Logarithm Properties - Business Calculus

1. Given that $\log_b(3) = 1.58$ and $\log_b(5) = 2.32$, what is the approximate value of $\log_b(\frac{b^2}{45})$?
   Answer: $-3.48$ Using the properties of logarithms, the expression can be expanded: $\log_b\left(\frac{b^2}{45}\right) = \log_b(b^2) - \log_b(45)$ Using the quotient rule. Then, we expand $\log_b(45)$: $\log_b(45) = \log_b(9 \cdot 5) = \log_b(3^2) + \log_b(5) = 2\log_b(3) + \log_b(5)$ Using product and power rules. The full expression is: $\log_b(b^2) - (2\log_b(3) + \log_b(5))$ We know $\log_b(b^2)=2$. Substituting the given values: $2 - (2(1.58) + 2.32) = 2 - (3.16 + 2.32) = 2 - 5.48 = -3.48$
2. The number of units $N$ produced per day by a new employee after $t$ days of training is modeled by $N(t) = 60 - 60e^{-0.08t}$. How many full days of training are required for the employee to produce at least 50 units per day?
   Answer: 23 We need to find the smallest integer $t$ for which $N(t) \ge 50$. Set up the inequality: $60 - 60e^{-0.08t} \ge 50$. Subtract 60 from both sides: $-60e^{-0.08t} \ge -10$. Divide by -60 and reverse the inequality sign: $e^{-0.08t} \le \frac{10}{60} = \frac{1}{6}$. Take the natural logarithm of both sides: $\ln(e^{-0.08t}) \le \ln(1/6)$. This simplifies to $-0.08t \le -\ln(6)$. Divide by -0.08 and reverse the inequality sign again: $t \ge \frac{\ln(6)}{0.08}$. Using a calculator, $t \ge \frac{1.7917}{0.08} \approx 22.39$. Since the question asks for the number of full days required to produce at least 50 units, we must round up to the next integer, which is 23.
3. A learning curve model suggests that the time $T(n)$ in minutes to produce the $n$-th unit is given by $T(n) = A - B \ln(n)$ for $n \ge 1$, where $A$ and $B$ are positive constants. Which of the following represents the time saved by producing the 10th unit instead of the 2nd unit?
   Answer: $B \ln(5)$ Learning curve problems test your ability to interpret function notation and calculate differences between function values. When you see "time saved," you're looking for the difference between two production times. The time to produce the 2nd unit is $T(2) = A - B \ln(2)$, and the time to produce the 10th unit is $T(10) = A - B \ln(10)$. The time saved by producing the 10th unit instead of the 2nd unit is the difference: $T(2) - T(10)$. Calculating this difference: $T(2) - T(10) = (A - B \ln(2)) - (A - B \ln(10)) = A - B \ln(2) - A + B \ln(10) = B \ln(10) - B \ln(2)$. Using the logarithm property $\ln(a) - \ln(b) = \ln(a/b)$, we get: $B \ln(10) - B \ln(2) = B(\ln(10) - \ln(2)) = B \ln(10/2) = B \ln(5)$. This matches choice D. Choice A gives $A - B \ln(5)$, which incorrectly keeps the constant $A$ that should cancel out. Choice B shows $B \ln(8)$, which comes from incorrectly calculating $10 - 2 = 8$ and using that inside the logarithm. Choice C gives $B \ln(20)$, which results from incorrectly multiplying $10 \times 2 = 20$ instead of dividing. Remember: when finding "time saved" or any difference between function values, always subtract in the right direction and be careful with logarithm properties. The difference $\ln(a) - \ln(b) = \ln(a/b)$ is crucial for learning curve calculations.
4. The sales $S_A$ of product A are growing according to $S_A(t) = 100e^{0.12t}$, and the sales $S_B$ of product B are growing according to $S_B(t) = 150e^{0.08t}$, where $t$ is the number of years from the present. At what time $t$ will the sales of product A equal the sales of product B?
