1.
Write as a single logarithm: \(5\log_2 A-\log_2 B+3\).
Use the power rule: \(5\log_2 A=\log_2(A^5)\). Since \(3=\log_2(8)\), the expression becomes \(\log_2(A^5)-\log_2(B)+\log_2(8)\). Combine using log laws: \(\log_2\left(\frac{8A^5}{B}\right)\). Answer: D
2.
A cubic aquarium would contain \(39.304\text{ ft}^3\) of water if it was completely full. What is the length of one of its edges?
For a cube, \(V=s^3\). Since \(39.304=s^3\), take the cube root: \(s=\sqrt[3]{39.304}=3.4\). Therefore, the edge length is \(3.4\text{ ft}\). Answer: B
3.
Solve the radical equation and state the restrictions on \(x\): \(\sqrt{x}-1=\sqrt{x+5}\).
Restrictions: \(x\ge0\) and \(x+5\ge0\), so \(x\ge0\). Square both sides: \((\sqrt{x}-1)^2=x+5\), so \(x-2\sqrt{x}+1=x+5\). Then \(-2\sqrt{x}=4\), giving \(\sqrt{x}=-2\), which is impossible. Equivalently, the extraneous value \(x=4\) does not check because \(2-1\ne3\). Therefore, there is no solution. Answer: C
4.
Which restriction of \(\log_b a=c\) is not being obeyed in the logarithmic expression \(\log_1 4=4\)?
For \(\log_b a=c\), the base must satisfy \(b>0\) and \(b\ne1\). In \(\log_1 4=4\), the base is \(1\), so the restriction \(b\ne1\) is not obeyed. Answer: A
5.
Change to exponential form: \(\log_2 P=t\).
Use the conversion rule \(\log_b a=c \Longleftrightarrow a=b^c\). Therefore, \(\log_2 P=t\) becomes \(P=2^t\). Answer: A
6.
Solve: \(\log_2(3-x)+\log_2 x=1\).
Domain restrictions: \(3-x>0\Rightarrow x0\). 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: \((x-1)(x-2)=0\), giving \(x=1,2\). Both satisfy the domain. Answer: C
7.
Solve the equation for \(b\): \(a=3(5)^b\).
Take the logarithm of both sides: \(\log a=\log(3\cdot5^b)=\log3+\log(5^b)=\log3+b\log5\). Then \(\log a-\log3=b\log5\), so \(b=\frac{\log a-\log3}{\log5}\). Answer: A
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