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Thread: Solving Logarithmic Equations Help. Any Aid Appreciated

  1. #1
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    Solving Logarithmic Equations Help. Any Aid Appreciated

    Hi everyone,

    This forum provides such excellent help that I am back.

    This time I am attempting to solve basic logarithmic equations. As I was working my way through the questions, a couple of them stumped me.

    Here they are:

    Solve for x:

    1) 5 logx (625) = 10


    2) -log3x -1 (1/32) = 5

    So, x, and 3x -1 are the respective bases of the logarithm in the above examples. In the first question, I started by dividing both sides by "5" to get logx 625 = 2. And then I simplified to get 4 logx 5 = 2. I divided both sides by 4 again and achieved: logx 5 = 2/4 = 1/2

    But from there I am not sure how to go. And I was on an utterly wrong tangent for question 2.

    Any help at all will be much appreciated!

    Thanks,

    Nathaniel
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  2. #2
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    $\displaystyle 5\log_x{(625)} = 10$

    $\displaystyle \log_x{(625)} = 2$

    $\displaystyle \log_x{(25^2)} = 2$

    $\displaystyle 2\log_x{(25)} = 2$

    $\displaystyle \log_x{(25)} = 1$

    $\displaystyle x^{\log_x{(25)}} = x^1$

    $\displaystyle 25 = x$.
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  3. #3
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    $\displaystyle -\log_{3x-1}\left(\frac{1}{32}\right) = 5$

    $\displaystyle \log_{3x - 1}\left[\left(\frac{1}{32}\right)^{-1}\right] = 5$

    $\displaystyle \log_{3x - 1}(32) = 5$

    $\displaystyle \log_{3x - 1}(2^5) = 5$

    $\displaystyle 5\log_{3x - 1}(2) = 5$

    $\displaystyle \log_{3x - 1}(2) = 1$

    $\displaystyle (3x - 1)^{\log_{3x - 1}(2)} = (3x - 1)^1$

    $\displaystyle 2 = 3x - 1$

    $\displaystyle 3 = 3x$

    $\displaystyle x = 1$.
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  4. #4
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    Thank you so much, Prove It!

    Also, two more things: with the first equation, when you reach logx 25 = 1, why is it that no further simplification takes place upon the 25 numeral? I know x = 25 is the correct answer, but I am just curious as to why no further simplification takes place.

    In the second equation, how come you don't divide 5 by -1 to yield -5?

    Cheers,

    Nathaniel
    Last edited by BinaryBoy; Sep 20th 2010 at 03:50 AM.
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  5. #5
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    Hello, BinaryBoy!

    If you use the identity: .$\displaystyle \log_b(N) \:=\:x \quad\Longleftrightarrow\quad b^x \:=\:N$

    . . you can avoid a lot of manipulation.



    $\displaystyle 1)\;5\log_x(625) \:=\: 10$

    We have: .$\displaystyle \log_x(625) \:=\:2$

    Then: .$\displaystyle x^2 \:=\:625 \quad\Rightarrow\quad x \:=\:\pm25$

    Since the base of a log must be positive: .$\displaystyle \boxed{x \,=\,25}$




    $\displaystyle 2)\; -\log_{3x -1}\!\left(\dfrac{1}{32}\right) \:=\: 5$

    We have: .$\displaystyle \log_{3x-1}\!\left(\dfrac{1}{32}\right) \:=\:-5 $

    Then: .$\displaystyle (3x-1)^{-5} \:=\:\dfrac{1}{32} \;=\;\dfrac{1}{2^5}$


    So we have: .$\displaystyle (3x-1)^{-5} \;=\;2^{-5} $


    Therefore: .$\displaystyle 3x-1 \:=\:2 \quad\Rightarrow\quad 3x \:=\:3 \quad\Rightarrow\quad \boxed{x \:=\:1}$

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  6. #6
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    Ah. Thanks a heap, Soroban.
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  7. #7
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    Quote Originally Posted by BinaryBoy View Post
    Thank you so much, Prove It!

    Also, two more things: with the first equation, when you reach logx 25 = 1, why is it that no further simplification takes place upon the 25 numeral? I know x = 25 is the correct answer, but I am just curious as to why no further simplification takes place.
    He could have said $\displaystyle log_x(25)= log_x(5^2)= 2log_x(5)= 1$ so that $\displaystyle log_x(5)= \frac{1}{2}$ but does that really simplify anything?

    $\displaystyle log_x(25)= 1$ leads to 25= x^1 and $\displaystyle log_x(5)= \frac{1}{2}$ leads to 5= x^{1/2} which then gives x= 25.

    The crucial point is: $\displaystyle log_a(b)= c$ if and only if $\displaystyle a^c= b$.

    In the second equation, how come you don't divide 5 by -1 to yield -5?
    If you divide both sides of $\displaystyle -log_{3x-1}(\frac{1}{32})= 5$ by -5, you get $\displaystyle log_{3x-1}(\frac{1}{32})= -5$ which (again changing $\displaystyle log_a(b)= c$ to $\displaystyle a^c= b$ becomes $\displaystyle (3x-1)^{-5}= \frac{1}{32}$ and probably the best thing to do would be to recognize that $\displaystyle (3x-1)^{-5}= \frac{1}{(3x-1)^5}= \frac{1}{32}$ so that $\displaystyle (3x-1)^5= 32= 2^5$.

    Instead, seeing that there was already a fraction inside the logarithm, Prove It chose to use the law of logarithms, taking the -1 inside the logarithm to "cancel" that fraction: $\displaystyle -log_{3x-1}\left(\frac{1}{32}\right)= log_{3x-1}\left(\left(\frac{1}{32}\right)^{-1}\right)= log_{3x-1}(32)$

    Once again, the crucial point for all these problems is that $\displaystyle log_a(b)= c$ is exactly the same as $\displaystyle a^c= b$.
    Cheers,

    Nathaniel
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