Saturday, March 21, 2009

log set 3

  1. log [2] x^2 + log [2] (x - 1) = 1 + log[2] (5x + 4)
  2. log [2] (x (x - 1)) = 1
  3. log z = log (y + 2) - 2 log y, find z in terms of y
  4. 2 log x = 1 + log ((4x -15 )/2)
  5. log (2x + 5) = 1 - log x
  6. log (x -8) + log (9/5) = 1 + log (x/4)
  7. log (2 + 2y) = log 5 + log 3
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29 comments:

  1. wattlessbulbMarch 21, 2009 at 3:45 PM

    question 2
    log[2](x(x-1))=1

    in order to equate, the base on both sides have to be the same in order to cancel out the log:

    if log[x]x=1...in this case, log[2]2=1
    therefore log[2](x(x-1))=log[2]2

    cancelling log[2] on both sides leaves:
    x(x-1)=2
    x^2-x=2
    x^2-x-2=0

    factorizing x^2-x-2=0
    gives (x-2)(x+1)=0
    therefore x=2 or x=-1

    ReplyDelete
  2. tacobellMarch 22, 2009 at 4:49 AM

    watles bulb ur a bit confusing can u plz explain a little clearer

    ReplyDelete
  3. mysticwings*March 22, 2009 at 3:29 PM

    no 1

    we can change the 1 to logs so that the problem will be easier to solve, by applying the rule log[x]x = 1

    log[2] x^2 + log[2](x-1) = log[2]2 + log[2](5x+4)

    log[2] x^2(x-1) = log[2] 2(5x+4)

    now we can drop logs on both sides:

    x^2(x-1) = 2(5x+4)

    x^3 - x^2 = 10x + 8

    x^3 - x^2 - 10x -8 = 0

    now its just simple algebra and we can solve for x

    ReplyDelete
  4. mysticwings*March 22, 2009 at 3:46 PM

    no 5

    this question is just like no 1, the approach is the same

    we know that when log is by itself this implies that it is to base 10

    log(2x+5) = l - log x

    now we can change 1 into log form to make the problem easier:

    log(2x+5)= log 10 - log x

    log(2x+5)= log(10/x)

    now we can drop logs on both sides:

    2x+5 = 10/x

    from here its jus algebra

    (2x+5)x = 10

    2x^2 + 5x - 10 = 0

    now we can solve the quadratic and find x

    ReplyDelete
  5. mysticwings*March 22, 2009 at 4:21 PM

    we can do no 6 the exact same way as 1 and 5.anyone wants to give it a try???????

    ReplyDelete
  6. Chun-LiMarch 22, 2009 at 5:16 PM

    ok i will give no.6 a try but i dont think its d same exact approach so wherever i go wrong sum1 please correct me cuz it kinda confusing lol...ok here goes:

    log(x-8)+log(9/5)=1+log(x/4)
    ...so here we make the 1..log10 to simplify it

    log(x-8)+ log(9/5)=log10 + log(x/4)
    ...i think here we can use the rule
    log a + log b = log(ab)
    therefore we get:

    log[(x-8)(9/5)] = log [(10)(x/4)]

    since both log are base 10 we can cancel them
    so we get: (x-8)(9/5) = (10)(x/4)
    (x-8)(1.8) = (10)(x/4)
    ... 1.8x -14.4 = 10x/4
    ... 1.8x -14.4 = 5x/2
    ... 2(1.8x -14.4) = 5x
    ... 3.6x -28.8 = 5x
    ... 3.6x -5x -28.8 =0
    ... -1.4x = 28.8
    ... x = 28.8 / 1.4
    therefore x = 20.6
    ... x = 28.8 / 1.4

    ReplyDelete
  7. Chun-LiMarch 22, 2009 at 5:17 PM

    ignore the last line in d comment before...dont know where that appear from loll...the answer is
    x = 20.6

    ReplyDelete
  8. BladeMarch 23, 2009 at 1:53 AM

    i'll try seven

    log (2 + 2y) = log 5 + log 3

    using the rule
    log a + log b = log(ab)
    log(2+2y)=log(5*3)
    log(2+2y)=log15
    same bases so i can drop logs
    2+2y=15
    2y=15-2
    y=13/2
    y=6.5

    ReplyDelete
  9. Dill on!March 23, 2009 at 6:41 PM

    using the rule.. log [x]x = 1;

    log[2] x^2(x-1) = log[2] 2(5x+4)

    then we drop logs on both sides to get:

    x^3 - x^2 = 10x + 8

    x^3 - x^2 - 10x -8 = 0

    ReplyDelete
  10. ~*DJ HalfTime*~March 24, 2009 at 3:19 AM

    for 3. log z = log (y + 2) - 2 log y, find z in terms of y

    logz = log(y+2) - 2logy

    when no base is given, put base 10,
    log[10]z = log[10](y+2) - 2log[10]y
    log[10]z = log[10](y+2) - log[10](y^2)
    log[10]z = log[10](y+2/y^2)

    antilog,
    z = (y+2)/y^2
    z in terms of y.

