Solving Exponential & Log Equations - Business Calculus

1. The solution to the equation $2^{x+2} = 3^{x-1}$ can be written in the form $x = \frac{\ln(A)}{\ln(B)}$. What are the values of $A$ and $B$?
   Answer: $A=12$, $B=1.5$ To solve $2^{x+2} = 3^{x-1}$, take the natural logarithm of both sides: $\ln(2^{x+2}) = \ln(3^{x-1})$. Using the power rule for logarithms, this becomes $(x+2)\ln(2) = (x-1)\ln(3)$. Distribute $\ln(2)$ and $\ln(3)$: $x\ln(2) + 2\ln(2) = x\ln(3) - \ln(3)$. Rearrange the terms to group $x$ on one side: $2\ln(2) + \ln(3) = x\ln(3) - x\ln(2)$. Factor out $x$: $2\ln(2) + \ln(3) = x(\ln(3) - \ln(2))$. Use log rules to simplify the expressions in parentheses: $\ln(2^2) + \ln(3) = x(\ln(3/2))$, which is $\ln(4) + \ln(3) = x\ln(1.5)$. Further simplification gives $\ln(4 \cdot 3) = x\ln(1.5)$, so $\ln(12) = x\ln(1.5)$. Solving for $x$ yields $x = \frac{\ln(12)}{\ln(1.5)}$. Comparing this to the form $x = \frac{\ln(A)}{\ln(B)}$, we have $A=12$ and $B=1.5$.
2. The decay of a radioactive substance used in medical imaging is modeled by $A(t) = A_0 e^{-kt}$. A hospital's policy requires disposing of the substance when it has decayed to 15% of its original amount. If the substance's half-life is 8 days, what is the disposal time, rounded to the nearest day?
   Answer: 22 days First, use the half-life information to find the decay constant $k$. Half-life means that at $t=8$, $A(8) = 0.5 A_0$. So, $0.5 A_0 = A_0 e^{-k(8)}$. Dividing by $A_0$ gives $0.5 = e^{-8k}$. Taking the natural logarithm, $\ln(0.5) = -8k$. So, $k = \frac{\ln(0.5)}{-8} = \frac{-\ln(2)}{-8} = \frac{\ln(2)}{8}$. Next, find the time $t$ when the substance has decayed to 15% of its original amount, i.e., when $A(t) = 0.15 A_0$. The equation is $0.15 A_0 = A_0 e^{-kt}$. Dividing by $A_0$ gives $0.15 = e^{-kt}$. Taking the natural logarithm, $\ln(0.15) = -kt$. Solving for $t$ gives $t = \frac{\ln(0.15)}{-k}$. Substitute the expression for $k$: $t = \frac{\ln(0.15)}{-(\ln(2)/8)} = \frac{-8 \ln(0.15)}{\ln(2)}$. Using a calculator, $t \approx \frac{-8(-1.897)}{0.693} \approx 21.9$ days. Rounded to the nearest day, the disposal time is 22 days.
3. A company's revenue growth follows the model $R(t) = 50,000 \cdot 1.08^t$ where $t$ is years since 2020. If the company wants to determine when their revenue will first exceed $100,000, which equation should they solve, and what is the approximate solution?
   Answer: Solve $50,000 \cdot 1.08^t = 100,000$; solution is approximately $t = 9.01$ years To find when revenue first exceeds $100,000, we solve for equality: $50,000 \cdot 1.08^t = 100,000$. Dividing both sides by 50,000 gives $1.08^t = 2$. Taking the natural logarithm: $t \ln(1.08) = \ln(2)$, so $t = \frac{\ln(2)}{\ln(1.08)} \approx \frac{0.693}{0.077} \approx 9.01$ years. Choice A uses the wrong value. Choice B incorrectly suggests solving an inequality rather than finding the boundary point. Choice D uses an incorrect calculation method.
4. A company's daily profit $P(x)$ from selling $x$ units is given by $P(x) = 1000(1 - e^{-0.02x})$. The company's operations are considered sustainable only if the profit exceeds a fixed daily cost of $500. For which range of units $x$ are the operations sustainable?
