Math / Chem

Math

Chem

Chem 7.31 - 7.45

HW 4.4 Q3

$6^{3x-13}=36$

$6^{3x-13}=6^2$

$3x-13=2\\3x=2+13\\3x=15\\\color{blue}{x=5}$

HW 4.4 Q4

$6^{x-1}=\frac{1}{7,776}$

$6^{x-1}=\frac{1}{6^5}$

$6^{x-1}=6^{-5}$

$x-1=-5\\x=-5+1\\\color{blue}{x=-4}$

HW 4.4 Q5

$\frac{x-3}{6^{10}}=\sqrt{6}$

$\frac{x-3}{6^{10}}=6^{\frac{1}{2}}$

$\frac{x-3}{10}=\frac{1}{2}$

$\frac{x-3}{10}=\frac{5}{10}$

$x-3=5\\x=5+3\\\color{blue}{x=8}$

HW 4.4 Q6

$e^{x+6}=\frac{1}{e}$

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

$x+6=-1\\x=-1-6\\=\color{blue}{-7}$

HW 4.4 Q7

$4e^x=27$

$e^x=27/4$

$\ln(\frac{27}{4})$

$=1.91$

HW 4.4 Q8

$2e^{2x}=610$

$e^{x2}=610/2$

$e^{x2}=305$

$2x \ln e=\ln 305$

$x =\frac{\ln 305}{2}$

$=2.86$

HW 4.4 Q9

$3^{(4x-2)}=5^{(x+2)}$

$(4x-2) \ln 3 = (x+2) \ln 5$

$4x \ln 3 - 2 \ln 3 = x \ln 5 + 2 \ln 5$

$4x \ln 3 - x \ln 5 = 2 \ln 3 + 2 \ln 5$

$x(4 \ln 3 - \ln 5) = 2 \ln 3 + 2 \ln 5$

$x = \frac{2 \ln 3 + 2 \ln 5}{4 \ln 3 - \ln 5}\\\approx1.94$

HW 4.4 Q10

$e^{2x}-14e^x+13=0$

$(e^x-13)(e^x-1)$

$e^x-13=0\\e^x=13\\\ln e^x=\ln 13\\x=\color{blue}{\ln 13\\\approx2.56}$

$e^x-1=0\\e^x=1\\\ln e^x=\ln 1\\x=\color{blue}{0}$

HW 4.4 Q11

$2^{2x}+2^x-90=0$

$u=2^x$

$u^{2}+u-90=0$

$(u+10)(u-9)$

$2^x+10 = 0$
$2^x-9=0$

$\ln 2^x = \ln 9$
$x \ln 2 = \ln 9$

$x = \frac{\ln 9}{\ln 2}$

$=3.17$

HW 4.4 Q12

$\log_5x=2$

$5^2=x$

$x=25$

HW 4.4 Q13

$\ln x=10$

$\log_ex=10$

$e^{10}=x$

$e^{10} \approx 22026.47$

HW 4.4 Q14

$\ln \sqrt{x+4}=2$

$e^2=\sqrt{x+4}$

$e^4=x+4$

$x=e^4-4$
$\approx 50.60$

HW 4.4 Q15

$\log_4(x+13)-\log_4(x-2)=2$

Quotient Rule

$\log_bM=c$

$b^c=M$

$\log_b(\frac{M}{N}) = \log_bM-\log_bN$

(If $b$ , $M$ , and $N$ are positive real numbers and $b\ne1$ )

$\log_4(\frac{x+13}{x-2}) = 2$

$4^2=\frac{x+13}{x-2}$

$16 = \frac{x+13}{x-2}$

Multiply both side by $x-2$

$16(x-2) = x+13$

$16x-32=x+13$

$16x-x=13+32\\15x=45\\x=3$

HW 4.4 Q16

$\log _6 (x-6) + \log _6 (x-5) - \log _6x = 1$

$\log _6 ((x-6)(x-5)) - \log _6x = 1$

$\log _6 (x^2-5x-6x+30) - \log _6x = 1$

$\log _6 (x^2-11x+30) - \log _6x = 1$

$\log _6 (\frac{x^2-11x+30}{x}) = 1$

$\frac{x^2-11x+30}{x} = 6^1$

$\frac{x^2-11x+30}{x} = 6$

$x^2-11x+30 = 6x$

$x^2-11x-6x+30 = 0$

$x^2-17x+30 = 0$

Factor:

$( x - 2 ) ( x - 15 ) = 0$

$x=2, x=15$

$\log _6 (x-6) + \log _6 (x-5) - \log _6x = 1$

$\log _6 (2-6) + \log _6 (2-5) - \log _62 = 1$

$\log _6 (-4) + \log _6 (-3) - \log _62 = 1$

Negative Logs: $x\ne2$

Solution set: $\brace15$

HW 4.4 Q17

$\log ( 7x-2 ) = \log ( x+3 ) + \log8$

$\log ( 7x-2 ) = \log 8( x+3 )$

$\log ( 7x-2 ) = \log ( 8x+24 )$

$\log_bM=\log_bN$
$M=N$

$7x-2=8x+24$
$7x-2-8x-24=0$
$( 8x - 7x ) + ( 2+24 ) =0$
$x+26=0$
$x=-26$

HW 4.4 Q18

$2\log x=\log36$

$\log x^2 = \log36$

$x^2=36$

$x=6$

HW 4.4 Q19

$\ln (x-2) - \ln (x+9) = \ln (x-1) - \ln ( x+14 )$

$\ln ( \frac{x-2}{x+9} ) = \ln ( \frac{x-1}{x+14} )$

$\frac{x-2}{x+9} = \frac{x-1}{x+14}$

LCD is $(x+9)(x+14)$ . Multiply both sides.

$(x-2)(x+14) = (x-1)(x+9)$

$x^2+14x-2x-28=x^2+9x-x-9$

$x^2 +12x -28 = x^2 + 8x - 9$

$x^2+12x-28 - x^2 -8x + 9=0$

$4x+19=0$
$x=19/4

HW 4.4 Q21

$A=24.2e^{0.0778t}$

$t=0$
$A=24.2e^{0.0778(0)}$
$A = 24.2$

$A=31.4$

$31.4 = 24.2e^{0.0778t}$

$\frac{31.4}{24.2} = e^{0.0778t}$

$\ln ( \frac{31.4}{24.2} ) = \ln e^{0.0778t}$

$\ln ( \frac{31.4}{24.2} ) = 0.0778t \ln e$

$\ln ( \frac{31.4}{24.2} ) = 0.0778t$

$\frac{ \ln ( \frac{31.4}{24.2} ) }{ 0.0778 } = t$

$\frac{ 0.26045525975 } { 0.0778 } = t$

$t + 2000 = 2003$

HW 4.4 Q22

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

$A=16,000$
$n=4$
$r=0.065$
$P=12,000$

$16000=12000(1+\frac{0.065}{4})^{4t}$

$16000=12000(1.01625)^{4t}$

$(1.01625)^{4t} = \frac{16000}{12000}$

$(1.01625)^{4t} = \frac{4}{3}$

$\ln (1.01625)^{4t} = \ln \frac{4}{3}$

$t = \frac{ \ln \frac{4}{3} } { 4 \ln (1.01625) }$

$\approx 4.5$

HW 4.4 Q23

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

$A=1,700$
$n=180$
$r=0.306$
$P=500$

$1700=500(1+\frac{0.306}{180})^{180t}$

$1700=500(1.0017)^{180t}$

$(1.0017)^{180t} = \frac{1700}{500}$

$(1.0017)^{180t} = \frac{17}{5}$

$\ln (1.0017)^{180t} = \ln \frac{17}{5}$

$t = \frac{ \ln \frac{17}{5} } { 180 \ln (1.0017) }$

$\approx 4.0$

HW 4.4 Q24

$A=Pe^{rt}$

$A=18,000$
$P=9,000$
$r=0.1$

$18000=9000e^{0.1t}$

$\frac{18000}{9000}=e^{0.1t}$

$e^{0.1t}=2$

$\ln e^{0.1t}= \ln 2$

$0.1t = \ln 2$

$t = \frac{\ln 2}{0.1}$

$\approx6.93$

HW 4.4 Q25