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Question 1. Tanville had a population of 50555 in 1990. The town grows according to the function \(A=Pe^{kt}\). If the value of \(k\) is \(0.045\), what will the population of Tanville be by the year 2030?
Step 1. Use the continuous growth formula \(A=Pe^{kt}\). Step 2. The initial population is \(P=50555\). Step 3. From 1990 to 2030, \(t=40\). Step 4. Substitute: \(A=50555e^{0.045(40)}\). Step 5. Calculate: \(A=50555e^{1.8}\approx302482\). Answer: C. \(302482\)
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Question 2. Radium-226 has a half-life of 1620 years. A lab has a \(235\)-gram sample that has been decaying for \(100\) years. Using \(A=Pe^{kt}\), and a continuous rate of decay of \(-4.28\times10^{-4}\), how many grams of radium-226 were in the original sample?
Step 1. Use \(A=Pe^{kt}\), where \(A=235\), \(k=-4.28\times10^{-4}\), and \(t=100\). Step 2. Substitute: \(235=Pe^{(-4.28\times10^{-4})(100)}\). Step 3. Simplify the exponent: \((-4.28\times10^{-4})(100)=-0.0428\). Step 4. Solve for \(P\): \(P=\frac{235}{e^{-0.0428}}\approx245\). Answer: D. \(245\)
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Question 3. A certain strain of bacteria doubles every \(5\) minutes. Assuming you start with \(100\) bacteria, using \(A=Pe^{kt}\), and a continuous growth rate of \(0.1386\), how many bacteria could be present at the end of \(30\) minutes?
Step 1. Use the continuous growth formula \(A=Pe^{kt}\). Step 2. Here, \(P=100\), \(k=0.1386\), and \(t=30\). Step 3. Substitute: \(A=100e^{0.1386(30)}\). Step 4. Calculate: \(A=100e^{4.158}\approx6394\). Answer: D. \(6394\)
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Question 4. A pine tree seedling has an initial height of \(30\) cm. It continually grows at a fixed rate \(k\) per year. If the tree is \(603\) cm tall in \(10\) years, based on the graphs below, what is the growth rate \(k\), to the nearest whole percent?
Step 1. Use the graph and locate \(t=10\) years on the horizontal axis. Step 2. Find the curve that has height approximately \(603\) cm at \(t=10\). Step 3. The curve matching this point is labeled \(k=0.30\). Step 4. Convert \(0.30\) to a percent: \(30\%\). Answer: B. \(30\%\)
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Question 5. When solving the logarithmic equation \(\log(x-6)+\log(x+3)=1\) for \(x\), the solution \(x=□\) and an extraneous solution is \(x\ne□\).
Step 1. Use the product rule: \(\log(x-6)+\log(x+3)=\log((x-6)(x+3))\). Step 2. Since the base is \(10\), \(\log((x-6)(x+3))=1\) means \((x-6)(x+3)=10\). Step 3. Expand: \(x^2-3x-18=10\). Step 4. Rearrange: \(x^2-3x-28=0\). Step 5. Factor: \((x-7)(x+4)=0\), so \(x=7\) or \(x=-4\). Step 6. Check the domain: \(x-6>0\) and \(x+3>0\), so \(x>6\). Thus \(-4\) is extraneous. Answer: A. \(7,-4\)
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Question 6. When solving the logarithmic equation \(\log_x(3x+1)+\log_x(x-4)=2\), an equation that could arise in the solution steps is:
Step 1. Use the product rule: \(\log_x(3x+1)+\log_x(x-4)=\log_x((3x+1)(x-4))\). Step 2. Since \(\log_x((3x+1)(x-4))=2\), rewrite in exponential form: \((3x+1)(x-4)=x^2\). Step 3. Expand: \(3x^2-12x+x-4=x^2\). Step 4. Simplify: \(3x^2-11x-4=x^2\). Step 5. Move all terms to one side: \(2x^2-11x-4=0\). Answer: A. \(2x^2-11x-4=0\)
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Question 7. Given \(\log a-2\log b+\frac{1}{4}\log c\), determine an expression for \(x\).
