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Math Help - solving exponents with logs

  1. #1
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    solving exponents with logs

    solve:
    2^{2x} = 5^x

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

    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

    2^{2x} = 5^x \implies \left( {\frac{4}<br />
{5}} \right)^x = 1.

    Now, what's the next step?
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  3. #3
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    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
    2^{2x} = 5^x \implies \left( {\frac{4}<br />
{5}} \right)^x = 1.
    doesnt this just turn every problem of its kind into x = 0?

    for instance with random numbers:

    12345^{67x} = 5432^x \implies \left( {\frac{12345^{67}}<br />
{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
    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 4^x = 5^x

    take the log of both sides

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

    \Rightarrow x \log 4 = x \log 5

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

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

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

    this second option is an impossibility, thus 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:

    12345^{67x} = 5432^x \implies \left( {\frac{12345^{67}}<br />
{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)

    4^{3x} = 64^x
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  7. #7
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    Isomorphism's Avatar
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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 4^{3x} = 64^x, we can see we cant use the above idea.
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