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\begin{document}

\title{Advanced functions, unit 2}



\author[1]{matt}%
\affil[1]{Affiliation not available}%


\vspace{-1em}



  
  \date{May 07, 2024}


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\sloppy




1.

a) 
\begin{align*}
(4a^3)(-9a^5)&=(4\times-9)(a^3\times a^5)\\
&=-36a^8
\end{align*}
b)
\begin{align*}
\frac{9x^3y^8z^17}{-3x^5y^5z^{10}}&=(9\times-3)(x^3 \times x^{-5})(y^8 \times y^{-5})(z^{17}\times z^{-10})\\
&=-27x^{-2}y^{3}z^{7} \\
&=\frac{-27y^3z^7}{x^2}
\end{align*}
c)
\begin{align*}
(-7k^8)^3&=(-7)^3(k^{8\times3})\\
&=-343k^{24}
\end{align*}
d)
\begin{align*}
\frac{(2a^3b^6)}{(-3a^4b^4)^2}&=\frac{(2^5)(a^{3\times5})(b^{6\times5})}{(-3)^2(a^{4\times2})(b^{4\times2})} \\
&=\frac{32a^{15}b^{30}}{9a^8b^8} \\
&=(32/9)(a^{15}\times a^{-8})(b^{30}\times b^{-8}) \\
&=\frac{32}{9} a^7 b^{22}
\end{align*}
e)
\begin{align*}
\frac{(b^{2a})^{2a+3}}{(b^{3a})^{a-6}}&=\frac{b^{2a(2a+3)}}{b^{3a(a-6)}} \\
&=\frac{b^{4a^2+6a}}{b^{3a^2-18a}} \\
&= b^{4a^2+6a}b^{-3a^2+18a} \\
&=b^{a^2+24a}
\end{align*}


2. 
a) 
\begin{align*}
\log_3(1/81)&=x\\
\log_3{1}-\log_3(81)&=x \\
0-4&=x \\
x&=-4
\end{align*}
b)
\begin{align*}
\log_71&=x \\
0&=x
\end{align*}
c)
\begin{align*}
\log_3{x}&=-2\\
3^{-2}&=x \\
1/9&=x
\end{align*}
d)
\begin{align*}
\log_x(0.0625)&=2 \\
x^2&=0.0625 \\
x&=\sqrt{0.0625}\\
x&=1/4
\end{align*}
e)
\begin{align*}
2\log_4 4^7&=x \\
2(7)&=x \\
14&=x
\end{align*}
f) 
\begin{align*}
\log_9 9^{\log_3 3}&=x \\
\log_3 3&=x \\
1&=x
\end{align*}
3. 
a) 
\begin{align*}
2\log_6 3 + \log_6 4 &= \log_6 3^2 +\log_6 4 \\
&=\log_6 9 + \log_6 4 \\
&= \log_6(9\times4)\\
&=\log_6 36 \\
&=2
\end{align*}
b) 
\begin{align*}
\log_2(16\sqrt{8})&=\log_2 16 + \log_2 \sqrt{8} \\
&=\log_2 16 + \log_2\sqrt{2^3} \\
&=\log_2 16 + \log_2(2^3)^{1/2} \\
&=\log_2 16 + \log_2 2^{3/2} \\
&= 4 + 3/2 \\
&= 4\frac{3}{2} \\
&=5\frac{1}{2}
\end{align*}
c) 
\begin{align*}
\log_8 1024 - \log_8 512 + 2\log_8 2 &= \log_8 1024 - \log_8 512 + \log_8 2^2 \\
&= \log_8 1024 - \log_8 512 + \log_8 4 \\
&= \log_8(\frac{1024\times4}{512}) \\
&= \log_8 8 \\
&=1
\end{align*}
4.
a)
\begin{align*}
16^{5a-1}\times256^{3a}&=128\\
(2^4)^{5a-1}\times(2^8)^{3a}&=128 \\
2^{20a-4}\times 2^{24a}&=128 \\
2^{44a-4}&=128 \\
\log_2(2^{44a-4})&=\log_2(128)\\
44a-4&=\log_2(2^7)\\
44a-4&=7 \\
44a&=11 \\
a&=11/44=1/4
\end{align*}
b)
\begin{align*}
3^{x+4}-5(3^x)&=684 \\
(3^x)(3^4)-5(3^x)&=684 \\
81(3^x)-5(3^x)&=684 \\
76(3^x)&=684 \\
3^x&=684/76 \\
3^x&=9 \\
x&=\log_3 9 \\
x=2
\end{align*}
c)
\begin{align*}
7(4^{2x})&=28(4^x) \\
\log_4(7(4^{2x})&=\log_4(28(4^x)) \\
\log_4(7)+log_4 4^{2x}&=log_4 28 + log_4 4^x \\
\log_4 7 + 2x &= \log_4(28)+x \\
\log_4 7 + 2x -\log_4 7 &= \log_4 28 + x -\log_4 7 \\
2x &= \log_4 28 - \log_4 7 + x \\
2x&=\log_4(4)+x \\
2x&=1+x\\
x&=1
\end{align*}