Step 1. Use the power rule: \(2\log b=\log(b^2)\) and \(\frac{1}{4}\log c=\log(c^{1/4})\). Step 2. Rewrite: \(\log a-\log(b^2)+\log(c^{1/4})\). Step 3. Use product and quotient rules: \(\log\left(\frac{a c^{1/4}}{b^2}\right)\). Step 4. Since \(c^{1/4}=\sqrt[4]{c}\), the expression is \(\frac{a\sqrt[4]{c}}{b^2}\). Answer: B. \(\frac{a\sqrt[4]{c}}{b^2}\)
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Question 8. How much better is the return on a \(4\%\) yearly interest rate investment compounded \(3\) times per year as opposed to compounded yearly?
Step 1. Use \(A=P\left(1+\frac{i}{n}\right)^{nt}\). Let \(P=1\) and \(t=1\). Step 2. Compounded 3 times per year: \(A=1\left(1+\frac{0.04}{3}\right)^3\approx1.0405\). Step 3. Compounded yearly: \(A=1(1+0.04)=1.04\). Step 4. Compare the difference as a percent of the compounded 3 times result: \(\frac{1.0405-1.04}{1.0405}\times100\%\approx1.23\%\). Step 5. This is between \(1.0\%\) and \(1.5\%\). Answer: B. Between \(1.0\%\) and \(1.5\%\) better.
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Question 9. Which one of the following equations shows how to solve for \(x\) in the equation \(30(2)^x=7\)?
Step 1. Start with \(30(2)^x=7\). Step 2. Divide both sides by \(30\): \(2^x=\frac{7}{30}\). Step 3. Take logarithms: \(\log(2^x)=\log\left(\frac{7}{30}\right)\). Step 4. Use the power rule: \(x\log 2=\log\left(\frac{7}{30}\right)\). Step 5. Solve: \(x=\frac{\log\left(\frac{7}{30}\right)}{\log 2}\). Answer: B. \(x=\frac{\log\left(\frac{7}{30}\right)}{\log 2}\)
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Question 10. Solve for \(x\): \(5^x=615\). Round to one decimal place.
Step 1. Start with \(5^x=615\). Step 2. Take logarithms of both sides: \(\log(5^x)=\log(615)\). Step 3. Use the power rule: \(x\log 5=\log 615\). Step 4. Solve: \(x=\frac{\log 615}{\log 5}\approx3.99\). Step 5. Rounded to one decimal place, \(x=4.0\). Answer: A. \(4.0\)
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Question 11. Which of the following is an expression for the inverse of \(y=8^x\)?
Step 1. Start with \(y=8^x\). Step 2. Switch \(x\) and \(y\): \(x=8^y\). Step 3. Rewrite in logarithmic form: \(y=\log_8 x\). Answer: D. \(y=\log_8 x\)
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Question 12. Express \(6^2=36\) in logarithmic form:
Step 1. Use the conversion rule \(a^b=c\) means \(\log_a c=b\). Step 2. Here, \(6^2=36\). Step 3. Therefore, the logarithmic form is \(\log_6 36=2\). Answer: C. \(\log_{6}36=2\)
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Question 13. Evaluate \(\log_7 37\) to two decimal places.
Step 1. Use the change of base formula: \(\log_7 37=\frac{\log 37}{\log 7}\). Step 2. Calculate: \(\frac{\log 37}{\log 7}\approx1.855\). Step 3. Round to two decimal places: \(1.86\). Answer: D. \(1.86\)
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Question 14. What is the equation of the exponential function graphed below?
Step 1. The graph is increasing, so the base must be greater than \(1\). This eliminates decreasing models with \(\frac{1}{2}\). Step 2. The graph has a horizontal asymptote at \(y=1\), so the function has a vertical shift of \(+1\). Step 3. Check the \(y\)-intercept: when \(x=0\), \(3(2)^0+1=3+1=4\), matching the graph. Answer: D. \(3(2)^x+1\)
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Question 15. Scientists measured the temperature on planet Xenon and found it grew according to the exponential function \(T=800(500)^t-600\). What was the initial temperature?
Step 1. The initial temperature occurs when \(t=0\). Step 2. Substitute \(t=0\): \(T=800(500)^0-600\). Step 3. Since \((500)^0=1\), \(T=800(1)-600\). Step 4. Calculate: \(T=200\). Answer: B. \(200\)
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Question 16. The graph below shows the value of certain investments depreciating over time. If the value \(A\) can be represented by the exponential function \(y=Ar^t+d\), which investment has the smallest value of \(A+d\)?