   Answer: $x > 50 \ln(2)$ To find the range of $x$ for sustainable operations, we must solve the inequality $P(x) > 500$: $1000(1 - e^{-0.02x}) > 500$. Divide by 1000: $1 - e^{-0.02x} > 0.5$. Subtract 1 from both sides: $-e^{-0.02x} > -0.5$. Multiply by -1 and reverse the inequality sign: $e^{-0.02x} < 0.5$. Take the natural logarithm of both sides: $\ln(e^{-0.02x}) < \ln(0.5)$, which simplifies to $-0.02x < \ln(0.5)$. Using the log property $\ln(0.5) = \ln(1/2) = -\ln(2)$, the inequality is $-0.02x < -\ln(2)$. To solve for $x$, divide by -0.02 and reverse the inequality sign again: $x > \frac{-\ln(2)}{-0.02} = \frac{\ln(2)}{0.02}$. Since $1/0.02 = 50$, the solution is $x > 50 \ln(2)$.
5. If $\log_2(x) + \log_4(x) = 6$, what is the value of $x$?
   Answer: $x = 16$, after converting to a common base and solving the resulting equation Convert $\log_4(x)$ to base 2 using the change of base formula: $\log_4(x) = \frac{\log_2(x)}{\log_2(4)} = \frac{\log_2(x)}{2}$. The equation becomes: $\log_2(x) + \frac{\log_2(x)}{2} = 6$. Factoring: $\log_2(x)\left(1 + \frac{1}{2}\right) = 6$, so $\log_2(x) \cdot \frac{3}{2} = 6$. Therefore: $\log_2(x) = 6 \cdot \frac{2}{3} = 4$. Converting to exponential form: $x = 2^4 = 16$. Choice B would result from solving $\log_2(x) = 5$. Choice C would result from solving $\log_2(x) = 6$. Choice D would result from solving $\log_2(x) = 7$.
6. A pharmaceutical company models drug concentration in the bloodstream as $C(t) = 15e^{-0.23t}$ mg/L, where $t$ is hours after injection. When will the concentration drop to exactly 25% of its initial value?
   Answer: After approximately $t = 6.02$ hours, using the equation $15e^{-0.23t} = 3.75$ Initial concentration is $C(0) = 15e^0 = 15$ mg/L. Twenty-five percent of this is $0.25 \times 15 = 3.75$ mg/L. We solve $15e^{-0.23t} = 3.75$. Dividing by 15: $e^{-0.23t} = 0.25$. Taking natural logarithm: $-0.23t = \ln(0.25) = \ln(1/4) = -\ln(4)$. Thus $t = \frac{\ln(4)}{0.23} \approx \frac{1.386}{0.23} \approx 6.02$ hours. Choice B incorrectly calculates 75% instead of 25%. Choice C sets up the wrong equation by omitting the coefficient 15. Choice D uses 0.25 as the absolute concentration rather than 25% of the initial value.
7. A company measures its market influence on a logarithmic scale called the Market Influence Index (MII), given by $M = \log(S/S_0)$, where $S$ is the company's annual sales and $S_0$ is a baseline sales figure. If Company X has an MII of 4.5 and Company Y has an MII of 6.0, how many times larger are Company Y's sales than Company X's sales?
   Answer: 31.6 times larger For Company X, $4.5 = \log(S_X/S_0)$, which means $S_X = S_0 \cdot 10^{4.5}$. For Company Y, $6.0 = \log(S_Y/S_0)$, which means $S_Y = S_0 \cdot 10^{6.0}$. To find how many times larger Company Y's sales are, we compute the ratio $S_Y/S_X$: $\frac{S_Y}{S_X} = \frac{S_0 \cdot 10^{6.0}}{S_0 \cdot 10^{4.5}} = 10^{6.0 - 4.5} = 10^{1.5}$. Calculating the value, $10^{1.5} = 10^1 \cdot 10^{0.5} = 10\sqrt{10} \approx 10 \cdot 3.162 \approx 31.6$.
8. An initial investment in Fund A grows according to the model $A(t) = 10000 e^{0.05t}$, while an initial investment in Fund B grows according to $B(t) = 15000 e^{0.03t}$, where $t$ is the time in years. At approximately what time $t$ will the value of Fund A equal the value of Fund B?