5.

$a=-1/3$, $k=1/8$, $c=0$, $d=4$ \\
Therefore $y=-\frac{1}{3}\log_2(\frac{1}{8}(x-4))+0$\\

6.
a) The first method to graph a function's inverse is to make a table of the $x$ and $y$ values of a function, the inverse of the function can be found by swapping the values in the $x$ and $y$ column.

The second method is to swap the $x$ and $y$ variable in the equation of the function and then rearrange for $y^{-1}$.

b) 

\begin{align*}
y&=(1/3)^x \\
x&=(1/3)^y \\
\log_{\frac{1}{3}} x &= \log_{\frac{1}{3}}(1/3)^y\\
\log_{\frac{1}{3}} x &= y\\
\text{Therefore}~ y^{-1}&=\log_\frac{1}{3} x ~\text{is the inverse of } y=(1/3)^x
\end{align*}

INSERT GRAPH

c) 

$y=(1/3)^x$\\
Domain: $\{x\in \mathbb{R}\}$ \\
Range: $\{y>0,y \in \mathbb{R}\}$ \\
$x$-intercept: Does not exist \\
$y$-intercept: $(0,1)$ \\
Vertical asymptote: Does not exist \\
Horizontal asymptote: $y=0$ \\
\\
$y^{-1}=\log_\frac{1}{3} x$\\
Domain: $\{x>0, x\in \mathbb{R}\}$ \\
Range: $\{y \in \mathbb{R}\}$ \\
$x$-intercept: $(1,0)$ \\
$y$-intercept: Does not exist \\
Vertical asymptote: $x=0$ \\
Horizontal asymptote: Does not exist\\

d) Yes, the inverse of $y=(1/3)^x$ is a function because a vertical line intersects the graph at only one point at a time.

7. 
a) INSERT PLOT

b) $y=-5^{-x}$ has been reflected across the $y$-axis and then relfected across the $x$-axis.

c) 
\begin{align*}
y&=5^x \\
x&=5^y \\
\log_5 x &= y \\
y^{-1}&=\log_5 x
\end{align*}
Therefore the inverse of $y=5^x$ is $y^{-1}=\log_5 x$.

d) INSERT PLOT

e) The transformed graph in part (a) is a graph of an exponential function, and the graph of the transformed equation in part (d) is of a logarithmic function. The two graphs are not exact inverses of each other, but their vertical asymptote and horizontal asymptote are switched with respect to their $x$ and $y$ values (e.g. the vertical asymptote of $y=\log_5 x$ is $x=0$ and the horizontal asymptote of $y=-5^{-x}$ is $y=0$). An exception of the exact switch is $x$-intercept of $y=\log_5(x)$ is $(1,0)$ and the $y$-intercept of $y=-5^{-x}$ is $(0,-1)$. The domain and range also aren't exactly switched as the domain of $y=\log_5 x$ is the positive real numbers and the range of $y=-5^{-x}$ is the negative real numbers. A full comparison between the two demonstrate their differences:

$y=-5^{-x}$\\
Domain: $\{x\in \mathbb{R}\}$ \\
Range: $\{y<0,y \in \mathbb{R}\}$ \\
$x$-intercept: Does not exist \\
$y$-intercept: $(0,-1)$ \\
Vertical asymptote: Does not exist \\
Horizontal asymptote: $y=0$ \\
\\
$y^{-1}=\log_5 x$\\
Domain: $\{x>0, x\in \mathbb{R}\}$ \\
Range: $\{y \in \mathbb{R}\}$ \\
$x$-intercept: $(1,0)$ \\
$y$-intercept: Does not exist \\
Vertical asymptote: $x=0$ \\
Horizontal asymptote: Does not exist\\

