Zie de uitleg. Probeer dit wel eerst zelf op te lossen!
`4*log(x) = 1 - log(x)` geeft `5*log(x)=1` en `log(x)=0,2` zodat `x=10^(0,2)~~1,58` .
Grafiek: `x gt 1,58` .
`\ ^2log(x) - 1 = 2* \ ^2log(x)` geeft `\ ^2log(x)=text(-)1` zodat `x=2^(text(-)1)=0,5` .
Grafiek: `0 lt x lt 0,5` .
`text(D)_(f)=langle 1, →rangle` en `text(B)_(f)=ℝ`
De verticale asymptoot is `x=1` .
Los eerst `3 *\ ^2log(x-1)+16 = 38` op:
`3 *\ ^2log(x-1)+16` |
`=` |
`38` |
|
`\ ^2log(x-1)` |
`=` |
`22/3` |
|
`x-1` |
`=` |
`2^ (22 /3) ≈ 161,27` |
|
`x` |
`~~` |
`162,27` |
Grafiek: `1 lt x lt 162,27`
`1 +4 *\ ^(0,5)log(x+5 )=text(-)3`
`\ ^(0,5)log(x+5 )` | `=` | `text(-)1` | |
`x+5` | `=` | `(1/2) ^(text(-)1)` | |
`x+5` | `=` | `2` | |
`x` | `=` | `text(-)3` |
`text(D)_(f)=langletext(-)5 ,→rangle` en `text(B)_(f)=ℝ`
De verticale asymptoot is `x=text(-)5` .
`f(x)=text(-)3` voor `x=text(-)3`
Voer in: `y_1=1+4*\ ^(0,5)log(x+5)` en `y_2=text(-)3`
Venster bijvoorbeeld: `[text(-)5, 5]xx[text(-)5, 5]`
Bekijk de grafiek. De uitkomst is: `text(-)5 < x≤text(-)3`
De verticale asymptoot van de grafiek van `f` is `x=0` .
De verticale asymptoot van de grafiek van `g` is `x=2` .
`text(D)_(f)=langle 0, →rangle` en `text(D)_(g)=langle ←, 2 rangle`
`x=2-x` geeft `x=1` .
`1 < x < 2`
`\ ^6log(x)+\ ^6log(x-1)` | `=` | `1` | |
`\ ^6log(x(x-1))` | `=` | `1` | |
`(x-3)(x+2)` | `=` | `0` | |
`x` | `=` | `text(-)2 vv x=3` |
Alleen `x=3` voldoet.
`log( (2 x) / (x-1) )` | `=` | `2` | |
`(2 x) / (x-1)` | `=` | `10^2` | |
`(2x)/(x-1)` | `=` | `100` | |
`2 x` | `=` | `100 x-100` | |
`x` | `=` | `100/98=50/49` |
`\ ^3log(x-2)` | `=` | `1 +5 *\ ^3log(2)` | |
`\ ^3log(x-2)` | `=` | `\ ^3log(3)+\ ^3log(2^5)` | |
`\ ^3log(x-2)` | `=` | `\ ^3log(3)+\ ^3log(32)` | |
`\ ^3log(x-2)` | `=` | `\ ^3log(96)` | |
`x-2` | `=` | `96` | |
`x` | `=` | `98` |
Kijk als je herleiding klaar is of je hetzelfde hebt gedaan als in het voorbeeld en of je dezelfde uitkomst hebt.
`p~~0,00002 *1,12^20 ~~ 19*10^(text(-)3)` Pa.
`L=20 *log((0,001)/ (0,00002) ) ~~ 34` dB.
`h = text(-)19log(p) + 57` geeft `log(p) = (h-57)/(text(-)19) ~~ text(-)0,0526h + 3` .
En dit betekent `p ~~ 10^(text(-)0,0526h + 3) = 10^3 * 10^(text(-)0,0526h) = 1000 * 0,886^h` .
`p ~~ 1000 * 10^(text(-)0,0526h)` .
`x=text(-)4`
`text(D)_(f)=langle text(-)4, →rangle`
`text(B)_(f)=ℝ`
`1 -3 *log(x+4 )` | `=` | `0` | |
`3*log(x+4)` | `=` | `1` | |
`log(x+4)` | `=` | `1/3` | |
`x+4` | `=` | `10^(1/3) ~~ 2,15` | |
`x` | `~~` | `text(-)1,85` |
Houd rekening met het domein. De oplossing van de ongelijkheid is `text(-)4 < x < text(-)1,85` .
`\ ^5log(x)` |
`=` |
`3 +4 *\ ^5log(x)` |
|
`text(-)3*\ ^5log(x)` |
`=` |
`3` |
|
`\ ^5log(x)` |
`=` |
`text(-)1` |
|
`x` |
`=` |
`5^(text(-)1) = 0,2` |
`m = 2/3 log(E/2) - 3` geeft `log(E/2) = 3/2*(m+3) = 1,5m + 4,5` , zodat `E = 2*10^(1,5m + 4,5) = 2*10^(4,5)*10^(1,5m) ~~ 63245*10^(1,5m)` .
