Logaritmen

Antwoorden bij de opgaven

    1. Voer in Yi=2^X en Y2=20000, stel het venster goed in (bijvoorbeeld `[0,20]xx[0,25000]`.
      Je vindt met behulp van de Intersect-optie het getal 14,29.
    2. Zie de Uitleg.
    3. Na 2`log(10000)` uur, dat is ongeveer 13,28 uur en dis 13 uur en 17 minuten.
    1. `3000 * 0,98^t = 2800`
    2. `0,98^t = 2800/3000 = 14/15`
    3. `t = `0,98`log(14/15) ~~ 3,415`
    1. `x = `2`log(7) ~~ 2,807`
    2. `x = `3`log(81) = 4`
    3. `x = `1/3`log(9) = -2`
    4. `x = `1/3`log(0,01) ~~ 4,192`
    5. `x = `10`log(1000000) = 6`
    6. `x = `0,001`log(0,1) = 1/3`
    7. `x = `0,001`log(100) = -2/3`
    1. 5`log(5^3) = 3`
    2. 5`log(5^(-2)) = -2`
    3. 4`log(4^3) = 3`
    4. 1/4`log((1/4)^(-3)) = -3`
    5. 1/3`log((1/3)^4) = 4`
    6. 2`log(2^(1/2)) = 1/2`
    1. `5^3 = 125` en `5^4 = 625`, dus `3 < `5`log(150) < 4`.
    2. `10^2 = 100` en `10^3 = 1000`, dus `2 < `10`log(758) < 3`.
    3. `2^5 = 32` en `2^6 = 64`, dus `5 < `2`log(60) < 6`.
    4. `2^(-3) = 1/8` en `2^(-2) = 1/4`, dus `-3 < ``log(17) < -2`.
    5. `(1/2)^(-4) = 16` en `(1/2)^(-5) = 32`, dus `-5 < `1/2`log(20) < -4`.
    6. `(1/3)^1 = 1/3` en `(1/3)^2 = 1/9`, dus `1 < `1/3`log(15) < 2`.
    1. 3,1
    2. 2,9
    3. 5,9
    4. -2,8
    5. -4,3
    6. 1,5
    1. `3^x = 600`, dus `x = `3`log(600) ~~ 5,8`.
    2. `1,7^t = 525`, dus `t = `1,7`log(525) ~~ 11,8`.
    3. `0,6^t = 30/572` , dus `t = `0,6log(30/572) ~~ 5,8`.
  8. Los op `10000 * 1,08^t = 15000`, oftewel `1,08^t = 1,5`, dus `t = `1,08`log(1,5) ~~ 5,268`.
    Dus na 5 jaar, 3 maanden en 7 dagen. Dat is april 2005.
    1. 4`log(4^3) = 3`
    2. 4`log(400) ~~ 4,3` (met de GR)
    3. 1/3`log(60) ~~ -3,7` (met de GR)
    4. 1/3`log((1/3)^(-4)) = -4`
    5. 1/3`log((1/3)^(4)) = 4`
    6. 1/10`log((1/10)^(-6)) = -6`
    1. `6^1 = 6` en `6^2 = 36`, dus `1 < `6`log(30) < 2`.
    2. `3^3 = 27` en `3^4 = 81`, dus `3 < `3`log(70) < 4`.
    3. `(1/2)^(-3) = 8` en `(1/2)^(-4) = 16`, dus `-4 < `1/2`log(10) < -3`.
    4. `(1/3)^4 ~~ 0,012` en `(1/3)^5 ~~ 0,004`, dus `4 < `1/3`log(0,001) < 5`.
    1. `5,026`
    2. `-8,399`
    3. `-3,597`
    4. `0,306`
    1. `10^x = 0,01`, dus `x = `10`log(0,01) = -2`.
    2. `2^x = 60`, dus `x = `2`log(60) ~~ 5,9`.
    3. `0,8^t = 0,5`, dus `t = `0,8log(0,5) ~~ 3,1`.
  13. `2^t = 3` geeft `t = `2`log(3) ~~ 1,58` uur.
    1. 2`log(2^(2,5)) = 2,5`
    2. 1/3`log((1/3)^(-3)) = -3`
    1. `2^9 = 512` en `2^(10) = 1024`, dus `9 < `2`log(513) < 10`.
      2`log(513) = 9,003`.
    2. `0,4^(-4) ~~ 29` en `0,4^(-3) = 15,625`, dus `-4 < `0,4`log(25) < -3`.
      0,4`log(25) = -3,513`.
    1. `4^x = 35/6`, dus `x = `4log(35/6) ~~ 1,27`.
    2. `1,08^t = 12/7`, dus `t = `1,08`log(12/7) ~~ 7,00`.
  17. `S(t) = 150 * 0,85^t = 10`, geeft `0,85^t = 1/15` en dus `t = `0,85log(1/15) ~~ 16,7`.
    Dus na 17 keer spoelen.