Step 1. In \(y=Ar^t+d\), the value \(A+d\) is the initial value because \(r^0=1\). Step 2. Therefore, compare the \(y\)-intercepts of the investment curves. Step 3. The curve with the smallest \(y\)-intercept on the graph is Investment C. Answer: A. Investment C
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Question 17. For the graph of \(y=0.3(5.19)^x+4.15\), which of the following statements is TRUE?
Step 1. The function has the form \(y=a b^x+d\). Step 2. The horizontal asymptote is \(y=d\), so the asymptote is \(y=4.15\). Step 3. Since \(0.3(5.19)^x\) is always positive, the graph is always above \(4.15\). Step 4. Therefore, the range is \(y>4.15\). Answer: A. Horizontal asymptote: \(y=4.15\), Range: \(y>4.15\)
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Question 18. Which of the following transformations would move the graph of \(f(x)=3(5)^x-4\) up \(8\) units?
Step 1. A vertical translation up by \(8\) adds \(8\) to the entire function. Step 2. Starting from \(y=f(x)\), moving up \(8\) units gives \(y=f(x)+8\). Answer: C. \(f(x)+8\)
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Question 19. Which of the following equations could describe the graph below?
Step 1. The graph is an exponential decay curve. Step 2. It passes through \((0,1)\), which is typical for \(a^x\) when there is no vertical shift. Step 3. Since the graph decreases as \(x\) increases, the base must be between \(0\) and \(1\). Step 4. The matching equation is \(\left(\frac{1}{5}\right)^x\). Answer: A. \(\left(\frac{1}{5}\right)^x\)
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Question 20. What is the \(y\)-intercept of the function \(y=6\log_{\frac{1}{7}}(x+4)-2\)? Round to \(2\) decimal places.
Step 1. The \(y\)-intercept occurs when \(x=0\). Step 2. Substitute \(x=0\): \(y=6\log_{\frac{1}{7}}(0+4)-2\). Step 3. Simplify: \(y=6\log_{\frac{1}{7}}4-2\). Step 4. Use change of base: \(\log_{\frac{1}{7}}4=\frac{\log 4}{\log(\frac{1}{7})}\approx-0.7124\). Step 5. Calculate: \(y=6(-0.7124)-2\approx-6.27\), which rounds to \(-6.3\). Answer: C. \(-6.3\)
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Question 21. What is the inverse of \(y=\log_3 x+6\)?
Step 1. Start with \(y=\log_3 x+6\). Step 2. Switch \(x\) and \(y\): \(x=\log_3 y+6\). Step 3. Subtract \(6\): \(x-6=\log_3 y\). Step 4. Rewrite in exponential form: \(y=3^{x-6}\). Answer: B. \(y=3^{x-6}\)
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Question 22. What is the half-life of the radioactive isotope with amount \(A\), whose rate of decay in \(t\) hours is shown in the graph beside?
Step 1. From the graph, the initial amount is \(A_0=20\) at \(t=0\). Step 2. The graph also shows \((35,5)\), so use \(A=A_0e^{-kt}\): \(5=20e^{-35k}\). Step 3. Divide by \(20\): \(0.25=e^{-35k}\). Step 4. Take natural logarithms: \(\ln(0.25)=-35k\), so \(k\approx0.0396\). Step 5. Half-life means \(A=10\): \(10=20e^{-0.0396t}\). Step 6. Divide by \(20\): \(0.5=e^{-0.0396t}\). Step 7. Solve: \(t=\frac{\ln(0.5)}{-0.0396}\approx17.5\). Answer: B. Between \(17.1\) and \(17.6\) hours
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Question 23. The expression \((125)^{\frac{x}{3}}\) is equivalent to:
Step 1. Rewrite \(125\) as a power of \(5\): \(125=5^3\). Step 2. Substitute: \((125)^{\frac{x}{3}}=(5^3)^{\frac{x}{3}}\). Step 3. Use the power of a power rule: \((5^3)^{\frac{x}{3}}=5^{3\cdot\frac{x}{3}}\). Step 4. Simplify the exponent: \(3\cdot\frac{x}{3}=x\). Step 5. Therefore, the expression is \(5^x\). Answer: C. \(5^x\)
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Question 24. If a computer depreciates at a rate of \(16\%\) per year, what is the monthly depreciation rate?
Step 1. A year has \(12\) months. Step 2. Divide the yearly depreciation rate by \(12\): \(\frac{16\%}{12}=1.333\ldots\%\). Step 3. Round to two decimal places: \(1.33\%\). Answer: C. \(1.33\%\)
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