Linear, Quadratic, and Exponential Models: Solving Exponential Equations with Logarithms (CCSS.F-LE.4)
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Common Core High School Functions › Linear, Quadratic, and Exponential Models: Solving Exponential Equations with Logarithms (CCSS.F-LE.4)
Context: A medication decays by half every 6 hours. The amount follows $A(t)=200\cdot 2^{-t/6}$. How long until only 30 mg remain? Solve for $t$ in hours. Round to three decimal places. What is the solution for $t$?
-29.443
16.422
-2.737
2.193
Explanation
Isolate: $2^{-t/6}=\dfrac{30}{200}=0.15$. Take logs: $(-t/6),\ln 2=\ln(0.15)$. Solve: $t=\dfrac{\ln(0.15)}{(-1/6),\ln 2}=\dfrac{\ln(30/200)}{(-1/6),\ln 2}\approx16.422$ hours. Choice C forgets to divide by $c=-\tfrac{1}{6}$. Choice A dropped the division by $a=200$. Choice D reverses change-of-base (puts $\ln b$ on top).
Equation: $5\cdot 2^{0.3t} = 40$. Give your answer in exact logarithmic form (no rounding). What is the solution for $t$?
$t=\frac{\ln 40}{0.3\ln 2}$
$t=\frac{\ln(8)}{\ln 2}$
$t=\frac{\ln(40/5)}{0.3\ln 2}$
$t=\frac{\ln 2}{0.3,\ln(8)}$
Explanation
Isolate the exponential: $2^{0.3t}=40/5=8$. Take logs: $0.3t,\ln 2=\ln 8$. Apply the power rule and solve: $t=\frac{\ln(8)}{0.3\ln 2}=\frac{\ln(40/5)}{0.3\ln 2}$. Choice A drops the division by $a$ (used $\ln 40$ instead of $\ln(40/5)$). Choice B forgets to divide by $c=0.3$. Choice D reverses the change-of-base structure by putting $\ln 2$ over $\ln(8)$.
An investment grows with continuous compounding as $A(t)=2000e^{0.06t}$ dollars after $t$ years. When will it reach 5000 dollars? Give your answer in exact logarithmic form (no rounding). What is the solution for $t$?
$t=\frac{\ln 5000}{0.06}$
$t=\ln(2.5)$
$t=\frac{\ln(5000/2000)}{0.06,\ln e}$
$t=\frac{\ln e}{0.06,\ln(2.5)}$
Explanation
Isolate: $e^{0.06t}=5000/2000=2.5$. Take logs: $0.06t=\ln(2.5)$. Solve: $t=\frac{\ln(2.5)}{0.06}=\frac{\ln(5000/2000)}{0.06\ln e}$. Choice A drops the division by $a$ (uses $\ln 5000$). Choice B forgets to divide by $c=0.06$. Choice D reverses the change-of-base by inverting the logs.
Equation: $0.2\cdot 10^{1.5t} = 50$. Give your answer in exact logarithmic form (no rounding). What is the solution for $t$?
$t=\frac{\ln(50)}{1.5\ln 10}$
$t=\frac{\ln(250)}{\ln 10}$
$t=\frac{\ln 10}{1.5\ln(250)}$
$t=\frac{\ln(50/0.2)}{1.5\ln 10}$
Explanation
Isolate: $10^{1.5t}=50/0.2=250$. Take logs: $1.5t,\ln 10=\ln 250$. Solve: $t=\frac{\ln(250)}{1.5\ln 10}=\frac{\ln(50/0.2)}{1.5\ln 10}$. Choice A misses dividing by $a$ (used $\ln 50$). Choice B forgets to divide by $c=1.5$. Choice C reverses change-of-base by inverting the logs.
Context: An investment grows continuously as $A(t)=150,e^{0.12t}$. How many years until it reaches 1000? Solve for $t$. Round to three decimal places. What is the solution for $t$?