8. 

\begin{align*}
y&=1/4 \log_2(8x-56)^{16} - 12 \\
&=16/4 \log_2(8x-56)-12 \\
&=4\log_2(8x-56) - 12 \\
&=4\log_2(8(x-7))-12 \\
&=4(\log_2 8 + \log_2 (x-7))-12 \\
&=4 \log_2 2^3 + 4 \log_2(x-7)-12 \\
&= 4\times3 + 4 \log_2(x-7)-12 \\
&=12+4\log_2(x-7)-12 \\
&=4\log_2(x-7)
\end{align*}

The simplified equation and the transformed equation share the same curve and are equivalent. The simplified equation, $y=4\log_2(x-7)$, represents how the original equation, $y=1/4 \log_2(8x-56)^{16} - 12$, is a translation of $y=\log_2 x$.

9.

$y=-4\log_3(3x+6)+7=-4\log_3(3(x+2))+7$ \\
Therefore $a=-4$, $k=3$, $d=-2$, $c=7$. The logarithmic equation in base $3$ is vertically stretched by a factor of $4$ and reflected in the $x$-axis, horizontally compressed by a factor of $1/3$, translated 2 units to the left and translated $7$ units upwards.

INSERT PLOT

10.
Starting from \\
$\log_7(9)+\log_7(x^2)=4\log_7(3)$ \\
you would first use the law of logarithms for powers $nlog_a p =\log_a p^n$, with $n=4$ and $p=3$, to rewrite the equation into the form\\
$\log_7(9)+\log_7(x^2)=\log_7(3^4)$ \\
You would then simplify $3^4$ \\
$\log_7(9)+\log_7(x^2)=\log_7(81)$\\
You can then use the law of logarithms for multiplication $\log_a p + \log_a q = \log_a(p\times q)$ on the left hand side \\
$\log_7(9x^2)=\log_7(81)$ \\
You would then cancel the bases by raising both sides to base $7$ \\
$7^{\log_7{9x^2}}=7^{\log_7(81)}$ \\
$9x^2=81$ \\
You would then divide both sides by 9 \\
$x^2 = 9$\\
and take the square root of both sides to isolate $x$ \\
$x=\sqrt{9}=3$

11. The domain of $y=\log_3(x)$ is $\{x>0,x\in \mathbb{R}\}$ as the equation has a vertical asymptote at $y=0$ and so there is no solution for $x=-4$ which is less than 0.

INSERT PLOT

12. A function is a relation where there is only one output value for each input value.

You can determine whether a relation is a function using the vertical line test. If a vertical line only crosses the graph once at every $x$ point then the relation is a function.

An example of a function is $y=2x$ and of a non-function is $y=\pm\sqrt{x}$. 

$y=2x$ is a function because there is a single value of $y$ for each value of $x$. $y=\pm\sqrt{x}$ is a non-function because there are 2 values of $y$ for each $x$ value.

13.

a)

b)

c)

d)

e)

14. $A=P(1+i)^n$ \\
with $P=1750$, $A=2\times1750=3500$, $i=4.8\%=0.048$.
\begin{align*}
3500&=1750(1+0.048)^n \\
2&=(1+0.048)^n \\
2&=1.048^n \\
\log_{1.048} 2 &= \log_{1.048} 1.048^n \\
\log_{1.048} 2 &= n \\
n&=14.78~\text{years}
\end{align*}
Therefore an investment of $\$1750$ would double in size after $14.78$ years with an interest rate of $4.8\%$. 

15. $A=P(1+i)^n$ \\
with $P=50$, $i=100\%=1$. \\
$n$ is the number of times the doubling occurs. If the doubling occurs every 40 minutes, then in $12$ hours the doubling with occur $n=(12\times60)/40=18$ times.
\begin{align*}
A&=50(1+1)^{18} \\
A&=50(2)^{18} \\
A&=50\times(262~144) \\
A&=13~107~200
\end{align*}
Therefore the population after $12$ hours is $13~107~200$ if the original population was $50$ and the population doubles every $40$ minutes.

