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Math Help - understanding logs

  1. #1
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    understanding logs

    Hello,

    Can you please help me understand how to solve this...

    log8(1/4) = x

    8^x = 1/4

    how would i solve this using algebra?

    This is simply review and i never really learned how to do it.
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  2. #2
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    Quote Originally Posted by l flipboi l View Post
    Hello,

    Can you please help me understand how to solve this...

    log8(1/4) = x

    8^x = 1/4

    how would i solve this using algebra?

    This is simply review and i never really learned how to do it.
    Notice that 8^{\frac{1}{2}} = 4.

    Therefore 8^{-\frac{1}{2}} = \frac{1}{4}.


    What do you think x equals?
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  3. #3
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    not sure =(
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  4. #4
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    Look at the question carefully.

    I have told you that 8^{-\frac{1}{2}} = \frac{1}{4}.


    You are told that 8^x = \frac{1}{4}.


    Therefore 8^x = 8^{-\frac{1}{2}}.


    What do you think x equals?
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  5. #5
    Super Member Bacterius's Avatar
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    Prove It : 8^{\frac{1}{2}} = \sqrt{8} \neq 4

    The x is already isolated, so you just have to evaluate. Do you need help about evaluating it ?
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  6. #6
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    the answer in the book is 2/3. I'm trying to figure out how to get this. so i looked up raising a power fractions and should it be

    sqrt(8) ?
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    yes, i need to see how to evaluate it.
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  8. #8
    Super Member Bacterius's Avatar
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    The change of base formula allows us to compute any base logarithms from another one. That is :

    \log_a{x} = \frac{\log_b{x}}{\log_b{a}}

    So you plug in your values : you have a = 8, and you want to be able to calculate this with your calculator. Your calculator (hopefully) has a "log" button, so we'll use this one. It's the decimal logarithm, so b = 10. And we want to compute x = \frac{1}{4}. So :

    \log_8{\frac{1}{4}} = \frac{\log_{10}{\frac{1}{4}}}{\log_{10}{8}}

    And this gives exactly \log_8{\frac{1}{4}} = - \frac{2}{3}

    This can be verified easily : it is equivalent to saying that \frac{1}{4} = 8^{- \frac{2}{3}}, which is true (you can check)
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    great, this is what i was looking for. Thanks!
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  10. #10
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    Quote Originally Posted by Bacterius View Post
    Prove It : 8^{\frac{1}{2}} = \sqrt{8} \neq 4

    The x is already isolated, so you just have to evaluate. Do you need help about evaluating it ?
    Hahaha oops. Of course it doesn't

    Sorry, massive brain fart there...
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  11. #11
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    Quote Originally Posted by l flipboi l View Post
    Hello,

    Can you please help me understand how to solve this...

    log8(1/4) = x

    8^x = 1/4

    how would i solve this using algebra?
    You wouldn't. You would use the definition of "logarithm to base a" which is that it is the "inverse function" to a^x. The logarithm is defined by "if y= log_a(x) then x= a^y".
    That's all you need.

    This is simply review and i never really learned how to do it.
    Learn the definitions first!

    Prove it, in addition to simply giving a wrong calculation, seems to think you were trying to solve 8^x= 1/4 but the solution to that is x= log_8 (1/4) which is your first equation. I assume you are really asking how to go from one equation to the other.

    If you are really asking how to write x= log_8(1/4) in terms your calculator can handle,use the fact that is 8^x= 1/4 then log(8^x)= xlog(8)= log(1/4) so that x= \frac{log(1/4)}{log(8)} where "log" can be either base 10 (common log) or base e (natural log), both of which should be on your calculator.
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  12. #12
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    Quote Originally Posted by Prove It View Post
    Notice that 8^{\frac{1}{2}} = 4.

    Therefore 8^{-\frac{1}{2}} = \frac{1}{4}.


    What do you think x equals?
    What? That's not true at all. 8^{1/2}= \sqrt{8}= 2\sqrt{2}, not 4! 8 times 1/2= 4.
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  13. #13
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    Quote Originally Posted by HallsofIvy View Post
    What? That's not true at all. 8^{1/2}= \sqrt{8}= 2\sqrt{2}, not 4! 8 times 1/2= 4.
    Yes, already mentioned and acknowledged in previous threads.
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  14. #14
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    understanding logs

    posted by flipboi

    log base8(1/4)=x find x
    8^x=1/4
    xlogbase2(8) =log base2(1/4)
    xtimes3=-2
    x=-2/3
    Easier.All logs bear a fixed relationship for all bases.You can pick your own to suit a problem

    bjh
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  15. #15
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    Quote Originally Posted by Prove It View Post
    Yes, already mentioned and acknowledged in previous threads.
    Hey, you expect me to read all the responses?
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