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Thread: Log/Exponential problem

  1. #1
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    Log/Exponential problem

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

    $\displaystyle 4^x + 6( 4 ^ {-x} ) = 5
    $

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

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

    If u can help, thanks!

    Last edited by Uncle6; Oct 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:

    $\displaystyle 4^x + 6( 4 ^ x ) = 5
    $

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

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

    If u can help, thanks!

    why would you do that?

    $\displaystyle 4^x + 6(4^x) = 5$

    $\displaystyle \Rightarrow 7(4^x) = 5$

    $\displaystyle \Rightarrow 4^x = \frac 57$

    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:

    $\displaystyle 4^x + 6( 4 ^ x ) = 5
    $

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

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

    If u can help, thanks!

    $\displaystyle 4^x+6(4^x)=5$

    take common, $\displaystyle 4^x$

    $\displaystyle 4^x(1+6)=5$

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

    Now, take natural log on both sides,

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

    now simplify further.
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    Whoops, my bad, the question is actually has a $\displaystyle 4^{-x}$. Which is why I found this question to be so hard =S
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    Quote Originally Posted by Uncle6 View Post
    Whoops, my bad, the question is actually has a $\displaystyle 4^{-x}$. Which is why I found this question to be so hard =S
    which of the $\displaystyle 4^x$'s should be $\displaystyle 4^{-x}$??
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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 $\displaystyle 4^x$, we get:

    $\displaystyle (4^x)^2 + 6 = 5(4^x)$

    or in other words,

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

    now do you see what to do?

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

    $\displaystyle a^2 - 5a + 6 = 0$

    now continue
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    WOW! that realy helps, thanks! I swear, I'd never see that trick..
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    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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    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
    $\displaystyle (4^x)^2 - 5(4^x) + 6 = 0$

    $\displaystyle (4^x-2)(4^x-3)=0$

    $\displaystyle 4^x=2$

    $\displaystyle 2^{2x}=2^1$

    2x=1

    x=1/2

    Also,

    $\displaystyle 4^x=3$

    $\displaystyle x \ln 4 = \ln 3$

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

    x = 1/2, 0.792

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