Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By Soroban

Math Help - If c=log_30 3, d=log_30 5. Then express log_30 8 in terms of c and d.

  1. #1
    Member
    Joined
    May 2014
    From
    India
    Posts
    78
    Thanks
    3

    Question If c=log_30 3, d=log_30 5. Then express log_30 8 in terms of c and d.

    log_{30} 3=c

    log_{30} 5=d

    Then log_{30} 8=

    A) 2(1-c-d)
    B) 3(1+c+d)
    C) 3(1+c-d)
    D) 3(1-c-d)

    I don't know how it could be expressed like this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,937
    Thanks
    785

    Re: If c=log_30 3, d=log_30 5. Then express log_30 8 in terms of c and d.

    Use rules of logarithms. Since it is multiple choice, there are only four possible answers. Check each one until you find one that works.

    \begin{align*}2(1-c-d) & = 2\left(1-\log_{30} 3 - \log_{30} 5\right) \\ & = 2\left(\log_{30} 30 - \log_{30} 3 - \log_{30} 5\right) \\ & = 2\left(\log_{30} \dfrac{30}{3} - \log_{30} 5\right) \\ & = 2\left(\log_{30} 10 - \log_{30} 5\right) \\ & = 2\left(\log_{30} \dfrac{10}{5}\right) \\ & = 2\log_{30} 2 = \log_{30}2^2 = \log_{30} 4 \neq \log_{30} 8\end{align*}

    So, the answer is not A. You try for B, C, and D.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    779

    Re: If c=log_30 3, d=log_30 5. Then express log_30 8 in terms of c and d.

    Hello, NameIsHidden!

    \text{Given: }\,\log_{30}3\,=\,c,\;\log_{30}5\,=\,d.\quad \text{Find }\,\log_{30}8

    . . (A)\;2(1-c-d) \qquad (B)\;3(1+c+d) \qquad (C)\;3(1+c-d) \qquad (D)\;3(1-c-d)

    Since \log(8) \,=\, \log(2^3) \,=\, 3\log(2),
    I would look for a way to get a "2", using 3 and 5 (and 30)
    with only multiplication and/or division.

    I see that: 2 \:=\:\frac{30}{3\cdot5}

    Then:- \log_{30}(2) \:=\:\log_{30}{\left(\frac{30}{3\cdot 5}\right) \:=\:\log_{30}(30) - \log_{30}(3\cdot5) \:=\:1 - (\log_{30}3 + \log_{30}5)

    . . . . . . =\;1 - \log_{30}3 - \log_{30}5\:=\:1 - c - d


    Therefore: . \log_{30}8 \;=\;3\log_{30}2 \;=\;3(1-c-d)\;\text{ . . . Answer (D)}
    Thanks from NameIsHidden
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2014
    From
    India
    Posts
    78
    Thanks
    3

    Re: If c=log_30 3, d=log_30 5. Then express log_30 8 in terms of c and d.

    Quote Originally Posted by Soroban View Post
    Hello, NameIsHidden!


    Since \log(8) \,=\, \log(2^3) \,=\, 3\log(2),
    I would look for a way to get a "2", using 3 and 5 (and 30)
    with only multiplication and/or division.

    I see that: 2 \:=\:\frac{30}{3\cdot5}

    Then:- \log_{30}(2) \:=\:\log_{30}{\left(\frac{30}{3\cdot 5}\right) \:=\:\log_{30}(30) - \log_{30}(3\cdot5) \:=\:1 - (\log_{30}3 + \log_{30}5)

    . . . . . . =\;1 - \log_{30}3 - \log_{30}5\:=\:1 - c - d


    Therefore: . \log_{30}8 \;=\;3\log_{30}2 \;=\;3(1-c-d)\;\text{ . . . Answer (D)}
    Thank you very much.

    Quote Originally Posted by SlipEternal View Post
    Use rules of logarithms. Since it is multiple choice, there are only four possible answers. Check each one until you find one that works.

    \begin{align*}2(1-c-d) & = 2\left(1-\log_{30} 3 - \log_{30} 5\right) \\ & = 2\left(\log_{30} 30 - \log_{30} 3 - \log_{30} 5\right) \\ & = 2\left(\log_{30} \dfrac{30}{3} - \log_{30} 5\right) \\ & = 2\left(\log_{30} 10 - \log_{30} 5\right) \\ & = 2\left(\log_{30} \dfrac{10}{5}\right) \\ & = 2\log_{30} 2 = \log_{30}2^2 = \log_{30} 4 \neq \log_{30} 8\end{align*}

    So, the answer is not A. You try for B, C, and D.
    But some questions consume a lot of time when doing like this checking each options. Anyway thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Express cos^2 (2x) in terms of cos 4x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 13th 2012, 07:00 AM
  2. [SOLVED] Given that y = (x-a)/(x+b), express in terms of a, b and y
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 19th 2011, 05:43 AM
  3. express in terms of x
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: August 24th 2009, 08:34 PM
  4. express y in terms of x
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 29th 2009, 10:02 PM
  5. express sin x/2 in terms of cos x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 15th 2007, 08:23 PM

Search Tags


/mathhelpforum @mathhelpforum