Math

• Notes
• HW 2.6
• HW 2.7
• HW 3.1
• HW 3.2
• HW 3.3
• HW 3.4
• HW 3.5
• HW 4.1
• HW 4.2
• HW 4.3
• HW 4.4
• HW 4.5
• Exam November 9th
• Exam November 9th Corrections
• Test Review 03-08-2016

• Chem 7.31 - 7.45

(43)

$f(x) = \frac{-4x}{x+8}$

Denominator cannot be $0$.

$x + 8 \ne 0$

$x \ne -8$

$(-\infty,-8)\cup(-8,\infty)$

(44)

$f(x)=\frac{1}{x-2} + \frac{4}{x-4}$

Take denominators and compare to $0$:

$x-2\ne0$
$x\ne2$

$x-4\ne0$
$x\ne4$

Domain: $(-\infty,2)\cup(2,4)\cup(4,\infty)$

(45)

$f(x)=9x^2-5x$

$g(x)=x^2-3x-10$

$f/g = \frac{9x^2-5x}{x^2-3x-10}$

Factor top and bottom (numerator and denominator).

$9x^2-5x=$
$x(9x-5)$

$x^2-3x-10=$
$(x -5)(x+2 )$

$f/g = \frac{x(9x-5)}{(x -5)(x+2 )}$

(46)

$f(x)=5x-2$

$g(x)=6x+3$

$fg=(5x-2)(6x+3)$ (multiply $f$ by $g$)

$=30x^2+15x-12x-6$

$=30x^2+3x-6$

(54)

$f(x) = \frac{6}{x+8}$

$g(x)=x + 1$

Find the domain of $f\cdot g$

$\frac{6}{x+8} (x+1)$

The equation contains $x+8$ in the denominator. Since any denominator must not equal zero, the domain is restricted to $x+8≠0$.

Compare the denominator to $0$:

$x+8\ne0$

Subtract $8$ from both sides:

$x\ne-8$

The domain in interval notation is:

$(-\infty,-8)\cup(-8,\infty)$

(55)

$f(x)=\sqrt{x}$

$g(x) = 3x+9$

$f\cdot g= (3x+9)\sqrt{x}$

The equation contains a square root. Since anything inside a square root must not be a negative value, the domain is restricted to $x≥0$ ($x$ is greater than or equal to $0$).

Domain: $[0,\infty)$

(56)

$f(x) = 3x + 12$

$g(x) = \sqrt{x}$

$f \cdot g = (3x+12)\sqrt{x}$

$x≥0$

Domain: $[0,\infty)$

(63)

$A=P(1+\frac{r}{n})^{nt}$

$A=3000(1+\frac{0.08}{1})^{9}$

$A=5,997.01$

(66)

$A=Pe^{rt}$

$A=3000e^{0.07(6)}$

$A=4565.88$

(68)

$P=10,000$
$r=8.75/100=0.0875$
$t=6$

$A=16904.59$

$P=10,000$
$r=8.9/100=0.089$
$n=2$
$t=6$

$A=16861.70$

$\$16904.59 > \$16861.70$

(73)

$5^2=25$
$\log_525=2$

(74)

$\sqrt[3]{125}=5$

$\color{blue}{\sqrt[3]{125}}$
is the same as
$\color{blue}{125^{1/3}}$

${125}^{\frac{1}{3}}=5$

$\log_{125}5=\frac{1}{3}$

(75)

$b^3=2744$

$\log_b2744=3$

(76)

$7^x=343$

$\log_7343=x$

(96)

$\log_525x$

Apply the product rule to split the $x$ from the $25$:

Product Rule
$\color{blue}{\log_b(MN) = \log_bM+\log_bN}$

$=\log_5x + \log_525$

$=\log_5x + \log_55^2$

Apply the following log rule:
$\color{blue}{\log_bb^n = n}$
(If the base is equal to the argument, and the argument is raised to the power of $n$ then the answer to the log is equal to $n$)

Therefore, $\log_55^2$ is equal to $2$

$\log_55^2=2$

$=\log_5x + \log_525$

$=\log_5x + \log_55^2$

$=\log_5x + 2$

$=2+\log_5x$

(97)

$\log_2(\frac{8}{x})$

Apply the quotient rule.