`E ~~ 63245*10^(1,5*5,2) ~~ 4,0*10^12` .
Dus ongeveer `4,0` TJ (TeraJoule).
`text(D)_(f)=langle 0, →rangle`
`text(B)_(f)=ℝ`
De verticale asymptoot van `f` is `x=0` .
`text(D)_(g)=langle←, 4 rangle`
`text(B)_(g)=ℝ`
De verticale asymptoot van `g` is `x=4` .
`log(x)` | `=` | `text(-)1 + log(4-x)` | |
`log(x) - log(4-x)` | `=` | `text(-)1` | |
`log(x/(4-x))` | `=` | `text(-)1` | |
`x/(4-x)` | `=` | `10^(text(-)1) = 0,1` | |
`x` | `=` | `0,4-0,1x` | |
`1,1x` | `=` | `0,4` | |
`x` | `=` | `(0,4)/(1,1) = 4/11` |
`0 < x≤4/11`
`4/11 < x < 4`
`D ~~ 5,62* 10^(0,25k) - 10`
`R=20*log((pi*125*400)/(1,3*340))-5=46` dB
Stel je gaat uit van een frequentie
`f_1 rArr R_1=20*log((pi*f_1*m)/(rho*c))-5`
.
Verdubbeling:
`f_2=2*f_1 rArr`
`R_2=20*log((pi*f_2*m)/(rho*c))-5=20*log((pi*2*f_1*m)/(rho*c))-5=20*log(2*(pi*f_1*m)/(rho*c))-5=`
`20*[log2+log((pi*f_1*m)/(rho*c))]-5=20*log2+20*log((pi*f_1*m)/(rho*c))-5=`
`20*10^(0,3)+20*log((pi*f_1*m)/(rho*c))-5=6+R_1 rArr`
Als
`f_2=2*f_1 rArr R_2=R_1+6`
dB.
Alternatieve oplossingsmethode:
`R=20*log((pi*f*m)/(rho*c))-5=20*log(2,84*f)-5`
`R=46 rArr 46=20*log(2,84*f)-5 rArr log(2,84*f)=2,55 rArr f=10^(2,55)/(2,84)~~124,9`
Hz
`R=52 rArr 52=20*log(2,84*f)-5 rArr log(2,84*f)=2,85 rArr f=10^(2,85)/(2,84)~~249,3`
Hz
`rArr`
De frequentie is verdubbeld.
`R=50 rArr 50=20*log((pi*f*m)/(rho*c))-5=20*log(2,84*f)-5 rArr f=204`
Hz.
Voor geluid met een frequentie lager dan
`204`
Hz isoleert de muur goed genoeg.
Stel
`(pi*f*m)/(rho*c)=x`
dan
`60=20*log(x)-5 rArr 20*log(x)=65 rArr`
`log(x)=(65)/(20) rArr x=10^((65)/(20))~~1778,3 rArr`
`(pi*f*m)/(rho*c)=1778,3 rArr m=(1778,3*rho*c)/(pi*f)=500,4`
kg/m2.
`d=m/rho~~0,208`
m
`=208`
mm.
`R_r = 20xx log((0,120*2000)/(2*0,100*1850) + (0,100*1850)/(2*0,120*2000)) ~~ 0,29` dB
`0,36 = 20*log((m_1)/(370) + (185)/(2m_1))` geeft `(0,36)/20 = 0,18 = log((2m_1^2 + 370*185)/(370*2*m_1))` .
Hieruit volgt `2m_1^2 + 68450 = 740m_1 * 10^(0,18) ~~ 771,302m_1` en dus `2m_1^2 - 771,302m_1 + 88450 = 0` .
Met de abc-formule vind je `m_1≈138,4` of `m_1≈247,2` kg/m2.
De breedte van de steen is `≈(138,4)/2000=0,069` m `=69` mm of `≈(247,2)/2000=0,124` m `=124` mm.
De tweede oplossing is het meest waarschijnlijk.
`x~~5,06` .
`x~~17,78` .
`x=sqrt(1/2)~~0,71`
`text(D)_(f)=langle 0 ,→rangle` en `text(B)_(f)=ℝ` .
Verticale asymptoot van `f` : `x=0` .
`text(D)_(g)=langle←,6 rangle` en `text(B)_(g)=ℝ` .
Verticale asymptoot van `g` : `x=6` .
`x=1/18` .
`x>9841,5`
`x=5` .
`x=2`
`2 ≤x < 6`
`G = 2,4*10^(0,008L)`
`G ≈ 26,3` kg.