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Thread: solving exponents with logs

  1. #1
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    solving exponents with logs

    solve:
    $\displaystyle 2^{2x} = 5^x$

    i went this far:
    $\displaystyle \log{2^{2x}} = \log{5^x}$

    $\displaystyle 2x\log{2} = x\log{5}$

    how would you finish it?
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  2. #2
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    You don't even need logs, just note that

    $\displaystyle 2^{2x} = 5^x \implies \left( {\frac{4}
    {5}} \right)^x = 1.$

    Now, what's the next step?
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  3. #3
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    $\displaystyle anything^0 = 1$

    but wouldnt you get a different answer with logs?
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    $\displaystyle 2^{2x} = 5^x \implies \left( {\frac{4}
    {5}} \right)^x = 1.$
    doesnt this just turn every problem of its kind into x = 0?

    for instance with random numbers:

    $\displaystyle 12345^{67x} = 5432^x \implies \left( {\frac{12345^{67}}
    {5432}} \right)^x = 1 \implies x = 0$
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by incyt3s View Post
    $\displaystyle anything^0 = 1$

    but wouldnt you get a different answer with logs?
    no you would not. Krizalid is right, logs are definitely not needed. however, if you do want to force the issue...

    we have $\displaystyle 4^x = 5^x$

    take the log of both sides

    $\displaystyle \Rightarrow \log 4^x = \log 5^x$

    $\displaystyle \Rightarrow x \log 4 = x \log 5$

    $\displaystyle \Rightarrow x \log 4 - x \log 5 = 0$

    $\displaystyle \Rightarrow x ( \log 4 - \log 5) = 0$

    $\displaystyle \Rightarrow x = 0 \mbox{ or } \log 4 - \log 5 = 0$

    this second option is an impossibility, thus $\displaystyle x = 0$ as we got from Krizalid's much simpler solution
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by incyt3s View Post
    doesnt this just turn every problem of its kind into x = 0?

    for instance with random numbers:

    $\displaystyle 12345^{67x} = 5432^x \implies \left( {\frac{12345^{67}}
    {5432}} \right)^x = 1 \implies x = 0$
    technically no, but i guess in general yes.

    here's one where x is not zero (well, at least that's not the only solution)

    $\displaystyle 4^{3x} = 64^x$
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  7. #7
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    Another way to see it is that a power of 2 can never have a factor of 5 or the other way.
    Thus technically x=0 is the only solution.
    But with $\displaystyle 4^{3x} = 64^x$, we can see we cant use the above idea.
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