   Answer: $t = \frac{\ln(3) - \ln(2)}{0.04}$ When you see exponential growth functions that need to be equal at some point in time, you're looking for the intersection of two exponential curves. This requires setting the functions equal and solving for the time variable. To find when sales are equal, set $S_A(t) = S_B(t)$: $100e^{0.12t} = 150e^{0.08t}$ Divide both sides by $100e^{0.08t}$: $e^{0.12t - 0.08t} = \frac{150}{100}$ $e^{0.04t} = 1.5$ Take the natural logarithm of both sides: $0.04t = \ln(1.5)$ $t = \frac{\ln(1.5)}{0.04}$ Since $1.5 = \frac{3}{2}$, we can use the logarithm property $\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$: $t = \frac{\ln(3) - \ln(2)}{0.04}$ This matches answer choice D. Choice A incorrectly uses 0.04 in the numerator's denominator rather than as the overall denominator. Choice B uses 0.20 as the denominator, which would come from incorrectly adding the exponents (0.12 + 0.08) instead of subtracting them. Choice C uses ln(50) in the numerator, which appears to come from multiplying the coefficients (100 × 150 ÷ 300 = 50) rather than finding their ratio. Remember: when solving exponential equations where the bases are the same, use division to combine the exponents, then apply natural logarithms to isolate the variable. Always subtract exponents when dividing exponential terms with the same base.
5. A student presents the following work to solve an equation. In which step does the first error appear? Equation: $\log_2(x-3)^2 = 4$ Step 1: $2 \log_2(x-3) = 4$ Step 2: $\log_2(x-3) = 2$ Step 3: $x-3 = 2^2 = 4$ Step 4: $x=7$ Conclusion: The only solution is $x=7$.
   Answer: The work is correct, but the conclusion is wrong because another solution was missed. The error lies not in the steps shown, but in an implicit assumption made in Step 1 that leads to a missed solution. The domain of $\log_2(x-3)^2$ is all $x \neq 3$, because $(x-3)^2$ is positive. However, the domain of $2\log_2(x-3)$ is $x>3$. Applying the power rule $\log_b(u^n) = n\log_b(u)$ is only valid if $u>0$. The correct way to apply the power rule here is $\log_2(x-3)^2 = 2\log_2|x-3|$. This leads to $\log_2|x-3|=2$, so $|x-3|=2^2=4$. This gives two possibilities: $x-3=4$ (so $x=7$) and $x-3=-4$ (so $x=-1$). Both solutions are valid in the original equation. The student's work missed the $x=-1$ solution. Therefore, the work shown is technically correct for finding one solution, but the conclusion is wrong because a solution was missed due to the subtle error in applying the power rule.
6. The sound intensity level of a running dishwasher is 50 decibels (dB). A garbage disposal is then turned on, and the combined sound level is measured at 56 dB. The sound level $L$ is defined as $L = 10 \log(I/I_0)$. By approximately what factor is the intensity of the garbage disposal greater than the intensity of the dishwasher?
   Answer: 3.0 Let $I_D$ be the dishwasher intensity and $I_G$ be the garbage disposal intensity. We have $50 = 10 \log(I_D/I_0)$, so $I_D = I_0 \cdot 10^5$. The combined intensity is $I_{total} = I_D + I_G$. The combined level is $56 = 10 \log((I_D+I_G)/I_0)$. This gives $5.6 = \log((I_D+I_G)/I_0)$, so $I_D+I_G = I_0 \cdot 10^{5.6}$. We want to find the ratio $I_G/I_D$. We can write $I_G = (I_D+I_G) - I_D = I_0 \cdot 10^{5.6} - I_0 \cdot 10^5 = I_0(10^{5.6} - 10^5)$. The ratio is $\frac{I_G}{I_D} = \frac{I_0(10^{5.6} - 10^5)}{I_0 \cdot 10^5} = \frac{10^{5.6}}{10^5} - 1 = 10^{0.6} - 1$. Since $10^{0.5} = \sqrt{10} \approx 3.16$ and $10^1=10$, $10^{0.6}$ is a bit more than 3.16. $10^{0.6} \approx 3.98$. So the factor is approximately $3.98 - 1 = 2.98$, which is closest to 3.0.
