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Exponents and logarithms

  1. Oct 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve the following equation: 2^3+2^3+2^3+2^3=2^x



    2. Relevant equations
    log(a)^x=x*log(a)



    3. The attempt at a solution
    What i attempted was to log both sides, bring down the exponents, and summarize them. This left me with 12*log(2)=x*log(2). I then divide both sides by log(2) and get x=12, which is wrong. Please note that all of this was with logarithms with base number 10.
     
  2. jcsd
  3. Oct 8, 2015 #2

    BvU

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    Hi,

    Where did you get your relevant equation ? It's dead (*) wrong, but visually close to the right one: ##\log (a^x) = x \log a##.

    Taking logarithms of a sum is generally a bad idea. For a product you are better off. So make a product of the lefthand side.
    And look at that. Maybe you don't need to take logarithms, but you might be able to use a simpler equation about exponentiation.

    PS the term "summarizing" isn't all that mathematically sound.... "sum" is what you mean. But you can't do that.

    (*) [edit] well, dangerous is a better expression. It raises confusion between ##(\log a)^x## and ##(\log (a^x))##
     
  4. Oct 8, 2015 #3

    HallsofIvy

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    I can see no reason to use logarithms. You are just trying to find x such that [itex]2^x= 32[/itex]. That should be elementary.
     
  5. Oct 8, 2015 #4

    RUber

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    I second what BvU says. If this is an exercise of the properties of exponents, you should notice that 2^3 is a repeated sum...how many times is it repeated?
    Rewrite it as a coefficient times 2^3, then write your coefficient as a power of 2.
    Now, you will be able to use properties of exponents to quickly solve for x.

    Otherwise, just add 8+8+8+8 and do what HallsofIvy suggested.

    And if you are really feeling like using logarithms ... use log base 2.
     
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