1.
Reduce completely: \(\frac{p^3q^8r^{-2}}{q^5r^8s^{-4}}\).
Subtract exponents when dividing: \(q^{8-5}=q^3,\ r^{-2-8}=r^{-10},\ s^{0-(-4)}=s^4\). Move negative exponent to denominator: \(\frac{p^3q^3s^4}{r^{10}}\). Answer: E
2.
State the restriction for \(-2\sqrt{3x+4}+6=-4\).
Require \(3x+4\ge0\Rightarrow x\ge-\frac43\). Answer: B
3.
A cube has volume \(39.304\text{ ft}^3\). Find its edge length.
\(s=\sqrt[3]{39.304}=3.4\text{ ft}\). Answer: A
4.
Which restriction is violated in \(\log_{0.1}(0.12589)=0.9\)?
Base \(0.1>0\), base \(\ne1\), argument \(>0\). All restrictions are obeyed. Answer: D
5.
Simplify: \(\left(\frac{6y}{5x^3y^{-3}}\right)^2\).
\(\frac{6y}{5x^3y^{-3}}=\frac{6y^4}{5x^3}\). Square: \(\frac{36y^8}{25x^6}\). Answer: B
6.
Evaluate \(\log_7 777\).
Use change of base: \(\log_7 777=\frac{\log777}{\log7}\approx3.420\). Answer: B
7.
Determine the range of the function \(y=-3^x+5\).
Since \(3^x>0\), \(-3^x<0\). Adding \(5\) gives \(y=-3^x+5<5\). The horizontal asymptote is \(y=5\), so the range is \(y<5\). Answer: B
8.
Solve \(\left(\frac23\right)^x=\frac{11}{12}\).
Take logs: \(x=\frac{\log(11/12)}{\log(2/3)}\approx0.215\). Answer: C
9.
Solve: \(3^x=9^3\).
\(9^3=(3^2)^3=3^6\). Therefore \(x=6\). Answer: D
10.
Determine an equation of the asymptote of \(y=2\log_3(x+4)-2\).
The vertical asymptote occurs where the logarithm argument equals zero: \(x+4=0\). Therefore, \(x=-4\). Answer: D
11.
For each increase of \(1\) unit in earthquake magnitude, there is a ten-fold increase in energy. An earthquake in Alaska measured \(3.5\), and one in Chile measured \(6.7\). About how many times more powerful was the Chile earthquake?
The difference in magnitude is \(6.7-3.5=3.2\). The energy ratio is \(10^{3.2}\approx1584.89\), so the Chile earthquake was about \(1585\) times more powerful. Answer: A
12.
The population of a type of bacteria triples every \(20\) hours. In how many hours will a population of \(30\) become \(1000\)?
Use \(P=P_0(3)^{t/20}\). Substitute: \(1000=30(3)^{t/20}\). Then \(\frac{1000}{30}=(3)^{t/20}\). Take logs: \(\log\left(\frac{100}{3}\right)=\frac{t}{20}\log3\). So \(t=\frac{20\log(100/3)}{\log3}\approx63.84\) hours. Answer: B
13.
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 brightness. Starting with \(100\%\), after \(8\) sheets the expression is \(100(0.8)^8\). Answer: C
14.
Given that \(\log a=4\) and \(\log b=6\), evaluate \(\log(a^2\sqrt{b})\).
Use log laws: \(\log(a^2\sqrt{b})=\log(a^2)+\log(b^{1/2})=2\log a+\frac12\log b\). Substitute \(\log a=4\) and \(\log b=6\): \(2(4)+\frac12(6)=8+3=11\). Answer: D
15.
Solve: \(\log_2(4-x)-\log_2 x=1\).
Domain: \(4-x>0\) and \(x>0\), so \(0<x<4\). Use the quotient rule: \(\log_2\left(\frac{4-x}{x}\right)=1\). Convert to exponential form: \(\frac{4-x}{x}=2\). Then \(4-x=2x\), so \(3x=4\), and \(x=\frac43\). Answer: A
16.
The decibel level is \(dB=10\log\left(\frac{I}{I_0}\right)\). A page being turned is \(10000\) times as intense as \(I_0\). What is the loudness?
Substitute \(\frac{I}{I_0}=10000\): \(dB=10\log(10000)=10(4)=40\). Answer: B
17.
Convert \(y=\log x\) to exponential form.
Common log has base 10: \(y=\log x\iff x=10^y\). Answer: B
18.
Solve for \(x\): \(\log_5 x=3\).
Convert to exponential form: \(\log_5 x=3\Rightarrow x=5^3\). Answer: B
19.
Equivalent to \(\log\sqrt{x}\).
\(\sqrt{x}=x^{1/2}\), so \(\log(x^{1/2})=\frac12\log x\). Answer: B
20.
Simplify as a single logarithmic expression: \(3\log_7 3+4\log_7 2\).
Use the power rule: \(3\log_7 3=\log_7(3^3)\) and \(4\log_7 2=\log_7(2^4)\). Then combine: \(\log_7(3^3)+\log_7(2^4)=\log_7(27\cdot16)=\log_7 432\). Answer: B
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