (d/a), (3) Use inverse property on left: ct = $\log_b$ (d/a), (4) Solve for variable: t = $\log_b$ (d/a)/c. This systematic approach works for any exponential equation in this form! To solve 3· $10^{2x}$ =75, first divide both sides by 3 to isolate the exponential: $10^{2x}$ =25, then take the common logarithm (base 10) of both sides: log( $10^{2x}$ )=log(25), which simplifies to 2x=log(25) using the inverse property, and finally divide by 2: x=log(25)/2. Choice A correctly isolates and takes the logarithm to get the exact solution as log(25)/2 and approximates it to 0.699 using technology. Choice C forgets to divide by 2 after taking the log, resulting in x=log(25) which is twice as large as it should be—remember to solve fully for the variable after applying the logarithm! 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!
What is $t$ in the equation $7e^{0.4t}=50$ ? Express $t$ exactly using a logarithm and then evaluate using technology.
Answer: $t=\dfrac{\ln(50/7)}{0.4}\approx 4.917$ 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 $7e^{0.4t}$ = 50 where the exponent contains the variable, logarithms are the tool that unlocks the solution: first isolate the exponential by dividing both sides by 7 to get $e^{0.4t}$ = 50/7, then take ln (natural log) of both sides to get ln( $e^{0.4t}$ ) = ln(50/7), and using the inverse property ln( $e^{0.4t}$ ) = 0.4t, we get 0.4t = ln(50/7), so t = ln(50/7)/0.4. To get a decimal approximation, use your calculator: ln(50/7) ≈ ln(7.143) ≈ 1.967, so t ≈ 1.967/0.4 ≈ 4.917. Choice A correctly isolates the exponential term, takes the natural logarithm, and divides by 0.4 to get t = ln(50/7)/0.4 ≈ 4.917. Choice B incorrectly uses ln(50) without first isolating the exponential by dividing by 7, while Choice C multiplies by 0.4 instead of dividing, and Choice D uses common log instead of natural log (which is most natural for base e). Isolation before logarithms: ALWAYS isolate the exponential expression $e^{0.4t}$ before taking logarithms. If you have 7· $e^{0.4t}$ = 50, first divide by 7 to get $e^{0.4t}$ = 50/7, THEN take ln. Taking ln of both sides of 7· $e^{0.4t}$ = 50 directly leads to ln(7· $e^{0.4t}$ ), which is more complex (requires log properties). Simple isolation first makes the logarithm application clean!
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.
Answer: $t=\dfrac{\log_2\left(\dfrac{100}{3}\right)}{2}\approx 2.529$ 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.
A bacteria culture is modeled by $P(t)=200\cdot 2^{t/6}$ . When does it reach $P=1500$ ? Solve for $t$ (in hours) using logarithms, and approximate using technology.
Answer: $t=6\log_2(7.5)\approx 17.44$ 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 P(t) = 200· $2^{t/6}$ = 1500 where the exponent contains the variable, logarithms are the tool that unlocks the solution: first isolate the exponential by dividing both sides by 200 to get $2^{t/6}$ = 7.5, then take log₂ of both sides to get log₂( $2^{t/6}$ ) = log₂(7.5), and using the inverse property log₂( $2^{t/6}$ ) = t/6, we get t/6 = log₂(7.5), so t = 6·log₂(7.5). To get a decimal approximation, use change of base: log₂(7.5) = ln(7.5)/ln(2) ≈ 2.015/0.693 ≈ 2.907, so t ≈ 6(2.907) ≈ 17.44 hours. Choice A correctly isolates the exponential term, takes log base 2, and multiplies by 6 to get t = 6·log₂(7.5) ≈ 17.44. Choice B incorrectly divides by 6 instead of multiplying, while Choice C uses common log instead of log base 2, and Choice D uses log₂(1500) without first isolating the exponential. Calculator strategy for non-standard bases: your calculator has log (base 10) and ln (base e) buttons, but what if you need log₂(7.5)? Use change of base: log₂(7.5) = ln(7.5)/ln(2) or log(7.5)/log(2)—both give the same answer ≈ 2.907. 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!
A bacteria culture follows $N(t)=200\cdot 2^{t/6}$ . When the culture reaches $N=1500$ , what is $t$ ? Express your answer as a logarithm and approximate using a calculator.
Answer: $t=6\log_{2}(7.5)\approx 17.438$ This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. Given the bacteria model N(t) = 200· $2^{t/6}$ and solving for when N = 1500: substitute to get 1500 = 200· $2^{t/6}$ , then isolate the exponential by dividing by 200: $2^{t/6}$ = 1500/200 = 7.5. Taking log base 2 of both sides: log₂( $2^{t/6}$ ) = log₂(7.5), using inverse property: t/6 = log₂(7.5), and multiplying by 6: t = 6·log₂(7.5). Using change of base: log₂(7.5) = ln(7.5)/ln(2) ≈ 2.015/0.693 ≈ 2.906, so t ≈ 6·2.906 ≈ 17.438. Choice A correctly isolates, takes log base 2, and multiplies by 6 to get t = 6·log₂(7.5) ≈ 17.438 (time units when population reaches 1500). Choice C incorrectly divides by 6 instead of multiplying, while Choice D uses the wrong fraction. In exponential growth models, when you have t/6 in the exponent, the 6 often represents a doubling time or characteristic time scale. Always verify: if t ≈ 17.438, then t/6 ≈ 2.906, and 200· $2^{2.906}$ ≈ 200·7.5 = 1500 ✓
Find $x$ in $4\cdot 2^{x/3}=30$ . Express the solution as a logarithm (base 2) and approximate using technology.
Answer: $x=3\log_{2}(7.5)\approx 8.719$ 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 equation 4· $2^{x/3}$ = 30 has a fractional coefficient in the exponent, requiring careful algebraic manipulation: (1) Isolate the exponential by dividing by 4: $2^{x/3}$ = 30/4 = 7.5, (2) Take log base 2 of both sides: log₂( $2^{x/3}$ ) = log₂(7.5), (3) Use inverse property: x/3 = log₂(7.5), (4) Multiply both sides by 3: x = 3·log₂(7.5). Using change of base: log₂(7.5) = ln(7.5)/ln(2) ≈ 2.015/0.693 ≈ 2.906, so x ≈ 3·2.906 ≈ 8.719. Choice C correctly isolates, takes log base 2, and multiplies by 3 to clear the fraction, giving x = 3·log₂(7.5) ≈ 8.719. Choice B incorrectly divides by 3 instead of multiplying, while Choice D forgets to multiply by 3 entirely. When you have x/3 in the exponent, remember that solving x/3 = something means x = 3·something. Always verify: if x ≈ 8.719, then x/3 ≈ 2.906, and $2^{2.906}$ ≈ 7.5, so 4·7.5 = 30 ✓

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