1.
Reduce completely with positive exponents: \(\frac{a^5b^8c}{a^3b^4c^3}\).
Subtract exponents when dividing: \(a^{5-3}=a^2\), \(b^{8-4}=b^4\), and \(c^{1-3}=c^{-2}\). Move \(c^{-2}\) to the denominator: \(\frac{a^2b^4}{c^2}\). Answer: D
2.
Simplify: \(\frac{12p^3q^{-7}}{15pq^6}\).
Reduce coefficients: \(\frac{12}{15}=\frac45\). For \(p\), \(p^{3-1}=p^2\). For \(q\), \(q^{-7-6}=q^{-13}\). Therefore, the simplified form is \(\frac{4p^2}{5q^{13}}\). Answer: D
3.
What is the radius of a sphere with a volume of \(343\text{ cm}^3\)? Use \(V=\frac{4}{3}\pi r^3\).
Use \(V=\frac{4}{3}\pi r^3\). Substitute \(343\): \(343=\frac{4}{3}\pi r^3\). Then \(r^3=\frac{3(343)}{4\pi}\), so \(r=\sqrt[3]{\frac{1029}{4\pi}}\approx4.34\text{ cm}\). Answer: A
4.
Solve for \(x\): \(\log_7 777=x\).
Use change of base: \(\log_7 777=\frac{\log777}{\log7}\approx3.420\). Answer: B
5.
Solve for \(x\): \(\log_{2x}78=3\).
Convert to exponential form: \(\log_{2x}78=3\Rightarrow (2x)^3=78\). Take the cube root: \(2x=\sqrt[3]{78}\), so \(x=\frac{\sqrt[3]{78}}{2}\). Answer: A
6.
The brightness of direct sunlight passing through a sheet of glass is reduced by \(20\%\). Determine an expression for the percent of the original brightness left after it passes through \(8\) sheets of glass.
Each sheet leaves \(80\%=0.8\) of the original brightness. After \(8\) sheets, the expression is \(100(0.8)^8\). Answer: C
7.
Solve for \(x\), giving your answer to three decimal places: \(10^x=12000\).
Convert to logarithmic form: \(x=\log_{10}(12000)=\frac{\log12000}{\log10}\approx4.079\). Answer: C
8.
Change \(b^a=c\) to logarithmic form.
Use the conversion rule \(b^a=c \Longleftrightarrow \log_b c=a\). Answer: A
9.
Write as a single logarithm: \(5\log_2 A-\log_2 B+3\).
Use log laws: \(5\log_2 A=\log_2(A^5)\). Also, \(3=\log_2(8)\). Therefore, \(5\log_2A-\log_2B+3=\log_2(A^5)-\log_2B+\log_2(8)=\log_2\left(\frac{8A^5}{B}\right)\). Answer: B
10.
Solve for \(x\): \(11^{6x-5}=11^{3x+4}\).
Since the bases are the same, set the exponents equal: \(6x-5=3x+4\). Then \(3x=9\), so \(x=3\). Answer: B
11.
For each increase of \(1\) unit in earthquake magnitude, there is a ten-fold increase in energy. An earthquake in Turkey measured \(2.7\), and one in Japan measured \(7.1\). About how many times more powerful was the Japan earthquake?
The magnitude difference is \(7.1-2.7=4.4\). The energy ratio is \(10^{4.4}\approx25118.86\), so the Japan earthquake was about \(25119\) times more powerful. Answer: C
12.
How many times more intense is the sound of a student talking \((60\text{ dB})\) than a student whispering \((30\text{ dB})\)?
Decibels use a base \(10\) logarithmic scale. A difference of \(60-30=30\) dB corresponds to an intensity ratio of \(10^{30/10}=10^3=1000\). Answer: B
13.
Solve: \(\log_2(3-x)+\log_2(x)=1\).
Domain: \(3-x>0\Rightarrow x0\). Combine logs: \(\log_2((3-x)x)=1\). Convert to exponential form: \((3-x)x=2\). Then \(3x-x^2=2\), so \(x^2-3x+2=0\). Factor: \((x-1)(x-2)=0\), so \(x=1,2\). Both satisfy the domain. Answer: C
14.
Simplify as a single logarithmic expression: \(\frac{1}{\log_2 5}-\frac{1}{\log_3 5}\).
Use the reciprocal identity: \(\frac{1}{\log_2 5}=\log_5 2\) and \(\frac{1}{\log_3 5}=\log_5 3\). Then \(\log_5 2-\log_5 3=\log_5\left(\frac23\right)\). Answer: E
15.
Give the domain of the function \(y=\log_4(x+7)-3\).
The argument of a logarithm must be positive: \(x+7>0\). Therefore, \(x>-7\). Answer: C
16.
Which restriction of \(\log_b a=c\) is not being obeyed in the logarithmic expression \(\log_{-2}64=-5\)?
For \(\log_b a=c\), the base must satisfy \(b>0\) and \(b\ne1\), and the argument must satisfy \(a>0\). Here \(b=-2\), so the restriction \(b>0\) is not obeyed. Answer: F
17.
Which graph best represents the function \(f(x)=\left(\frac14\right)^x\)?
The function \(f(x)=\left(\frac14\right)^x\) is an exponential decay function because the base is between \(0\) and \(1\). The correct graph is the decreasing exponential graph. Answer: C
18.
A culture has \(300\) bacteria. The number of bacteria doubles every \(4\) hours. How long will it take for the number of bacteria to reach \(72000\)?
Use \(P=P_0(2)^{t/4}\). Substitute \(72000=300(2)^{t/4}\). Then \(240=(2)^{t/4}\). Take logs: \(\log240=\frac{t}{4}\log2\), so \(t=\frac{4\log240}{\log2}\approx31.63\) hours. Answer: D
19.
Determine the restrictions on \(x\), if any: \(\frac12\sqrt{x^2-4}=6\).
The radicand must be nonnegative: \(x^2-4\ge0\). Then \(x^2\ge4\), so \(x\le-2\) or \(x\ge2\). Answer: A
20.
Solve for \(y\): \(x=\log(c^y)\).
Use the power rule: \(x=\log(c^y)=y\log c\). Divide both sides by \(\log c\): \(y=\frac{x}{\log c}\). Answer: D
1 out of 1