    ReplyDelete
  11. DarkyMarch 31, 2009 at 12:59 PM

    log (2 + 2y) = log 5 + log 3

    put in base 10

    log [10] 2+ 2y = log [10] 5 + log [10] 3
    =0.6 +0.477

    = 0.477

    log[10] 2+ 2y = 0.477

    applying 2^3=8 log [2]8=3
    log [10] 2 + 2y = 0.477
    10 ^0.477 = 2 + 2y
    2.99 -2 = 2y
    0.999 /2 = y
    0.499= y

    ReplyDelete
  12. dutchessApril 2, 2009 at 5:40 PM

    Ques.7
    Darky i don't agree with you!
    log(2+2y)= log5+log3
    log(2+2y)=log5*3
    since loga + logb =log(ab)
    log(2+2y)=log15
    2+2y=15
    2y=15-2
    2y=13
    y=13/2
    y=6.5

    ReplyDelete
  13. Low riderApril 2, 2009 at 6:39 PM

    I agree with dutchess solution for question 7. Darky i think you were just trying to make the question more complicated for yourself.

    ReplyDelete
  14. wattlessbulbApril 5, 2009 at 6:53 PM

    Darky's attempt was not wrong or complicated...there was jus 1 mistake that made the answer wrong

    after the line:
    log [10] 2+ 2y = log [10] 5 + log [10] 3
    it's spose to be:


    log [10] 2+ 2y = log [10] 5 + log [10] 3
    log[10] 2+ 2y =0.0699+0.477

    log [10] 2+ 2y = 1.176

    applying 2^3=8 log [2]8=3
    log [10] 2 + 2y = 1.176
    10 ^1.176 = 2 + 2y
    14.9 -2 = 2y
    12.9 /2 = y
    6.5= y

    and there...the same ans as dutchess

    ReplyDelete
  15. Low riderApril 6, 2009 at 7:28 AM

    Well ok i now see the error sorry about that.

    ReplyDelete
  16. empressApril 6, 2009 at 12:20 PM

    #2.
    log [2] (x (x - 1)) = 1
    log [2] (x^2 - x) = 1
    using => 2^3 = 8 <=> log[2]8 = 3
    we get: 2^1 = x^2 - x
    2 = x^2 - x
    WHAT DO I DO NEXT?
    SOME1 PLZ HELP

    ReplyDelete
  17. empressApril 6, 2009 at 12:37 PM

    #7.
    log (2 + 2y) = log 5 + log 3
    using log[a]b + log[a]c = log[a](bc)
    log(2 +2 y)=log(5 * 3)
    log(2 + 2y)=log15
    now, we can divide by log[10]
    2 +2y =1 5
    2y = 15 - 2
    2y = 13
    y = 13/2
    y = 6.5

    ReplyDelete
  18. WeezyApril 6, 2009 at 1:28 PM

    question..&..amswer

    log[2]9-log[2]X=log[2]45
    log[2]9/X=log[2]45
    9/X=45

    ReplyDelete
  19. WeezyApril 6, 2009 at 1:41 PM

    question..with
    answer

    logx^2-1=log(1-2X)
    logX^2-log10=log(1-2X)
    logX^2/10=log(1-2X)
    X^2=10-20X
    0=X^2+20X-10

    ReplyDelete
  20. DarkyApril 6, 2009 at 2:35 PM

    log (2 + 2y) = log 5 + log 3
    put in base 10

    log [10] (2+ 2y) = log [10]5 + log [10}3

    ReplyDelete
  21. saeedApril 7, 2009 at 12:54 PM

    lg[8]x+2=2-lg[8]
    lg[8]x+2=-lg[8]2^2
    x+2=-2^2
    x=4-2=2

    2.lg[4]x=12--------->4^12=x
    x=16777216

    3. 6^x+2=21
    lg6^x+2=lg21
    x+2=lg21/lg6n9
    x+2lg6=lg21
    x=1.69-2=0.30