   Answer: $t = \frac{\ln(1.5)}{0.02}$ To find when the values are equal, set $A(t) = B(t)$: $10000 e^{0.05t} = 15000 e^{0.03t}$. Divide both sides by $10000$ and by $e^{0.03t}$ to isolate the exponential terms: $\frac{e^{0.05t}}{e^{0.03t}} = \frac{15000}{10000}$. Using the exponent rule $e^a/e^b = e^{a-b}$, we get $e^{0.05t - 0.03t} = 1.5$, which simplifies to $e^{0.02t} = 1.5$. To solve for $t$, take the natural logarithm of both sides: $\ln(e^{0.02t}) = \ln(1.5)$, which gives $0.02t = \ln(1.5)$. Finally, divide by $0.02$: $t = \frac{\ln(1.5)}{0.02}$.
9. If $x$ and $y$ are positive numbers such that $\ln(x) - \ln(y) = 2$ and $\log_5(x) + \log_5(y) = 4$, what is the value of $x$?
   Answer: $25e$ We have a system of two equations with two variables. From the first equation, using the quotient rule for logarithms: $\ln(x/y) = 2$. Converting to exponential form gives $x/y = e^2$, so $x = ye^2$. From the second equation, using the product rule for logarithms: $\log_5(xy) = 4$. Converting to exponential form gives $xy = 5^4 = 625$. Now substitute the expression for $x$ from the first equation into the second: $(ye^2)y = 625$. This simplifies to $y^2e^2 = 625$. Taking the square root of both sides gives $ye = \sqrt{625} = 25$ (since $y>0$). From this, we find $y = 25/e$. Finally, substitute this value of $y$ back into the expression for $x$: $x = (25/e)e^2 = 25e$.
10. The number of active users for two competing social media platforms, A and B, are modeled by $U_A(t) = 10^6 \cdot \log_3(t)$ and $U_B(t) = 10^6 \cdot \log_9(4t)$, respectively, where $t$ is the number of months since launch ($t \ge 1$). At what value of $t$ will the number of active users be equal for both platforms?
    Answer: $t=4$ Set the number of users equal: $10^6 \cdot \log_3(t) = 10^6 \cdot \log_9(4t)$. The $10^6$ factor cancels, leaving $\log_3(t) = \log_9(4t)$. To solve this, convert both logarithms to the same base using the change of base formula, $\log_b(x) = \frac{\log_a(x)}{\log_a(b)}$. Let's use base 3. The equation becomes $\log_3(t) = \frac{\log_3(4t)}{\log_3(9)}$. Since $\log_3(9) = 2$, this simplifies to $\log_3(t) = \frac{\log_3(4t)}{2}$. Multiply by 2: $2\log_3(t) = \log_3(4t)$. Use the power rule for logarithms: $\log_3(t^2) = \log_3(4t)$. Now that the bases and coefficients are the same, we can equate the arguments: $t^2 = 4t$. This gives $t^2 - 4t = 0$, or $t(t-4) = 0$. The solutions are $t=0$ and $t=4$. Since the model is valid for $t \ge 1$, the only valid solution is $t=4$.
11. A product must be heated to a temperature of 180°C. It starts at 20°C in an oven set to 200°C. The temperature $T$ after $t$ minutes follows Newton's Law of Heating: $T(t) = T_a + (T_0 - T_a)e^{-kt}$, where $T_a$ is the ambient temperature and $T_0$ is the initial temperature. If the product reaches 100°C after 10 minutes, approximately how long will it take in total to reach 180°C?
    Answer: 37.4 minutes First, set up the model with the given values: $T_a = 200$ and $T_0 = 20$. So, $T(t) = 200 + (20 - 200)e^{-kt} = 200 - 180e^{-kt}$. Use the information $T(10) = 100$ to find $k$: $100 = 200 - 180e^{-10k} \implies -100 = -180e^{-10k} \implies e^{-10k} = 100/180 = 5/9$. So, $-10k = \ln(5/9)$, which gives $k = -\frac{\ln(5/9)}{10} = \frac{\ln(9/5)}{10} = \frac{\ln(1.8)}{10}$. Now, find the time $t$ when the temperature reaches 180°C: $180 = 200 - 180e^{-kt} \implies -20 = -180e^{-kt} \implies e^{-kt} = 20/180 = 1/9$. Take the natural logarithm: $-kt = \ln(1/9) = -\ln(9)$. So, $t = \frac{\ln(9)}{k}$. Substitute the value of $k$: $t = \frac{\ln(9)}{\frac{\ln(1.8)}{10}} = \frac{10 \ln(9)}{\ln(1.8)} \approx \frac{10(2.1972)}{0.5878} \approx 37.38$ minutes.
