First, multiply the outside term with each term within the parentheses:

$\displaystyle ab^{-2}(a^{2}b + a^{-1}b^{3}-a^{3}b^{-1})$

$\displaystyle a^{3}b^{-1} + b-a^{4}b^{-3}$

Rearranging the polynomial into fractional form, we get:

$\displaystyle \frac{a^{3}}{b}+b-\frac{a^{4}}{b^{3}}$

Since $\displaystyle \log_{10} x = A$ and $\displaystyle \log_{10} y = B$, it follows that $\displaystyle 10^{A} = x$ and $\displaystyle 10^{B} = y$

$\displaystyle 100^{2A-B} = \left( 10^{2} \right ) ^{2A-B} =10^{2 (2A-B)} =10^{4A-2B}=\frac{10^{4A}}{10^{2B}} =\frac{\left ( 10^{A}\right )^4}{\left (10^{B}\right )^2}=\frac{x^4}{y^2}$

$\displaystyle \textup{This can be re-written as an exponent problem: } 2^{x}=32$

$\displaystyle 2\times2\times2\times2\times2=32\:\textup{ so }32=2^{5}\:\:\:x=5$

Recall that by definition a logarithm is the inverse of the exponential function. Thus, our logarithm corresponds to the value of $\displaystyle x$ in the equation: 

$\displaystyle 2^{x} = 8$. 

We know that $\displaystyle 2^{3} = 8$ and thus our answer is $\displaystyle 3$.

$\displaystyle \log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

$\displaystyle \log_{2} \left[ \left ( x+4 \right )\left ( x+5 \right ) \right ] = \log_{2} 6$

$\displaystyle 2^{ \log_{2} \left[ \left ( x+4 \right )\left ( x+5 \right ) \right ] }=2^{ \log_{2} 6}$

$\displaystyle (x + 4)(x + 5) = 6$

FOIL: $\displaystyle x^{2} + 4x + 5x + 4 \cdot 5 = 6$

$\displaystyle x^{2} + 9x + 20 = 6$

$\displaystyle x^{2} + 9x + 20 - 6 = 6- 6$

$\displaystyle x^{2} + 9x + 14 = 0$

$\displaystyle (x + 2) (x + 7) = 0$

$\displaystyle x + 2 = 0 \textrm{ or }x + 7 = 0$

$\displaystyle x = -2 \textrm{ or } x = -7$

These are our possible solutions. However, we need to test them.

$\displaystyle x = -2$:

$\displaystyle \log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

$\displaystyle \log_{2}\left ( -2+4 \right ) + \log_{2}\left ( -2+5 \right ) = \log_{2} 6$

$\displaystyle \log_{2}2 + \log_{2}3 = \log_{2} 6$

$\displaystyle \log_{2}2 = 1$

$\displaystyle \log_{2}3 = \frac{\ln 3}{\ln 2} \approx \frac{1.10}{0.69} \approx 1.59$

$\displaystyle \log_{2}6 = \frac{\ln 6}{\ln 2} \approx \frac{1.79}{0.69} \approx 2.59$

The equation becomes $\displaystyle 1 + 1.59 = 2.59$. This is true, so $\displaystyle x = -2$ is a solution.

$\displaystyle x = -7$:

$\displaystyle \log_{2}\left ( x+4 \right ) + \log_{2}\left ( x+5 \right ) = \log_{2} 6$

$\displaystyle \log_{2}\left ( -7+4 \right ) + \log_{2}\left ( -7+5 \right ) = \log_{2} 6$

$\displaystyle \log_{2}\left (-3 \right ) + \log_{2}\left (-2 \right ) = \log_{2} 6$

However, negative numbers do not have logarithms, so this equation is meaningless. $\displaystyle x = -7$ is not a solution, and $\displaystyle x = -2$ is the one and only solution. Since this is not one of our choices, the correct response is "The correct solution set is not included among the other choices."

$\displaystyle \textup{Reduce the fraction by subtracting exponents, then halve all exponents to find the square root.}$

$\displaystyle \sqrt{\frac{x^{4}y^{11}z^{5}}{x^{2}y^{5}z^{7}}} =\sqrt{\frac{x^{2}y^{6}}{z^{2}}}\:\:=\frac{xy^{3}}{z}$

This sum can be determined using the formula for the sum of an infinite geometric series, with initial term $\displaystyle a_{0} = 1$ and common ratio $\displaystyle r = -\frac{2}{3}$:

$\displaystyle S = \frac{a_{0}} {1-r} = \frac{1} {1- \left ( -\frac{2}{3}\right ) } = \frac{1} {1+ \frac{2}{3} }= \frac{1} { \frac{5}{3} } =\frac{3}{5}$

An arithmetic sequence is one in which there is a common difference between consecutive terms. For example, the sequence {2, 5, 8, 11} is an arithmetic sequence, because each term can be found by adding three to the term before it. 

Let $\displaystyle a_{n}$ denote the nth term of the sequence. Then the following formula can be used for arithmetic sequences in general:

$\displaystyle a_n=a_1 +(n-1)d$, where d is the common difference between two consecutive terms. 

We are given the 4th and 8th terms in the sequence, so we can write the following equations:

$\displaystyle a_4=a_1+(4-1)d=a_1+3d=-20$

$\displaystyle a_8=a_1+(8-1)d=a_1+7d=-10$.

We now have a system of two equations with two unknowns:

$\displaystyle a_1+3d=-20$

$\displaystyle a_1+7d=-10$

Let us solve this system by subtracting the equation $\displaystyle a_1+7d=-10$ from the equation $\displaystyle a_1+3d=-20$. The result of this subtraction is

$\displaystyle -4d=-10$.

This means that d = 2.5.

Using the equation $\displaystyle a_1+7d=-10$, we can find the first term of the sequence.

$\displaystyle a_1+7(2.5)=-10$

$\displaystyle a_1=-27.5$

Ultimately, we are asked to find the hundredth term of the sequence.

$\displaystyle a_{100}=a_1+(100-1)d=-27.5+99(2.5)=220$

The answer is 220.

The formula for the summation of an infinite geometric series is

$\displaystyle S= \frac{a_1}{1-r}$,

where $\displaystyle a_1$ is the first term in the series and $\displaystyle r$ is the rate of change between succesive terms.  The key here is finding the rate, or pattern, between the terms.  Because this is a geometric sequence, the rate is the constant by which each new term is multiplied. 

Plugging in our values, we get:

$\displaystyle S = \frac{3}{1-\frac{1}{3}}$

$\displaystyle S=\frac{3}{\frac{2}{3}}$

$\displaystyle S = \frac{9}{2}$

The formula for the summation of an infinite geometric series is

$\displaystyle S= \frac{a_1}{1-r}$,

where $\displaystyle a_1$ is the first term in the series and $\displaystyle r$ is the rate of change between succesive terms in a series

Because the terms switch sign, we know that the rate must be negative. 

Plugging in our values, we get:

$\displaystyle S = \frac{8}{1-(\frac{-3}{4})}$

$\displaystyle S = \frac{8}{\frac{7}{4}}$

$\displaystyle S = \frac{32}{7}$