15.809
57.565
1.897
4.392
Explanation
Isolate: $e^{0.12t}=\dfrac{1000}{150}=\dfrac{20}{3}$. Take logs: $0.12t,\ln e=\ln!\left(\tfrac{20}{3}\right)$. Since $\ln e=1$, $t=\dfrac{\ln(1000/150)}{0.12}\approx15.809$ years. Choice B dropped $a=150$. Choice C forgets to divide by $c=0.12$. Choice D reverses change-of-base.
Context: A city's population follows $P(t)=1200\cdot 10^{0.02t}$. After how many years will the population reach 3000? Solve for $t$. Round to three decimal places. What is the solution for $t$?
173.867
0.398
125.705
19.897
Explanation
Isolate: $10^{0.02t}=\dfrac{3000}{1200}=2.5$. Take logs: $0.02t,\ln 10=\ln 2.5$. Solve: $t=\dfrac{\ln(2.5)}{0.02,\ln 10}\approx19.897$ years. Choice B forgets to divide by $c=0.02$. Choice A dropped $a=1200$. Choice C reverses change-of-base (puts $\ln b$ on top).
A medication decays according to $M(t)=120e^{-0.45t}$ milligrams after $t$ hours. When will the amount reach 30 mg? Solve for $t$. Round to three decimal places.
3.081
-7.558
1.603
-1.386
Explanation
Isolate the exponential: $e^{-0.45t}=30/120=0.25$. Take logs: $-0.45t=\ln(0.25)$. Solve: $t=\frac{\ln(0.25)}{-0.45}\approx 3.081$. Choice B misses dividing by $a$ (used $\ln 30$), giving a large negative time. Choice D forgets to divide by $c$ (it is just $\ln(0.25)$). Choice C effectively reverses the change-of-base structure, yielding $\frac{1}{c,\ln(d/a)}$.
A bacterial culture grows as $N(t)=300\cdot 2^{0.8t}$ cells after $t$ hours. When will it reach 4800 cells? Solve for $t$. Round to three decimal places.
4
5
0.312
15.289
Explanation
Isolate: $2^{0.8t}=4800/300=16$. Take logs: $0.8t,\ln 2=\ln 16$. Solve: $t=\frac{\ln 16}{0.8\ln 2}=\frac{4\ln 2}{0.8\ln 2}=5.000$. Choice A forgets to divide by $c$ (gives $\log_2 16=4$). Choice D misses dividing by $a$ (used $\ln 4800$). Choice C reverses the change-of-base structure.
Equation: Solve $7\cdot e^{0.5t}=56$. Give your answer in exact logarithmic form. What is the solution for $t$?
t = \dfrac{\ln 56}{0.5}
t = \dfrac{\ln(56/7)}{\ln e}
t = \dfrac{\ln(56/7)}{0.5}
t = \dfrac{\ln e}{0.5,\ln(56/7)}
Explanation
Isolate: $e^{0.5t}=\dfrac{56}{7}=8$. Take logs: $0.5t,\ln e=\ln 8$. Since $\ln e=1$, $0.5t=\ln 8$, so $t=\dfrac{\ln(56/7)}{0.5}$. Choice A ignored dividing by $a=7$. Choice B forgets to divide by $c=0.5$. Choice D reverses the change-of-base relationship.
Equation: Solve $5\cdot 10^{0.4t}=80$. Give your answer in exact logarithmic form. What is the solution for $t$?
t = \dfrac{\ln(80/5)}{0.4,\ln 10}
t = \dfrac{\ln(80/5)}{\ln 10}
t = \dfrac{\ln 80}{0.4,\ln 10}
t = \dfrac{\ln 10}{0.4,\ln(80/5)}
Explanation
Isolate the exponential: $10^{0.4t}=\dfrac{80}{5}=16$. Take logs: $0.4t,\ln 10=\ln 16$. Solve: $t=\dfrac{\ln(16)}{0.4,\ln 10}=\dfrac{\ln(80/5)}{0.4,\ln 10}$. Choice B forgets to divide by $c=0.4$. Choice C dropped the factor $a=5$ before taking logs. Choice D reverses the change-of-base relationship.