1.
Solve for \(x\): \(A^x-B=0\).
From \(A^x-B=0\), get \(A^x=B\). Take logs: \(\log(A^x)=\log B\). Then \(x\log A=\log B\), so \(x=\frac{\log B}{\log A}\). Answer: C
2.
Change \(b^a=c\) to logarithmic form.
Use the conversion rule \(b^a=c \Longleftrightarrow \log_b c=a\). Therefore, the logarithmic form is \(\log_b c=a\). Answer: B
3.
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: A
4.
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
5.
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 \(x-2\sqrt{x}+1=x+5\). This gives \(\sqrt{x}=2\), so \(x=4\). Check: \(\sqrt4-1=1\), but \(\sqrt{4+5}=3\). Since \(1\ne3\), \(x=4\) is extraneous. Therefore, there is no solution. Answer: D
6.
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 \(a>0\). In \(\log_{-2}64=-5\), the base is \(-2\), so \(b>0\) is not obeyed. Answer: B
7.
Express \(2\log k+\log5-\log p\) as a single logarithm.
Use the power rule: \(2\log k=\log(k^2)\). Then combine: \(\log(k^2)+\log5-\log p=\log(5k^2)-\log p=\log\left(\frac{5k^2}{p}\right)\). Answer: D
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