Results 1 to 4 of 4

Math Help - More logarithms to check (With Work Part 2 of 4)

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    113

    More logarithms to check (With Work Part 2 of 4)

    Please check the following logarithms and let me know if they are correct. If they are not, please point out the mistake I am making and what to fix You will be thanked!


    1. log4(x^2) - log4(4x - 4) = 0

    log4(x^2/4x-4) = 0
    log4(x^2/4(x)+4(-1)) = 0
    log4(x^2/4(x-1)) = 0
    4log4(x^2/4(x-1) = 4^0
    4log4(x^2/4(x-1) = 1
    x^2/4(x-1) = 1
    x^2/1 = 4(x-1)
    x^2 = 4(x-1)
    x^2 = (4(x) + 4(-1))
    x^2 = (4x -4)
    x^2 = 4x - 4
    x^2 - 4x = -4
    x^2 - 4x + 4 = 0
    (x - 2)(x - 2) = 0
    (x-2)^2 = 0
    x- 2 = 0
    x = 2


    2. log2(x) + log2(x + 2) = 3

    log2((x)(x + 2)) = 3
    log2(x(x + 2)) = 3
    2log2(x(x + 2)) = 2^3
    2log2(x(x + 2)) = 8
    x(x+2) = 8
    (x(x) + x(2)) = 8
    (x^2 + 2x) = 8
    x^2 + 2x = 8
    x^2 + 2x - 8 = 0
    (x + 4)(x - 2) = 0
    x + 4 = 0
    x - 2 = 0
    x = -4
    x = 2
    x = -4, 2


    3. 2^(2x + 2) = 64

    2^(2x + 2) = 2^6
    (2x + 2) = 6
    2x + 2 = 6
    2x = -2x + 6
    2x = 4
    2x/2 = 4/2
    x = 4/2
    x = 2


    4. 3^(x^2 + 2x) = 27

    3^(x^2 + 2x) = 3^3
    (x^2 + 2x) = 3
    x^2 + 2x = 3
    x^2 + 2x - 3 = 0
    (x + 3)(x-1) = 0
    x + 3 = 0
    x - 1 = 0
    x = -3
    x = 1
    x = -3, 1


    5. 4^x = 63

    ln(4^x) = ln(63)
    xln(4) = ln(63)
    xln(4)/ln(4) = ln(63)/ln(4)
    x = ln(63)/ln(4)


    6. 10^4x = 101

    ln(10^4x) = ln(101)
    4xln(10) = ln(101)
    4xln(10)/4ln(10) = ln(101)/4ln(10)
    x = ln(101)/4ln(10)[/LEFT]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by bobbyboy1111 View Post
    Please check the following logarithms and let me know if they are correct. If they are not, please point out the mistake I am making and what to fix You will be thanked!


    1. log4(x^2) - log4(4x - 4) = 0

    log4(x^2/4x-4) = 0
    log4(x^2/4(x)+4(-1)) = 0
    log4(x^2/4(x-1)) = 0
    4log4(x^2/4(x-1) = 4^0
    4^log4(x^2/4(x-1) = 1
    x^2/4(x-1) = 1
    x^2/1 = 4(x-1)
    x^2 = 4(x-1)
    x^2 = (4(x) + 4(-1))
    x^2 = (4x -4)
    x^2 = 4x - 4
    x^2 - 4x = -4
    x^2 - 4x + 4 = 0
    (x - 2)(x - 2) = 0
    (x-2)^2 = 0
    x- 2 = 0
    x = 2
    f(2) = 0 so 2 is the correct answer. You made a small typo which I've marked in red.


    2. log2(x) + log2(x + 2) = 3

    log2((x)(x + 2)) = 3
    log2(x(x + 2)) = 3
    (x(x + 2)) = 2^3
    (x(x + 2)) = 8
    x(x+2) = 8
    (x(x) + x(2)) = 8
    (x^2 + 2x) = 8
    x^2 + 2x = 8
    x^2 + 2x - 8 = 0
    (x + 4)(x - 2) = 0
    x + 4 = 0
    x - 2 = 0
    x = -4
    x = 2
    x = -4, 2
    x = 2 is a solution but x= -4 is not a solution. This is because you cannot take the log of a negative number or 0. In other words x > 0

    I've made two omissions at the start of lines three and 4 because they are not necessary.

    3. 2^(2x + 2) = 64

    2^(2x + 2) = 2^6
    (2x + 2) = 6
    2x + 2 = 6
    2x = -2 + 6
    2x = 4
    2x/2 = 4/2
    x = 4/2
    x = 2
    Correct apart from the x I removed in line 4 (you took away 2 not 2x so I'm guessing it was a typo)

    4. 3^(x^2 + 2x) = 27

    3^(x^2 + 2x) = 3^3
    (x^2 + 2x) = 3
    x^2 + 2x = 3
    x^2 + 2x - 3 = 0
    (x + 3)(x-1) = 0
    x + 3 = 0
    x - 1 = 0
    x = -3
    x = 1
    x = -3, 1
    Correct


    5. 4^x = 63

    ln(4^x) = ln(63)
    xln(4) = ln(63)
    xln(4)/ln(4) = ln(63)/ln(4)
    x = ln(63)/ln(4)
    Correct.

    6. 10^4x = 101

    ln(10^4x) = ln(101)
    4xln(10) = ln(101)
    4xln(10)/4ln(10) = ln(101)/4ln(10)
    x = ln(101)/4ln(10)
    Correct
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2009
    Posts
    113
    Quote Originally Posted by e^(i*pi) View Post
    f(2) = 0 so 2 is the correct answer. You made a small typo which I've marked in red.
    O.k.

    Quote Originally Posted by e^(i*pi) View Post
    x = 2 is a solution but x= -4 is not a solution. This is because you cannot take the log of a negative number or 0. In other words x > 0
    ????????So, should I keep x =-4, 2 in the answer and just write that the solution is x = 2 under it???????


    Quote Originally Posted by e^(i*pi) View Post
    I've made two omissions at the start of lines three and 4 because they are not necessary.

    Thanks for the fixed stuff too!
    Last edited by mr fantastic; August 9th 2009 at 03:57 AM. Reason: Closed the first quote. Added the quote tags for the second and third quotes.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by bobbyboy1111 View Post
    [snip]
    So, should I keep x =-4, 2 in the answer and just write that the solution is x = 2 under it?
    [snip]
    You write both values of x. Then you provide a reason why x = -4 is rejected as a solution to the original equation. Then you state x = 2 as your answer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: March 12th 2011, 04:05 AM
  2. Replies: 0
    Last Post: August 17th 2010, 08:29 AM
  3. Replies: 2
    Last Post: August 16th 2010, 07:09 AM
  4. Replies: 8
    Last Post: August 8th 2009, 05:01 AM
  5. Logarithms Part 2
    Posted in the Algebra Forum
    Replies: 5
    Last Post: October 28th 2007, 10:32 AM

Search Tags


/mathhelpforum @mathhelpforum