Targeted Review Solutions

Multiply.

${(a - 4 b)}^{2}$

$(a - 4 b) (a - 4 b) a^{2} - 4 a b - 4 a b + 16 b^{2}$

$a^{2} - 8 a b + 16 b^{2}$

${(a - 4 b)}^{3}$

$(a - 4 b) (a - 4 b) (a - 4 b) (a^{2} - 8 a b + 16 b^{2}) (a - 4 b) a^{3} - 4 a^{2} b - 8 a^{2} b + 32 a b^{2} + 16 a b^{2} - 32 b^{3}$

$a^{3} - 12 a^{2} b + 48 a b^{2} - 32 b^{3}$

${(a - 4 b)}^{4}$

$(a - 4 b) (a - 4 b) (a - 4 b) (a - 4 b) (a^{2} - 8 a b + 16 b^{2}) (a^{2} - 8 a b + 16 b^{2}) a^{4} - 8 a^{3} b + 16 a^{2} b^{2} - 8 a^{3} b + 64 a^{2} b^{2} - 128 a b^{3} + + 16 a^{2} b^{2} - 128 a b^{3} + 256 b^{4}$

$a^{4} - 16 a^{3} b + 96 a^{2} b^{2} - 256 a b^{3} + 256 b^{4}$

In problems 1–3, what patterns do you notice in the exponent of the first term, a, and the second term, b?

Sample: The exponents for a start with the power the binomial is raised to and decrease until $a = 0 .$ The exponents for b start with $b = 0,$ and increase until they end with the power the binomial is raised to.

Write $\log_{a} 4 + 2 \log_{a} x - \log_{a} y$ as a single logarithm.

$\log_{a} 4 + \log_{a} x^{2} - (\log_{a} y) \log_{a} 4 x^{2} - (\log_{a} y)$

$\log_{a} \frac{4 x^{2}}{y}$

Convert to logarithmic form: $2^{5} = 32$

$\log_{2} 32 = 5$

Use the figure to complete problems 7–8.

Determine the geometric probability of landing on the circle with the given radius drawn inside (inscribed in) the square.

$\begin{array}{rcl} r & = & x + 3 \\ b & = & h = 2 (x + 3) = 2 x + 6 \\ P (circle) & = & \frac{Area of circle}{Area of square} = \frac{π r^{2}}{b h} \\ = & \frac{π {(x + 3)}^{2}}{(2 x + 6) (2 x + 6)} = \frac{π (x + 3) (x + 3)}{2 \cdot 2 (x + 3) (x + 3)} \end{array}$

$\begin{array}{rcl} P (circle) & = & \frac{π}{4} \end{array}$

Write the geometric probability of landing on the shaded region in terms of π.

$\begin{array}{rcl} r & = & x + 3 \\ b & = & h = 2 (x + 3) = 2 x + 6 \\ P (shaded) & = & \frac{(Area of square) - (Area of circle)}{Area of square} = \frac{b h - π r^{2}}{b h} \\ = & \frac{{(2 x + 6)}^{2} - (π {(x + 3)}^{2})}{{(2 x + 6)}^{2}} = \frac{(4 x^{2} + 24 x + 36) - π (x^{2} + 6 x + 9)}{4 x^{2} + 24 x + 36} \\ = & \frac{4 x^{2} + 24 x + 36 - π x^{2} - 6 π x + 9 π}{4 x^{2} + 24 x + 36} \end{array}$

$\begin{array}{rcl} P (shaded) & = & \frac{(4 - π) x^{2} + (24 - 6 π) x + (36 - 9 π)}{4 x^{2} + 24 x + 36} \end{array}$

Multiple Choice

Select the equations that best match the description.
The natural log, $g (x),$ is reflected over $y = x,$ shifted right 3 units, and shifted up 2 units from $f (x)$

$f (x) = \ln (x - 3) + 2 g (x) = e^{x}$
$f (x) = e^{x} g (x) = \ln (x - 3) + 2$
$f (x) = e^{x} g (x) = \ln (x + 3) + 2$
$f (x) = e^{(x - 3)} + 2 g (x) = \ln x$

$e \to \ln = inverse f (x) : a = 1, h = 0, k = 0 g (x) : a = 1, h = 3, k = 2$

Note

The functions are reversed.
$g (x)$ is shifted left 3 units. 
This option translates $g (x)$ to $f (x) .$

What is the solution to the equation $7^{x} = 15$ ?

$x = 2.14$
$\log_{5} x = 17$
$x = \frac{\log 7}{\log 15}$
$x = \frac{\log 15}{\log 7}$

$\begin{array}{rcl} \log 7^{x} & = & \log 15 \\ x \cdot \log 7 & = & \log 15 \\ x & = & \frac{\log 15}{\log 7} \end{array}$

Note

In this option, 15 is divided by 7. 
In this option, the exponential equation is written as a logarithm. 
This option switches the numerator and denominator.

Evaluate: $e^{\ln 5 - \ln 8}$

$\ln - 3$
$\frac{5}{8}$
$e^{3}$
$\frac{8}{5}$

$e^{\ln \frac{5}{8}} \frac{5}{8}$

Note

The quotient rule needs to be used.
The natural log does not simplify to 3.
This option is the reciprocal of the answer.

Name the domain for $y = \ln (x - 4) + 2 .$

$\{x | x \in ℝ, x > 4\}$
$\{x | x \in ℝ, x < 2\}$
$\{x | x \in ℝ, x < 4\}$
$\{x | x \in ℝ\}$

$h = 4, k = 2$

Note

The value of k does not affect the domain for this equation.
The domain is greater than, not less than, 4.
The range of the equation is all real numbers.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L3	L3	L3	–	L40	L38	L8	L8	L43	L41	L42	L43

L = Lesson in this level, A1 = Algebra 1: Principles of Secondary Mathematics