12. The number of units produced, $x$ (in thousands), is related to the advertising spending, $a$ (in thousands of dollars), by the equation $\log_2(x+3) + \log_2(x-1) = 5$. If each unit sells for $20, what is the total revenue from the units produced?
    Answer: $100,000 First, solve the logarithmic equation for $x$. Using the product rule for logarithms, $\log_2((x+3)(x-1)) = 5$. Convert to exponential form: $(x+3)(x-1) = 2^5 = 32$. Expand the left side: $x^2 + 2x - 3 = 32$. Rearrange into a standard quadratic form: $x^2 + 2x - 35 = 0$. Factor the quadratic: $(x+7)(x-5) = 0$. The solutions are $x=-7$ and $x=5$. Since $x$ represents the number of units, it must be positive. Furthermore, the arguments of the logarithms, $(x+3)$ and $(x-1)$, must be positive, which requires $x > 1$. Thus, $x=5$ is the only valid solution. Since $x$ is in thousands of units, this corresponds to 5,000 units. The total revenue is $5,000 \text{ units} \times \$20/\text{unit} = \$100,000$.
13. The relationship between advertising spending $A$ and sales $S$ is modeled in two different ways by two marketing agencies. Agency 1 uses $S_1(A) = 100 + 5 \log_2(A)$. Agency 2 uses $S_2(A) = 150 + k \log_4(A)$. If both models predict the same sales when spending $A=16$, what is the value of the constant $k$?
    Answer: $-15$ First, calculate the sales predicted by Agency 1 when $A=16$: $S_1(16) = 100 + 5 \log_2(16)$. Since $2^4 = 16$, $\log_2(16) = 4$. So, $S_1(16) = 100 + 5(4) = 100 + 20 = 120$. Next, set Agency 2's prediction equal to this value: $S_2(16) = 120$. The equation is $120 = 150 + k \log_4(16)$. Since $4^2=16$, $\log_4(16) = 2$. The equation becomes $120 = 150 + k(2)$. Now, solve for $k$: $120 - 150 = 2k$, which gives $-30 = 2k$, so $k = -15$.
14. The population $P(t)$ of a certain bacteria culture (in thousands) at time $t$ (in hours) follows a model where $t$ must satisfy the equation $e^{2t} - 6e^t + 5 = 0$. The company harvests the bacteria at the latest valid time predicted by the model. What is the population, given by $P(t) = 100e^t$, at that time?
    Answer: 500,000 The equation $e^{2t} - 6e^t + 5 = 0$ is a quadratic in form. Let $y = e^t$. The equation becomes $y^2 - 6y + 5 = 0$. Factoring the quadratic gives $(y-1)(y-5) = 0$, so the solutions for $y$ are $y=1$ and $y=5$. Substitute back $e^t$ for $y$: $e^t = 1$ or $e^t = 5$. Solving for $t$ gives $t = \ln(1) = 0$ or $t = \ln(5)$. The latest valid time is $t = \ln(5)$. The question asks for the population $P(t) = 100e^t$ at this time. At $t = \ln(5)$, we have $e^t = 5$. Therefore, $P(\ln(5)) = 100(5) = 500$. Since $P(t)$ is in thousands, the population is 500,000.
15. A bacteria population grows according to $P(t) = 1200 \cdot 3^{t/4}$ where $t$ is in hours. After how many hours will the population reach exactly 32,400?
    Answer: $t = 12$ hours, since $3^{12/4} = 3^3 = 27$ and $1200 \times 27 = 32,400$ We need to solve $1200 \cdot 3^{t/4} = 32,400$. Dividing both sides by 1200: $3^{t/4} = \frac{32,400}{1200} = 27$. Since $27 = 3^3$, we have $3^{t/4} = 3^3$, which means $\frac{t}{4} = 3$, so $t = 12$ hours. Choice A correctly identifies this solution. Choice B calculates $3^4$ but this gives the wrong population value. Choice C uses an incorrect logarithmic approach. Choice D uses $3^2$ which gives a population of only 10,800, not 32,400.

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