7. Given that $\log_2(5) \approx 2.32$ and $\log_2(3) \approx 1.58$, what is the approximate value of $\log_3(50)$?
   Answer: 3.58 This problem requires using the change of base formula and other logarithm properties. We want to find $\log_3(50)$. Using the change of base formula with base 2: $\log_3(50) = \frac{\log_2(50)}{\log_2(3)}$. We can expand the numerator using the product rule: $\log_2(50) = \log_2(2 \cdot 25) = \log_2(2 \cdot 5^2) = \log_2(2) + \log_2(5^2) = 1 + 2\log_2(5)$. Now substitute the given approximate values: $\log_3(50) \approx \frac{1 + 2(2.32)}{1.58} = \frac{1 + 4.64}{1.58} = \frac{5.64}{1.58} \approx 3.569$. The closest answer is 3.58.
8. A company's production efficiency $E$ is modeled by the equation $E = 5 \left( 3\ln(R) - \frac{1}{2}\ln(M) - 2\ln(C) \right)$, where $R$ represents units of raw material, $M$ represents machinery age, and $C$ represents the cost of capital. Which of the following expressions is equivalent to the efficiency model?
   Answer: $E = \ln\left( \frac{R^{15}}{M^{5/2}C^{10}} \right)$ First, apply the power rule for logarithms inside the parentheses: $3\ln(R) = \ln(R^3)$, $\frac{1}{2}\ln(M) = \ln(M^{1/2})$, and $2\ln(C) = \ln(C^2)$. The expression becomes $E = 5 \left( \ln(R^3) - \ln(M^{1/2}) - \ln(C^2) \right)$. Next, combine the terms inside the parentheses using the quotient rule: $E = 5 \ln\left( \frac{R^3}{M^{1/2}C^2} \right)$. Finally, apply the power rule again to bring the outer coefficient $5$ inside as an exponent: $E = \ln\left( \left( \frac{R^3}{M^{1/2}C^2} \right)^5 \right) = \ln\left( \frac{R^{15}}{M^{5/2}C^{10}} \right)$. Note that choice B is a correct intermediate step, but not the fully simplified form equivalent to A.
9. An investment grows according to $A(t) = P \cdot e^{rt}$. If $\ln(A(5)) - \ln(A(2)) = 1.5$, and you need to find the same growth difference using base-2 logarithms, what is $\log_2(A(5)) - \log_2(A(2))$?
   Answer: $\frac{1.5}{\ln(2)}$ From ln(A(5)) - ln(A(2)) = ln(A(5)/A(2)) = 1.5, we have A(5)/A(2) = $e^{1.5}$. Using change of base formula: log₂(A(5)) - log₂(A(2)) = log₂(A(5)/A(2)) = log₂($e^1$.5) = 1.5·log₂(e) = 1.5/ln(2). Choice B incorrectly multiplies by ln(2). Choice C uses the reciprocal relationship incorrectly. Choice D would give (1.5·log₂(e))² which is incorrect.
10. An investment grows according to the formula $A = P \cdot 5^{rt}$. The time $t$ in years required for the principal $P$ to triple is given by the equation $3 = 5^{rt}$. If the annual interest rate is $r=0.08$, which of the following expressions correctly represents the time $t$?
    Answer: $t = \frac{\ln(3)}{0.08 \ln(5)}$ The equation to solve is $3 = 5^{0.08t}$. To solve for $t$, we take the natural logarithm of both sides: $\ln(3) = \ln(5^{0.08t})$. Using the power rule for logarithms, we get $\ln(3) = 0.08t \cdot \ln(5)$. To isolate $t$, we divide both sides by $0.08 \ln(5)$, which gives $t = \frac{\ln(3)}{0.08 \ln(5)}$. By the change of base formula, this is also equal to $\frac{\log_5(3)}{0.08}$.
