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Math Help - Log/Exponential problem

  1. #1
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    Log/Exponential problem

    I have a log/exponential problem I need help with:

    4^x + 6( 4 ^ {-x} ) = 5<br />

    So far the only thing i've thought of is to change 6 and 5 into a base of 4.

     6 = 4^\frac {\log_6}{log_4}

    If u can help, thanks!

    Last edited by Uncle6; October 31st 2008 at 04:29 PM.
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  2. #2
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    Quote Originally Posted by Uncle6 View Post
    I have a log/exponential problem I need help with:

    4^x + 6( 4 ^ x ) = 5<br />

    So far the only thing i've thought of is to change 6 and 5 into a base of 4.

     6 = 4^\frac {\log_6}{log_4}

    If u can help, thanks!

    why would you do that?

    4^x + 6(4^x) = 5

    \Rightarrow 7(4^x) = 5

    \Rightarrow 4^x = \frac 57

    can you finish?
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  3. #3
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    Quote Originally Posted by Uncle6 View Post
    I have a log/exponential problem I need help with:

    4^x + 6( 4 ^ x ) = 5<br />

    So far the only thing i've thought of is to change 6 and 5 into a base of 4.

     6 = 4^\frac {\log_6}{log_4}

    If u can help, thanks!

    4^x+6(4^x)=5

    take common, 4^x

    4^x(1+6)=5

    4^x=\frac{5}{7}

    Now, take natural log on both sides,

    \ln \;(4^x)=\ln \left(\frac{5}{7}\right)

    now simplify further.
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  4. #4
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    Whoops, my bad, the question is actually has a 4^{-x}. Which is why I found this question to be so hard =S
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Uncle6 View Post
    Whoops, my bad, the question is actually has a 4^{-x}. Which is why I found this question to be so hard =S
    which of the 4^x's should be 4^{-x}??
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  6. #6
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    uhh, I fixed it up there.
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    Quote Originally Posted by Uncle6 View Post
    uhh, I fixed it up there.
    ok, here's the trick

    multiply through by 4^x, we get:

    (4^x)^2 + 6 = 5(4^x)

    or in other words,

    (4^x)^2 - 5(4^x) + 6 = 0

    now do you see what to do?

    to make it a bit clearer, what if we made a = 4^x, then we have

    a^2 - 5a + 6 = 0

    now continue
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    WOW! that realy helps, thanks! I swear, I'd never see that trick..
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Uncle6 View Post
    WOW! that realy helps, thanks! I swear, I'd never see that trick..
    tell us your solution when finished
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  10. #10
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    x= 1/2 or around 0.792
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  11. #11
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    Quote Originally Posted by Uncle6 View Post
    x= 1/2 or around 0.792
    (4^x)^2 - 5(4^x) + 6 = 0

    (4^x-2)(4^x-3)=0

    4^x=2

    2^{2x}=2^1

    2x=1

    x=1/2

    Also,

    4^x=3

    x \ln 4 = \ln 3

    x = \frac{\ln 3}{\ln 4}=0.792

    x = 1/2, 0.792

    your answers are right
    Last edited by Shyam; November 1st 2008 at 04:50 PM.
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