    4.e^3x=9
    lne^3x=ln9
    3xlne=ln9
    3x=ln9
    x=1/3ln9
    x=0.732

    5.3^x[2^2x]=7[5^x]
    [3*4]^x=7{5^x]
    12^x=7{5^x]
    7=[12/5]^x
    xlg[12/5]=lg7
    x=lg7/lg12/5
    x=2.22

    6.6^x+2=21---------->x+2=lg21
    =lg21/lg6=0.301

    7.lg[7]5*lg[5]9*lg[9]7=lg5/lg7 *lg9/lg5 *lg7/lg9
    =0.82*1.3*0.88=0.93

    8.lg[2]3+lg[2]5=lg[2][3*5]
    =lg[2]15
    =lg15/lg2=3.90

    9.lg[6]3+lg[6]2=lg[6][3*2]
    =lg[6]6
    =lg[6]/lg[6]=0.77/0.77=1

    10.lg[2]40+lg[2]0.1+lg[2]0,25=lg[2]40*0.01*0.25
    =lg[2]1
    =lg1/lg2=0

    11.lg[4]8-lg[4]2=lg[4][8/4]
    =lg[4]4=1

    12.lg[2]2sq.rt.2=lg[2][2*2^1/2]
    =lg[2]2^3/2=3/2

    13.lg[2]x^4-lg[2]x^3=lg[x4/x3]
    =lg[2]x

    14.lg[a]8/lg[a]4=lg[a]2^3/lg[a]2^2=
    3lg[a]^2/2lg[a]^2=3/2


    MY OWN QUESTIONS: PRODUCT;QUOTIENT RULE;POWERS;
    NATURAL LOGS.


    =lg

    ReplyDelete
  22. WeezyApril 10, 2009 at 5:43 PM

    logs`questions

    2^x = 128
    log 128/ log 2 = x
    7 = x

    ReplyDelete
  23. WeezyApril 10, 2009 at 5:43 PM

    logs`questions

    3^y = 1/9
    log(0.11)/ log 3 = y
    -2 = y

    ReplyDelete
  24. KenApril 12, 2009 at 11:06 AM

    7) log (2 + 2y) = log 5 + log 3

    asuming this question is to the base 10

    applying the rules

    log[a]b + log[a]c = log[a](bc)

    we get

    log(2 + 2y)=log15
    2 +2y =1 5
    2y = 15 - 2
    y = 13/2

    ReplyDelete
  25. KenApril 12, 2009 at 11:11 AM

    5.log (2x + 5) = 1 - log x

    log(2x+5)= log 10 - log x

    log(2x+5)= log(10/x)

    i stuck up to this point

    ReplyDelete
  26. KenApril 12, 2009 at 11:14 AM

    3ln2 + ln(x-1) = ln 24

    could some one please explain the steps involved in doing this question?

    ReplyDelete
  27. Chun-LiApril 24, 2009 at 11:51 AM

    ok i think i can give it a try but first there are some things u hada remember -
    * ln = log e so u work it jus like a log problem
    * ln a - ln b = ln a/b
    * always get rid of the coefficient first...so eg. 5 ln 3 = ln 3^5

    ReplyDelete
  28. Chun-LiApril 24, 2009 at 12:02 PM

    so d answer is -

    3ln2 + ln(x-1) = ln 24

    get rid of coefficient...

    ln 2^3 + ln(x-1) = ln24
    so, ln(x-1) = ln24 - ln2^3
    remember the rule above..

    ln(x-1) = ln 24/ 2^3

    since 2^3 = 8 we get..
    ln(x-1) = ln 24/8
    so, ln(x-1) = ln3
    at this point we can calculate ln3 in the calculator to get ln3= 1.098 lets say 1.1

    so, ln(x-1) = 1.1
    x-1 = ln inverse 1.1
    so, x-1 = 3.0

    x = 3+1....hence x=4

    ReplyDelete
  29. Chun-LiApril 24, 2009 at 12:13 PM

    i noticed two approaches to question 2...the person who ans. the first comment is one approach and the other approach is :


    log2 [x (x - 1)] = 1
    gives, log2 (x^2 - x) = 1
    using the rule --> 2^3 = 8 <-> log[2]8 = 3
    we get... 2^1 = x^2 - x or

    2 = x^2 - x
    hence,
    x^2-x=2 at this point we equate it to zero and solve
    so, x^2-x-2=0

    factorising x^2-x-2=0
    we get, (x-2)(x+1)=0
    hence x=2 or x=-1

    ReplyDelete

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