16. $A=P(1+i)^n$ \\
with $A=324.5$, $i=-15\%=-0.15$, $n=5$. \\
\begin{align*}
324.5&=P(1-0.15)^5 \\
324.5&=P(0.85)^5 \\
324.5&=P\times0.4437 \\
P&=\frac{324.5}{0.4437}=731.34
\end{align*}
Therefore if a computer is worth $\$324.50$ when it is $5$ years old and it depreciated at $5\%$ per year then it was worth $\$731.34$ when it was new.

17. $pH=-\log(H^+)$ \\
in exponential form $H^+=10^{-pH}$.
Therefore the ratio of $H^+$ ions in a $pH=2.7$ and $pH=3$ solution is \\
\begin{equation*}
\frac{10^{-3}}{10^{-2.7}}=0.50
\end{equation*}

The $pH=3$ solution is $1/2$ the strength of the original solution. Therefore 1 part water must be added to 1 part original solution. As the original solution was $2$ L, then $2$ L of water must be added to dilute the $pH=2.7$ solution to a $pH=3$ solution.

18. 
$\frac{I}{I_0}=10^{M}$ \\
\\
$M_1=6.3$, $I_1=25I_2$ \\

\begin{align*}
I_1&=I_0 10^{M_1}~&~I_2&=I_0 10^{M_2} \\
25I_2&=I_0 10^{6.3} \\
\end{align*}
\begin{align*}
\frac{I_1}{I_2}&=\frac{I_0 10^{M_1}}{I_0 10^{M_2}} \\
\frac{25I_2}{I_2}&=\frac{I_0 10^{6.3}}{I_0 10^{M_2}} \\
25&=\frac{10^{6.3}}{10^{M_2}} \\
25&=10^{6.3-M_2} \\
\log(25)&=\log(10^{6.3-M_2}) \\
\log(25)&=6.3-M_2 \\
M_2&=6.3-\log(25)=4.9
\end{align*}
Therefore the magnitude of the aftershock is $4.9$ if the original earthquake had a magnitude of $6.3$ and was $25$ times as intense. 

19. $L=10\log(\frac{I}{I_0})$ \\
$I_2=100I_1$, $L_1=45$.
\begin{align*}
L_1&=10\log\frac{I_1}{I_0}~&~L_2&=10\log\frac{I_2}{I_0} \\
45&=10\log\frac{I_1}{I_0}~&~L_2&=10\log\frac{100I_1}{I_0} \\
4.5&=\log\frac{I_1}{I_0}~&~\frac{L_2}{10}&=\log\frac{100I_1}{I_0} \\
10^{4.5}&=10^{\log\frac{I_1}{I_0}}~&~10^{\frac{L_2}{10}}&=10^{\log\frac{100I_1}{I_0}} \\
10^{4.5}&=\frac{I_1}{I_0}~&~10^\frac{L_2}{10}&=\frac{100I_1}{I_0} \\
I_010^{4.5}&=I_1~&~I_0 10^\frac{L_2}{10}&=100I_1 \\
\end{align*}
\begin{align*}
\frac{I_0 10^\frac{L_2}{10}}{I_010^{4.5}}&=\frac{100I_1}{I_1}\\
\frac{10^\frac{L_2}{10}}{10^{4.5}}&=\frac{100I_1}{I_1}\\
\frac{10^\frac{L_2}{10}}{10^{4.5}}&=100\\
10^\frac{L_2}{10}&=100\times10^{4.5}\\
10^\frac{L_2}{10}&=10^2\times10^{4.5}\\
10^\frac{L_2}{10}&=10^{2+4.5}\\
10^\frac{L_2}{10}&=10^{6.5}\\
\log10^\frac{L_2}{10}&=\log10^{6.5}\\
\frac{L_2}{10}&=6.5\\
L_2&=65\\
\end{align*}
Therefore if restaurant conversation is $45$ dB and ordinary conversation is $100$ times greater, then ordinary conversation is $65$ dB.

20. $A=P(1+i)^n$\\
with $A=500$, $P=2$ and $i=200\%=2$.
\begin{align*}
500&=2(1+2)^n \\
500&=2(3)^n\\
250&=3^n\\
\log_3(250)&=\log_3(3^n)\\
\log_3(250)&=n \\
n&=5.026 \\
2n&=10~\text{months}
\end{align*}
Therefore, there are $5.026$ triplings of the rabbit population from $2$ rabbits to $500$ rabbits which takes $2\times5.026=10$ months.

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