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Math Help - exponent

  1. #1
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    exponent

    hi,

    if we have: 2^10 = 1024

    what is 10^? = 1024

    Since this is a small number I can do rote trying and come to 10^3=1000 ~1024 but how I can approximate this with some formula or some algorithms when I have really big number.

    For example to convert from base 2 to 10 and play with the exponents to get some number?

    Thanks!
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  2. #2
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    Quote Originally Posted by problady View Post
    hi,

    if we have: 2^10 = 1024

    what is 10^? = 1024

    Since this is a small number I can do rote trying and come to 10^3=1000 ~1024 but how I can approximate this with some formula or some algorithms when I have really big number.

    For example to convert from base 2 to 10 and play with the exponents to get some number?

    Thanks!
    10^x = 2^{10}

    log (base 10) both sides ...

    \log(10^x) = \log(2^{10})<br />

    power property of logs ...

    x\log{10} = 10\log{2}<br />

    x = 10\log{2}

    note that \log{2} \approx 0.30103
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  3. #3
    Senior Member nikhil's Avatar
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    Lightbulb

    xlog_2[10]=log_2[1024]
    x=log_10[2]*10
    note: log_a[b]=1/log_b[a]
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  4. #4
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    Quote Originally Posted by skeeter View Post
    10^x = 2^{10}

    log (base 10) both sides ...

    \log(10^x) = \log(2^{10})<br />

    power property of logs ...

    x\log{10} = 10\log{2}<br />

    x = 10\log{2}

    note that \log{2} \approx 0.30103
    But what if I have the opposite?

    10^3=2^x or 10^3=2^?

    How I can solve this case?

    And what if I would like to manipulate other exponents not just 2 to 10 or 10 to 2?

    Any help...?

    Thanks!
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  5. #5
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    Quote Originally Posted by problady View Post
    But what if I have the opposite?

    10^3=2^x or 10^3=2^?

    How I can solve this case?

    And what if I would like to manipulate other exponents not just 2 to 10 or 10 to 2?

    Any help...?

    Thanks!
    You can use any base you like and then use algebra to simplify.

    For example I will solve 10^3 = 2^x using base e

    3\ln(10) = x\ln(2)

    x = \frac{3\ln(10)}{\ln(2)}


    More generally if a^b = c^x where a,b and c are positive constants and a \neq c then x = b\log_c(a) or x = \frac{b\log_e(a)}{\log_e(c)}
    Last edited by e^(i*pi); May 31st 2010 at 07:43 AM. Reason: latex
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  6. #6
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    Quote Originally Posted by problady View Post
    But what if I have the opposite?

    10^3=2^x or 10^3=2^?

    How I can solve this case?

    And what if I would like to manipulate other exponents not just 2 to 10 or 10 to 2?

    Any help...?

    Thanks!
    same procedure for the general case ...

    a^b = b^x

    use either a base 10 or base e log because those values are readily available from any scientific calculator ...

    \log(a^b) = \log(b^x)

    b\log{a} = x\log{b}

    x = \frac{b\log{a}}{\log{b}}
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  7. #7
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    Note that in neither of those cases, 2^{10}= 1024 so 10^?= 1024 nor 10^3= 2^? can you expect the answer to be an integer. That is because 2 to any power can have only 2 as a prime factor while 10 to a power greater than 0 will have prime factors of both 2 and 5.
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