Results 1 to 9 of 9

Math Help - logs. is my work correct?

  1. #1
    Junior Member
    Joined
    May 2007
    Posts
    31

    logs. is my work correct?

    are these right?

    3^2x=75 ---> i got log75/2log3

    solve for x
    log(2)x = 5 ---> i got x = 32

    solve for x
    e^3x-2=5 ----> i got x ~ 0.6486

    solve for x
    2^2x=8^x-4

    2^2x=(2^3)x-4
    2^2x=2^3x-4
    2x+4=3x <--- as far as i could get
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    Hmm somebody check me here but I'm pretty confident.

    1) 3^2x=75 ---> i got log75/2log3
    I didn't get that.

    I'm assuming you mean: 3^{2x}\,=\,75

    Change to logarithmic form: log_3\,75\,=\,2x

    You can get the answer now.

    2) log(2)x = 5 ---> i got x = 32
    If you mean: log_{2}\,x\,=\,5 then yes.

    Someone else help on 3.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by trancefanatic View Post

    solve for x
    2^2x=8^x-4

    2^2x=(2^3)x-4
    2^2x=2^3x-4
    2x+4=3x <--- as far as i could get
    all were fine except for the above. i don't see how you got stuck at that last line though, it's simple algebra from there. to be able to solve logarithmic equations and not be able to solve that is like knowing how to run without knowing how to walk, but anyway, let's see where you made the mistake.

    first of all, is the -4 a part of the power of 8 or is it separate? i will do both, choose which is right. Please make sure to type your questions clearly next time.

    2^{2x} = 8^{x-4}

    \Rightarrow 2^{2x} = (2^3)^{x-4}

    \Rightarrow 2^{2x} = 2^{3x - 12}

    equating the powers we get:

    2x = 3x - 12

    \Rightarrow 0 = x - 12 .........subtracted 2x from both sides

    \Rightarrow x = 12



    Or, if the power of 8 was actually x, and the -4 was separate, the problem becomes a bit more interesting
    2^{2x} = 8^{x} - 4

    \Rightarrow 2^{2x} = 2^{3x} - 4

    Let y = 2^x, we get:

    y^2 = y^3 - 4

    \Rightarrow y^3 - y^2 - 4 = 0

    we see that y = 2 is a root, so y - 2 is a factor.
    Dividing the equations by y - 2 using long division (or synthetic division) we obtain:

    y^3 - y^2 - 4 = (y - 2)(y^2 + y + 2) = 0

    \Rightarrow y - 2 = 0 \mbox{ or } y^2 + y + 2 = 0

    only y - 2 = 0 has real roots, so

    y = 2

    But y = 2^x

    \Rightarrow 2^x = 2

    \Rightarrow x = 1 by equating the coefficients
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Jonboy View Post
    Hmm somebody check me here but I'm pretty confident.

    I didn't get that.

    I'm assuming you mean: 3^{2x}\,=\,75

    Change to logarithmic form: log_3\,75\,=\,2x
    You are also right. the difference is he used \log_{10}, it makes the answer a bit messier, but it's right nonetheless
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2007
    Posts
    31
    Quote Originally Posted by Jhevon View Post
    all were fine except for the above. i don't see how you got stuck at that last line though, it's simple algebra from there. to be able to solve logarithmic equations and not be able to solve that is like knowing how to run without knowing how to walk, but anyway, let's see where you made the mistake.
    i admit that i got lazy and didn't see the next steps. baby steps i know. thats just me makin it harder then it really is. but thank you.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by trancefanatic View Post
    i admit that i got lazy and didn't see the next steps. baby steps i know. thats just me makin it harder then it really is. but thank you.
    what was the actual power of 8? was it just x, or was it x - 4 ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    May 2007
    Posts
    31
    Quote Originally Posted by Jhevon View Post
    what was the actual power of 8? was it just x, or was it x - 4 ?
    the power was x-4.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by trancefanatic View Post
    the power was x-4.
    ok, so you want the first problem i did. do you see your mistake? you did not distribute the 3 correctly. you multiplied the x by 3 but not the - 4, this made your answer incorrect

    also you equatied the powers incorrectly
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    May 2007
    Posts
    31
    Quote Originally Posted by Jhevon View Post
    ok, so you want the first problem i did. do you see your mistake? you did not distribute the 3 correctly. you multiplied the x by 3 but not the - 4, this made your answer incorrect

    also you equatied the powers incorrectly
    yeah, i realized i needed to distribute it to the whole thing. thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Is this correct or should I use logs?
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: May 24th 2011, 03:30 AM
  2. Is my work correct?
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: March 19th 2010, 01:21 PM
  3. Is my work correct?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 13th 2009, 01:11 PM
  4. Is this work correct?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 26th 2009, 09:15 PM
  5. logs help i think its correct??
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 24th 2008, 08:37 AM

Search Tags


/mathhelpforum @mathhelpforum