11. If $\log_c(x) = 4$ and $\log_c(y) = -1$, what is the value of $\log_c\left(\frac{\sqrt{xy^2}}{c^2}\right)$?
    Answer: $-1$ When you encounter logarithmic expressions with multiple operations, the key is systematically applying logarithm properties to break down complex expressions into simpler components you can evaluate. Start by using the given information: $\log_c(x) = 4$ and $\log_c(y) = -1$. To find $\log_c\left(\frac{\sqrt{xy^2}}{c^2}\right)$, apply the quotient property: $\log_c\left(\frac{A}{B}\right) = \log_c(A) - \log_c(B)$. This gives you: $\log_c(\sqrt{xy^2}) - \log_c(c^2)$ For the first term, use the product property and power property. Since $\sqrt{xy^2} = (xy^2)^{1/2}$: $\log_c(\sqrt{xy^2}) = \frac{1}{2}\log_c(xy^2) = \frac{1}{2}[\log_c(x) + \log_c(y^2)] = \frac{1}{2}[\log_c(x) + 2\log_c(y)]$ Substituting the given values: $\frac{1}{2}[4 + 2(-1)] = \frac{1}{2}[4 - 2] = \frac{1}{2}(2) = 1$ For the second term: $\log_c(c^2) = 2\log_c(c) = 2(1) = 2$ Therefore: $\log_c\left(\frac{\sqrt{xy^2}}{c^2}\right) = 1 - 2 = -1$ Answer choice A (1) represents just the numerator calculation without subtracting the denominator. Choice C (-0.5) likely comes from incorrectly handling the square root operation. Choice D (3) appears to result from adding terms instead of properly applying quotient and power rules. Remember: always work systematically through logarithm properties—quotient, product, and power rules—and double-check that you're applying operations (especially subtraction for quotients) in the right direction.
12. If $\log_b(x) = p$ and $\log_b(y) = q$, which of the following expressions is equivalent to $\log_b(\sqrt[3]{x^2y})$?
    Answer: $\frac{1}{3}(2p+q)$ Using logarithm properties, we expand the expression. First, rewrite the cube root as a fractional exponent: $\log_b((x^2y)^{1/3})$. Apply the power rule: $\frac{1}{3}\log_b(x^2y)$. Next, apply the product rule inside the logarithm: $\frac{1}{3}(\log_b(x^2) + \log_b(y))$. Apply the power rule again: $\frac{1}{3}(2\log_b(x) + \log_b(y))$. Finally, substitute the given values $\log_b(x) = p$ and $\log_b(y) = q$ to get $\frac{1}{3}(2p+q)$. Note that choice A, $\frac{2}{3}p + \frac{1}{3}q$, is algebraically equivalent to choice D when the distributive property is applied.
13. A company's revenue model follows the equation $R(t) = 50000 \cdot 3^{2t-1}$ where $t$ is time in years. To analyze growth patterns, the finance team needs to express this using base-10 logarithms. If $\log_{10}(3) \approx 0.477$, what is $\log_{10}(R(3))$?
    Answer: $\log_{10}(50000) + 5\log_{10}(3)$ First, substitute t=3: R(3) = 50000·$3^{2·3-1}$ = 50000·$3^5$. Taking log₁₀ of both sides: log₁₀(R(3)) = log₁₀($50000·3^5$) = log₁₀(50000) + log₁₀($3^5$) = log₁₀(50000) + 5log₁₀(3). Choice A incorrectly uses 4.699 instead of log₁₀(50000). Choice B makes the same error and incorrectly calculates the exponent as 10. Choice D incorrectly calculates 2t-1 as 6 instead of 5.
