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Thread: logarithmic equation

  1. #1
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    logarithmic equation

    hi again,
    i have a question

    $ \ -6log_{3} (x-3) = -24$
    i rearrange into $ 3^{-24} = (x-3)^{-6} $

    but this gives me some numbers like


    $ 3.540706161*10^{-12}= \frac{1}{x^6-18x^5+135x^4-540x^3+1215x^2-1458x+729} $


    at this point im starting to doubt im doing this right .. did i missed something here ?

    on the other note, what are the ways to get x from this ^

    thanks in advance for help
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  2. #2
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    Re: logarithmic equation

    Yes, that is correct but it is not at all a good way of approaching this problem. Your equation is 3^{-24}= (x- 3)^{-6}.
    The first thing I would do is "invert" both sides: 3^{24}= (x- 6)^6. Then I would notice that 24= 6*4 so I can write this as (3^4)^6= (x- 3)^6 which reduces to either x- 3= 3^4= 81 or x- 3= -3^4= -81 since the power, 6, is even. If you allowing complex solutions there are another 4 complex roots.
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  3. #3
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    Re: logarithmic equation

    thank you , i totally forgot about the index
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    Re: logarithmic equation

    Work smart.

    $-\ 6 log_3(x - 3) = -\ 24 \implies log_3(x - 3) = 4 \implies x - 3 = 3^4 = 81 \implies x = 84.$

    Let's check.

    $-\ 6 log_3(84 - 3) = -\ 6 log_3(81) = -\ 6 log_3(3^4) = -\ 6 * 4 = -\ 24.$ It checks.

    You CAN do it your way.

    $-\ 6 log_3(x - 3) = -\ 24 \implies log_3\{(x - 3)^{-6}\} = -\ 24 \implies (x - 3)^{-6} = 3^{-24} \implies$

    $\dfrac{1}{(x - 3)^6} = \dfrac{1}{3^{24}} \implies \left ( \dfrac{1}{x - 3} \right )^6= \left ( \dfrac{1}{3} \right)^{24} = \left \{ \left (\dfrac{1}{3} \right )^4 \right \}^6 \implies$

    $\sqrt[6]{\left ( \dfrac{1}{(x - 3} \right )^2} = \sqrt[6]{ \left \{ \left (\dfrac{1}{3} \right )^4 \right \}^6} \implies \dfrac{1}{x - 3} = \left ( \dfrac{1}{3} \right )^ 4 = \dfrac{1}{81} \implies$

    $x - 3 = 81 \implies x = 84.$

    But why would you bother with that method?
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  5. #5
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    Re: logarithmic equation

    But x-3 can't equal -81 because you can't take the log of a negative number. So the only solutions will come from x-3 = 81.
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    Re: logarithmic equation

    Quote Originally Posted by AbYz View Post


    $ \ -6log_{3} (x-3) = -24$
    i rearrange into $ 3^{-24} = (x-3)^{-6} $

    ** This equation is wrong, because it does not follow. See below.
    No, those are not equivalent equations. Look at the argument, (x - 3).

    x must be greater than 3.

    Here is the standard solution to this (type of) problem:


    -6log_3(x - 3) \ = \ -24

    \dfrac{-6log_3(x - 3)}{-6} \ = \ \dfrac{-24}{-6}


    log_3(x - 3) \ = \ 4


    Change it to exponential form:

    3^4 \ = \ x - 3

    81 \ = \ x - 3

    84 \ = \ x

    or

    x \ = \ 84


    That is the only real solution.


    ** That equation, the one in HallsofIvy's post, and the equation given starting in the fifth line of JeffM's post are incorrect.
    They do not follow from the original logarithmic equation because of the domain.
    Last edited by greg1313; Apr 4th 2017 at 01:42 PM.
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  7. #7
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    Re: logarithmic equation

    Quote Originally Posted by HallsofIvy View Post
    Yes, that is correct but it is not at all a good way of approaching this problem. Your equation is 3^{-24}= (x- 3)^{-6}.
    The first thing I would do is "invert" both sides: 3^{24}= (x- 6)^6.
    Typo- of course this should be 3^{24}= (x- 3)^6.

    Then I would notice that 24= 6*4 so I can write this as (3^4)^6= (x- 3)^6 which reduces to either x- 3= 3^4= 81 or x- 3= -3^4= -81 since the power, 6, is even. If you allowing complex solutions there are another 4 complex roots.
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  8. #8
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    Re: logarithmic equation

    Quote Originally Posted by greg1313 View Post
    No, those are not equivalent equations. Look at the argument, (x - 3).

