`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` geeft:

`\ ^(0,5)log(x+5 )`	`=`	`text(-)1`
`x+5`	`=`	`(1/2) ^(text(-)1)`
`x+5`	`=`	`2`
`x`	`=`	`text(-)3`

`text(D)_(f) = langle text(-)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 lt x le 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`

`log((2x)/(x-1))`	`=`	`2`
`(2x)/(x-1)`	`=`	`10^2`
`(2x)/(x-1)`	`=`	`100`
`2x`	`=`	`100x - 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` .

`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 lt x le 4/11`

`4/11 lt x lt 4`

`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*(log(2)+log((pi*f_1*m)/(rho*c))) - 5 = 20*log(2) + 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`
Dus 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
Dus 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.

`X = 10*log((P_u)/(P_i)) = 10*log(20/(10^(text(-)3))) ~~ 43` dB.

`46 = 10*log((P_u)/(10^(text(-)3))) rArr 4,6 = log((P_u)/(10^(text(-)3))) rArr (P_u)/(10^(text(-)3)) = 10^(4,6) rArr P_u = 10^(1,6) ~~ 39,8` Watt.

`X = 10*log((P_u)/(P_i)) = 10*log((U_u^2)/(U_i^2)) = 10*log(((U_u)/(U_i))^2) = 10*2*log((U_u)/(U_i)) = 20*log((U_u)/(U_i))` .

`text(-)20 = 20*log((U_u)/1) rArr text(-)1 = log(U_u) rArr U_u = 10^(text(-)1) = 0,1` Volt (verzwakker).

`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 gt 9841,5`

`x=5` .

`x = 2`

`2 le x lt 6`

`G = 2,4*10^(0,008L)`

`G ≈ 26,3` kg.