1.
Express \(2\log a-\log(3b)\) as a single logarithm.
Use the power rule: \(2\log a=\log(a^2)\). Then use the quotient rule: \(\log(a^2)-\log(3b)=\log\left(\frac{a^2}{3b}\right)\). Answer: A
2.
Which restriction of \(\log_b a=c\) is not being obeyed in the logarithmic expression \(\log_{0.1}(0.12589)=0.9\)?
For \(\log_b a=c\), the restrictions are \(b>0\), \(b\ne1\), and \(a>0\). Here \(b=0.1>0\), \(b\ne1\), and \(a=0.12589>0\). Therefore, all restrictions are obeyed. Answer: A
3.
The Zorb Ball has a volume of \(524\text{ ft}^3\). What is its radius? Note: the formula to calculate the volume of a sphere is \(V=\frac{4}{3}\pi r^3\).
Use \(V=\frac{4}{3}\pi r^3\). Substitute \(V=524\): \(524=\frac{4}{3}\pi r^3\). Then \(r^3=\frac{3(524)}{4\pi}\). Taking the cube root gives \(r\approx5\). Therefore, the radius is \(5\text{ ft}\). Answer: B
4.
Solve exactly for \(x\): \(3^x=18\).
Take the logarithm of both sides: \(\log(3^x)=\log18\). By the power rule, \(x\log3=\log18\). Therefore, \(x=\frac{\log18}{\log3}\). Answer: B
5.
Change \(y=\log_3 x\) to exponential form.
Use the rule \(\log_b a=c \Longleftrightarrow a=b^c\). Therefore, \(y=\log_3 x\) becomes \(x=3^y\). Answer: A
6.
Solve the radical equation: \(2\sqrt{x+3}=\sqrt{3x}\).
Restrictions: \(x+3\ge0\) and \(3x\ge0\), so \(x\ge0\). Square both sides: \(4(x+3)=3x\). Then \(4x+12=3x\), so \(x=-12\). Since \(x=-12\) does not satisfy the restriction \(x\ge0\), it is extraneous. Therefore, there is no solution. Answer: B
7.
Solve: \(\log_2 x+\log_2(x-2)=3\).
Domain restrictions: \(x>0\) and \(x-2>0\), so \(x>2\). Combine logs: \(\log_2[x(x-2)]=3\). Convert to exponential form: \(x(x-2)=2^3=8\). Then \(x^2-2x-8=0\), so \((x-4)(x+2)=0\). Thus \(x=4\) or \(x=-2\), but only \(x=4\) satisfies \(x>2\). Answer: D
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