    ]
    but the original equation is equivalent to $ \log_{3} (x-3)^{-6} = -24$ right ? and i went from here , so do you mean that i could get some other solutions that way that will not be correct ?
    Last edited by AbYz; Apr 4th 2017 at 02:28 PM.
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    Re: logarithmic equation

    Quote Originally Posted by AbYz View Post
    but the original equation is equivalent to $ \log_{3} (x-3)^{-6} = -24$ right ?
    No, it is not equivalent to that. The (x - 3) in your argument of what you're taking the logarithm of, could be negative, because it is being
    raised to an even exponent, -6. But, in the original equation, the argument, (x - 3) has to be positive. x must be greater than 3.

    Quote Originally Posted by HallsofIvy View Post
    Typo- of course this should be 3^{24}= (x- 3)^6.
    And this is still incorrect (as in it does not follow from the restricted domain), because of what I stated in post #6, as well as here immediately above.
    Last edited by greg1313; Apr 4th 2017 at 03:09 PM.
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    Re: logarithmic equation

    Quote Originally Posted by HallsofIvy View Post
    Typo- of course this should be 3^{24}= (x- 3)^6.
    Quote Originally Posted by greg1313 View Post
    , the argument, (x - 3) has to be positive. x must be greater than 3.
    And this is still incorrect (as in it does not follow from the restricted domain), because of what I stated in post #6, as well as here immediately above.
    @greg1313, Why do you assume that a PhD mathematician, such as Prof. HallofIvy, would not consider an assumed domain?
    Come to think about it, why do you assume that the domain requires $x>3~?$ It could be a complex logarithm and there thus could be at least six different complex answers. Of course, because this is posted as a Pre-Caculus question it is reasonable to assume a real domain.

    But look at this solution.logarithmic equation-wolframalpha.gif
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    Re: logarithmic equation

    Quote Originally Posted by Plato View Post
    @greg1313, Why do you assume that a PhD mathematician, such as Prof. HallofIvy, would not consider an assumed domain?
    Come to think about it, why do you assume that the domain requires $x>3~?$ It could be a complex logarithm and there thus could be at least six different complex answers.
    Of course, because this is posted as a Pre-Caculus question it is reasonable to assume a real domain.
    His math is wrong. I don't care to know what degree he/she has. It's irrelevant. Certain PhD's in Mathematics display much ignorance in certain math steps/topics
    on down through high school algebra and geometry, so he gets no pass. No, of course the domain requires x > 3, especially given that it's in the Pre-Calculus section.
    The existence of your post makes no sense.
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  12. #12
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    Re: logarithmic equation

    Quote Originally Posted by Debsta View Post
    But x-3 can't equal -81 because you can't take the log of a negative number. So the only solutions will come from x-3 = 81.
    The only real number solutions come from x- 3= 81.
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  13. #13
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    Re: logarithmic equation

    Quote Originally Posted by greg1313 View Post
    ** ... the equation given starting in the fifth line of JeffM's post are incorrect.
    They do not follow from the original logarithmic equation because of the domain.
    I apologize if I am being dense, but I am not following your argument.
    Quote Originally Posted by greg1313 View Post
    No, those are not equivalent equations. Look at the argument, (x - 3).

    x must be greater than 3.

    I agree. Are you objecting to that not being explicitly demonstrated? You yourself give no explicit demonstration, but the demonstration is quite obvious.
    If the problem is that x > 3 is not formally demonstrated, I admit I assumed the obvious demonstration as trivial.

    $x \le 3 \implies log_3(x - 3) \not \in \mathbb R \implies -\ 6log_3(x - 3) \not \in \mathbb R \implies -\ 6log_3(x - 3) \ne -\ 24.$

    But by hypothesis $-\ 6log_3(x - 3) = -\ 24 \implies x > 3.$

    Now my logical error may be here.

    $a,\ b,\ c \in \mathbb R\ and\ b,\ c > 0\ and\ b \ne 1 \implies a log_b(c) \equiv log_b(c^a).$

    If that is not in error, then does it not follow that

    $-\ 6,\ 3,\ x \in \mathbb R\ and\ 3 > 0\ and\ x > 3\ and\ 3 \ne 1 \implies -\ 6log_3(x - 3) = log_3\{(x - 3)^{-6}\}.$

    What am I missing?

    Or my logical error may be here.

    $p,\ q,\ r \in \mathbb R\ and\ p,\ q > 0\ and\ p \ne 1 \implies log_p(q) = r \iff q = p^r.$

    If that is not in error, then does it not follow that

    $x > 3 \implies (x - 3) > 0 \implies (x - 3)^{-6} = \dfrac{1}{(x - 3)^6} > 0\ and$

    $3,\ (x - 3)^{-6},\ -\ 24 \in \mathbb R\ and\ 3,\ (x - 3)^{-6} > 0 \implies $

    $log_3\{(x - 3)^{-6}\} = -\ 24 \implies 3^{-24} = (x - 3)^{-6}.$

    What am I missing?

    I have no problem with your assertion that x > 3. Given that assumption, what is incorrect in the logic given? I truly do not understand. (And don't worry: I have no PhD.) I greatly prefer to avoid errors but cannot if I do not know what they are.
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