14. What is the value of the expression $2\log_6(3) + \log_6(4)$?
    Answer: $2$ This question tests your understanding of logarithm properties, specifically how to simplify expressions using the power rule and product rule for logarithms. Let's work through this step-by-step using logarithm properties. First, apply the power rule: $a\log_b(x) = \log_b(x^a)$. This means $2\log_6(3) = \log_6(3^2) = \log_6(9)$. Now your expression becomes: $\log_6(9) + \log_6(4)$ Next, use the product rule: $\log_b(x) + \log_b(y) = \log_b(xy)$. Therefore: $\log_6(9) + \log_6(4) = \log_6(9 \times 4) = \log_6(36)$ Since $36 = 6^2$, we have $\log_6(36) = \log_6(6^2) = 2$. Looking at the wrong answers: Choice A gives 3, which you might get if you incorrectly thought $\log_6(36) = \log_6(6 \times 6) = 1 + 1 + 1$ by misapplying logarithm rules. Choice C, $\log_6(10)$, results from incorrectly adding the arguments: $3 + 3 + 4 = 10$, forgetting to apply logarithm properties entirely. Choice D, $\log_6(24)$, comes from properly applying the power rule to get $\log_6(9)$ but then incorrectly adding arguments instead of multiplying: $9 + 4 + 4 + 4 + 3 = 24$. Study tip: When simplifying logarithmic expressions, always look for opportunities to use the power rule first ($a\log_b(x) = \log_b(x^a)$), then the product/quotient rules. Remember that addition of logs means multiplication of their arguments.
15. A financial analyst is modeling the price of two competing stocks. The price of Stock A is given by $P_A(t) = 120(1.03)^t$ and the price of Stock B is given by $P_B(t) = 150(1.015)^t$, where $t$ is the number of years from today. Which of the following equations must be solved to find the number of years, $t$, until the stocks have the same price?
    Answer: $t = \frac{\ln(1.25)}{\ln(1.03) - \ln(1.015)}$ To find when the stocks have the same price, set $P_A(t) = P_B(t)$: $120(1.03)^t = 150(1.015)^t$. First, isolate the exponential terms: $\frac{(1.03)^t}{(1.015)^t} = \frac{150}{120}$, which simplifies to $\left(\frac{1.03}{1.015}\right)^t = 1.25$. To solve for $t$, take the natural logarithm of both sides: $t \ln\left(\frac{1.03}{1.015}\right) = \ln(1.25)$. Using the quotient rule for logarithms, this becomes $t(\ln(1.03) - \ln(1.015)) = \ln(1.25)$. Finally, divide to solve for $t$: $t = \frac{\ln(1.25)}{\ln(1.03) - \ln(1.015)}$.
16. A pharmaceutical company models drug concentration using $C(t) = C_0 \cdot 2^{-t/h}$ where $h$ is the half-life. Given that $\log_3(C(4)) - \log_3(C(1)) = -2$, what is the half-life $h$?
    Answer: $\frac{3\log_3(2)}{2}$ log₃(C(4)) - log₃(C(1)) = log₃(C(4)/C(1)) = log₃($2^{-4/h}$/$2^{-1/h}$) = log₃($2^{-4/h+1/h}$) = log₃($2^{-3/h}$) = (-3/h)log₃(2) = -2. Solving: (-3/h)log₃(2) = -2, so 3log₃(2) = 2h, giving h = 3log₃(2)/2. Choice A ignores the logarithm conversion. Choice B incorrectly inverts the log₃(2) term. Choice D has the wrong fraction structure.
17. Let $f(x) = \ln(x^2-4)$ and $g(x) = 2\ln(x-2)$. Which of the following expressions is equivalent to $e^{f(x)-g(x)}$ for $x > 2$?
    Answer: $\frac{x+2}{x-2}$ First, simplify the exponent $f(x)-g(x)$. Using logarithm properties: $f(x)-g(x) = \ln(x^2-4) - 2\ln(x-2) = \ln((x-2)(x+2)) - \ln((x-2)^2)$. Applying the product rule gives $\ln(x-2) + \ln(x+2) - \ln((x-2)^2)$. Applying the quotient rule: $\ln\left(\frac{(x-2)(x+2)}{(x-2)^2}\right) = \ln\left(\frac{x+2}{x-2}\right)$. So, the original expression becomes $e^{\ln((x+2)/(x-2))}$. Since $e^x$ and $\ln(x)$ are inverse functions, $e^{\ln(u)} = u$. Therefore, the expression simplifies to $\frac{x+2}{x-2}$.