Quotient Rule
$\color{blue}{\log_b(\frac{M}{N}) = \log_bM-\log_bN}$

(When the argument is a fraction, you can rewrite the equation using subtraction)

$=\log_28-\log_2x$

Expand:

$=\log_2(2\cdot4)-\log_2x$

Apply the product rule:

$=\log_22 + \log_24-\log_2x$

Expand:

$=\log_22 + \log_22^2-\log_2x$

If the base is equal to the argument, the answer is $1$:
$\color{blue}{\log_a(a)=1}$

$\log_22=1$
$\log_22^2=2$

$=1 + 2-\log_2x\\=3+\log_2x$

(98)

$\ln(\frac{e^2}{9})$

$\ln =$ natural log

$=\ln e^2-\ln9$

Use the power rule to move the $2$

Power Rule
$\color{blue}{\log_bM^p = p \log_bM}$

$=2 \ln e - \ln 9$

The natural log of $e$ ($\ln e$) is always equal to $1$

$=2 (1) - \ln 9\\=2-\ln 9$

(99)

$\log_57^{-3}$

Use the power rule to move the power to the front:

$-3 \log_57$
(Can’t expand further)

(100)

$\log_5\sqrt{3x}$

Use the radical product rule to split the $x$ from the $3$:

$\color{blue}{(\sqrt[n]{a})\times( \sqrt[n]{b}) = \sqrt[n]{ab}}$

$=\log_5(\sqrt{3}\cdot\sqrt{x})$

Apply the logarithm product rule:

$=\log_5\sqrt{3} + \log_5\sqrt{x}$

Use the following rule to convert the radicals to exponential form:

$\color{blue}{\sqrt[n]{a}=a^{\frac{1}{n}}}$

$=\log_53^{\frac{1}{2}} + \log_5x^{\frac{1}{2}}$

Use the power rule to move the exponent to the front:

$\color{blue}{\log_bM^p = p \log_bM}$

$=\frac{1}{2}\log_53 + \frac{1}{2}\log_5x$

(101)

$\log_5(\frac{125}{\sqrt{x-1}})$

Apply the quotient rule:

$=\log_5125-\log_5\sqrt{x-1}$

$=\log_55^3-\log_5\sqrt{x-1}$

$\color{blue}{\log_bb^n = n}$

$\log_55^3 = 3$

$3-\log_5\sqrt{x-1}$

Apply the radical to exponent rule:

$\color{purple}{3-\log_5(x-1)^\frac{1}{2}}$

(102)

$\log( \frac{3x^3 \cdot \sqrt[4]{4-x}}{4(x+4)^2} )$

$\log(x)$ is $\log_{10}(x)$ implied

Apply quotient rule:

$(\log_{10}3x^3 \cdot \sqrt[4]{4-x}) - (\log_{10}4(x+4)^2)$

Apply product rule:

$\color{blue}{\log_{10}3x^3} + \color{red}{\log_{10} \sqrt[4]{4-x}} -\color{green}{\log_{10}4(x+4)^2}$

$=\color{blue}{\log3x^3} + \color{red}{\log \sqrt[4]{4-x}} -\color{green}{\log4(x+4)^2}$

Apply product rule and split $3x^3$:

$\color{blue}{\log3x^3}\\=\log3 + \log x^3$
$=\log3 + 3\log x$ (Apply the power rule)

Apply the radical to exponent rule and log power rule:

$\color{red}{\log \sqrt[4]{4-x}}$
$=\log(4-x)^{1/4}$
$=\frac{1}{4}\log(4-x)$

Expand the last segment and apply power rule:

$\color{green}{\log4(x+4)^2}$
$=\log4 + \log(x+4)^2$
$=\log4 + 2\log(x+4)$

Put segments back (I think the answer on the review sheet is wrong. I think the last log is positive and not negative like it shows on the sheet):

$=\log3 + 3\log x + \frac{1}{4}\log(4-x) - \log4 + 2\log(x+4)$

(103)

$\log_3(x+3)-\log_3(x+6)$

Use the quotient rule to condense:

$=\log_3(\frac{x+3}{x+6})$

(104)

$2\ln x - \frac{1}{4} \ln y$

Use the power rule:

$=\ln x^2 - \ln \sqrt[4]{y}$

Use the quotient rule:

$=\ln \frac{x^2}{\sqrt[4]{y}}$

(105)

$3 \log_42+\frac{1}{6}\log_4(r-3)-\frac{1}{2}\log_4r$

Apply the power rule:

$\log_42^3+\log_4(r-3)^\frac{1}{6}-\log_4r^\frac{1}{2}$

$2^3=8$

Exponent to radical:

$\log_48+\log_4\sqrt[6]{r-3}-\log_4\sqrt{r}$

Product rule:

$\log_48\sqrt[6]{r-3}-\log_4\sqrt{r}$

Quotient rule:

$\log_4(\frac{8\sqrt[6]{r-3}}{\sqrt{r}})$

(110)

$8^{5x}=4.4$

Take the natural log $\ln$ of both sides:

$\ln 8^{5x}=\ln 4.4$

Apply the power rule:

$5x \ln 8=\ln 4.4$

Apply divide both side by $5 \ln 8$:

$x =\frac{\ln 4.4}{5\ln 8}$

(111)

$e^{3x}=8$

Take the natural log of both sides:

$\ln e^{3x}=\ln 8$

Apply the power rule:

$3x \ln e=\ln 8$

The natural log of $e$ is always $1$:

$\ln e = 1$

$3x = \ln 8$

Divide both sides by 3:

$x=\frac{\ln8}{3}$

(116)

$4^{x+4}=5^{2x+5}$

$\ln$ of both sides:

$\ln4^{x+4}=\ln5^{2x+5}$

Apply power rule:

$(x+4)\ln4=(2x+5)\ln5$

Expand $(x+4)\ln4$:

$(x+4)\ln4\\=(x+4)\ln2^2\\=2\ln2(x+4)$

$=2x\ln2+8\ln2$

Expand $(2x+5)\ln5$:

$(2x+5)\ln5$
$=2x\ln5+5\ln5$

$2x\ln2+8\ln2=2x\ln5+5\ln5$

Subtract $8\ln2$ from both sides:

$2x\ln2=2x\ln5+5\ln5-8\ln2$

Subtract $2x\ln5$ from both sides:

$2x\ln2-2x\ln5=5\ln5-8\ln2$

Factor to isolate $x$:

$x(2\ln2-2\ln5)=5\ln5-8\ln2$

Divide both sides by $(2\ln2-2\ln5)$:

$x=\frac{5\ln5-8\ln2}{2\ln2-2\ln5}$

(117)

$\log_2x=5$

$2^5=x$

$x=32$

(118)

$\log_6(x+2)=-1$

$6^{-1}=x+2$

$\frac{1}{6}=x+2$

Multiply both sides by 6:

$1=6x+12$

Subtract $1$ from both sides:

$0=6x+12-1$
$0=6x+11$

Subtract $11$ from both sides:

$-11=6x$
$6x=-11$

Divide both sides by $6$:

$x=-\frac{11}{6}$

(122)

$\ln x + \ln(x-1)=\ln 56$

Use the product rule:

$\ln(x(x-1))=\ln 56$

Cancel $\ln$ from both sides:

$x(x-1)=56$

$x^2-x=56$

Factor:

$x^2-x-56=0$
$(x-8)(x+7)=0$

$x=8, x=-7$

Replace $x$ in original equation and test $x=8$:

$\ln 8 + \ln(8-1)$

$\ln 8 + \ln7$

$\ln(8\cdot7)=\ln56$

Replace $x$ in original equation and test $x=-7$:

$\ln 8 + \ln(-7-1)$

$\ln 8 + \ln(-8)$

$\ln(8\cdot-8)\ne\ln 56$

$x\ne-7$
$x=8$

(124)

$A=P(1+\frac{r}{n})^{nt}$

$P=2800$
$A=2800\times2=5600$
$r=9/100=0.09$
$n=4$

$5600=2800(1+\frac{0.09}{4})^{4t}$

$5600=2800(1.0225)^{4t}$

Divide both sides by $2800$:

$\frac{5600}{2800}=(1.0225)^{4t}$

Simplify and take the natural log of both sides:

$5600/2800=2$
$\ln2=\ln(1.0225)^{4t}$

Apply power rule:

$\ln 2= 4t\ln(1.0225)$

Divide both sides by $4\ln(1.0225)$:

$\frac{\ln 2}{4\ln(1.0225)} = t$

Use calculator to solve for $t$:

$(ln(2)) ÷ (4 ln(1.0225))\\=7.78795742838\\=7.8$

(125)

$A = 173e^{0.046t}$

$A=329$

$329=173e^{0.046t}$

Divide sides by $173$:

$\frac{329}{173}=e^{0.046t}$

Take $\ln$ of both sides:

$\ln \frac{329}{173}= \ln e^{0.046t}$

Apply power rule:

$\ln \frac{329}{173}= 0.046t \ln e$

$\ln e$ is always equal to $1$:

$\ln \frac{329}{173}= 0.046t (1)$

$\ln \frac{329}{173}= 0.046t$

Divide both sides by $0.046$:

$\frac{\ln \frac{329}{173}}{0.046}= t$

Use calculator to solve $t$:

$((ln(329 ÷ 173)) ÷ 0.046)\\=13.9731773102\\=14$

$1998+14=2012$

(127)

$A=P(1+\frac{r}{n})^{nt}$

$A=3000$
$P=2000$
$r=11/100=0.11$
$n=12$

$3000=2000(1+\frac{0.11}{12})^{12t}$

Divide by 2000:

$\frac{3000}{2000}=(1+\frac{0.11}{12})^{12t}$

$\frac{3}{2}=(1+\frac{0.11}{12})^{12t}$

Take $\ln$:

$\ln 1.5=\ln(1+\frac{0.11}{12})^{12t}$

Apply power rule:

$\ln 1.5=12t\ln(1+\frac{0.11}{12})$

Divide by $12\ln(1+\frac{0.11}{12})$:

$\frac{ \ln 1.5 }{ 12\ln(1+\frac{0.11}{12}) }=t$

Use calculator:

$\ln(1.5)$
$=0.4054651081$
$\div (12 \ln((1 + (0.11 ÷ 12))))$
$=3.70291512361\\=3.703$

Be sure to use brackets when entering into calculator!!

$A=Pe^{rt}$

$A=3000$
$P=2000$
$r=11/100=0.11$

$3000=2000e^{0.11t}$

Divide sides by 2000:

$\frac{3000}{2000}=e^{0.11t}$

$\frac{3}{2}=e^{0.11t}$

Take $\ln$:

$\ln\frac{3}{2}=\ln e^{0.11t}$

Apply power rule:

$\ln\frac{3}{2}=0.11t \ln e$

$\ln e=1$

$\ln\frac{3}{2}=0.11t$

Divide by 0.11:

$\frac{\ln\frac{3}{2}}{0.11}=t$

Use calculator:

$\ln(3 ÷ 2)\\=0.4054651081\\\div0.11\\=3.68604643735\\=3.686$

(128)

$y = y_0e^{0.012t}$

$y_0=10,184,000$

$t=2007-1999=8$

$y=10,184,000e^{0.012(8)}$

$y=10,184,000e^{0.096}$

Use calculator to solve $y$:

$10184000e^{0.096}\\=11,210,130\\=11,210,000$

(129)

$A = A_0e^{-0.0077x}$

$A_0 = 400$
$x=80$

$A = 400e^{-0.0077(80)}$

$A = 400e^{-0.616}$

$A=216$

(130)

$A =22e^{kt}$

$A=30$
$t=1993-1984=9$

$30 =22e^{9k}$

Divide both sides by 22:

$\frac{30}{22} =e^{9k}$

$\frac{15}{11} =e^{9k}$

Take natural log of both sides:

$\ln \frac{15}{11} =\ln e^{9k}$

Use power rule:

$\ln \frac{15}{11} =9k (1)$
($\ln e = 1$)

Divide by 9:

$\frac{\ln \frac{15}{11}}{9} =k$

Use calc:

$(\ln((15 ÷ 11))) ÷ 9\\=0.0344616587\\=0.034$

(131)

$A = 5100e^{0.057t}$

$A_0=5,100$

Answer: $\$5,100$

(132)

$A = 1500e^{0.045t}$

$r=0.045\times100=4.5\%$

(133)

$A = A_0e^{-0.01155x}$

$A_0=700$
$x=170$

$A = 700e^{-0.01155(170)}$

$A=700e^{(-0.01155 × 170)}\\=98.25639474\\=98$

(134)

$\color{blue}{A/2 = Ae^{kx}}$

$A=30$
$x=710$

$30/2=30e^{710k}$

$15=30e^{710k}$

Divide by 30:

$15/30=e^{710k}$

Take natural log:

$\ln \frac{15}{30}=\ln e^{710k}$

Apply power rule:

$\ln 0.5=710k$

Divide by 710:

$\frac{ \ln 0.5 }{710}=k$

Use calc:

$(\ln0.5)\div710\\=-0.00097626363$

$k=-0.00097626363$

$\color{blue}{A = A_0e^{kx}}$
$A_0=30$t=300k=-0.00097626363$

$A = 30e^{-0.00097626363(300)}$

$A = 30e^{-0.292879089}$

Use calc:

$A=22.3833706545\\=22.383$

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