18. Let $x = \log_3(80)$ and $y = 2 + \log_3(10)$. Which of the following correctly describes the relationship between $x$ and $y$?
    Answer: $y > x$ When you encounter logarithmic comparison problems, the key is to manipulate the expressions using logarithm properties to make the relationship clear, rather than trying to estimate decimal values. Let's work with the given expressions: $x = \log_3(80)$ and $y = 2 + \log_3(10)$. To compare them effectively, we need to rewrite $y$ using logarithm properties. Since $2 = \log_3(3^2) = \log_3(9)$, we can write: $y = \log_3(9) + \log_3(10)$ Using the logarithm property $\log_b(m) + \log_b(n) = \log_b(mn)$: $y = \log_3(9 \times 10) = \log_3(90)$ Now we're comparing $x = \log_3(80)$ and $y = \log_3(90)$. Since the logarithm function is increasing (larger inputs give larger outputs), and $90 > 80$, we have $\log_3(90) > \log_3(80)$, which means $y > x$. Choice A ($x = y$) is wrong because $\log_3(80) \neq \log_3(90)$. Choice B ($x > y$) is incorrect because $80 < 90$, so $\log_3(80) < \log_3(90)$. Choice D suggests we need a calculator, but logarithm properties allow us to solve this algebraically without decimal approximations. Study tip: When comparing logarithmic expressions, use properties like $\log_b(m) + \log_b(n) = \log_b(mn)$ and $n = \log_b(b^n)$ to rewrite expressions with the same base, then compare the arguments directly.
19. Let $A = \log_2(15)$, $B = \log_3(30)$, and $C = \log_4(60)$. Which of the following correctly orders these values from greatest to least?
    Answer: $A > B > C$ When comparing logarithms with different bases, you can't simply compare the arguments (the numbers inside the logs). Instead, you need to convert them to a common base or use properties of logarithms to make meaningful comparisons. Let's convert each logarithm to base 10 using the change of base formula: $\log_b(x) = \frac{\ln(x)}{\ln(b)}$ For $A = \log_2(15)$: $A = \frac{\ln(15)}{\ln(2)} = \frac{2.708}{0.693} ≈ 3.91$ For $B = \log_3(30)$: $B = \frac{\ln(30)}{\ln(3)} = \frac{3.401}{1.099} ≈ 3.10$ For $C = \log_4(60)$: $C = \frac{\ln(60)}{\ln(4)} = \frac{4.094}{1.386} ≈ 2.95$ This gives us $A > B > C$, confirming answer choice C. Option A ($C > B > A$) reverses the correct order entirely. This might trap students who incorrectly assume that larger bases always produce larger logarithmic values. Option B ($B > A > C$) correctly places $C$ last but switches $A$ and $B$. Students might fall for this if they don't carefully calculate the values or make arithmetic errors. Option D ($A > C > B$) correctly identifies $A$ as largest but reverses $B$ and $C$. This could result from miscalculating the decimal approximations. Study tip: When comparing logarithms with different bases, always convert to a common base or use a calculator to find decimal approximations. Don't assume that larger arguments or smaller bases automatically mean larger values.
20. If $\ln(x+5) + \ln(x-2) = \ln(8)$, which of the following is a valid solution for $x$?
    Answer: $x = 3$ Using the product rule for logarithms, the equation becomes $\ln((x+5)(x-2)) = \ln(8)$. This implies $(x+5)(x-2) = 8$. Expanding the left side gives $x^2 + 3x - 10 = 8$, which simplifies to the quadratic equation $x^2 + 3x - 18 = 0$. Factoring gives $(x+6)(x-3) = 0$, so the potential solutions are $x=-6$ and $x=3$. However, the domain of a logarithm requires its argument to be positive. For $x=-6$, the term $\ln(x+5)$ becomes $\ln(-1)$, which is undefined. For $x=3$, the arguments are $\ln(8)$ and $\ln(1)$, which are both defined. Therefore, $x=3$ is the only valid solution.

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