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Anyone good with logarithms?

  1. Nov 7, 2004 #1
    Hi! I'm not sure how I would tackle this exponential equation:

    4^x + 4^x+1 = 40

    I was using logs to try and solve it but I'm getting nowhere. I don't know what to do exponentially either. Please help! :cry:
     
  2. jcsd
  3. Nov 7, 2004 #2

    arildno

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    Dearly Missed

    Hint:
    [tex]4^{x}+4^{x}=2*(4^{x})[/tex]
     
  4. Nov 7, 2004 #3
    My problem isn't that simple though.
     
  5. Nov 7, 2004 #4

    Galileo

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    Is your equation
    [tex]4^x + 4^x+1 = 40[/tex]
    or
    [tex]4^x + 4^{x+1} = 40[/tex]???
    In both cases you can write [itex]4^x[/itex] as a factor.
     
    Last edited: Nov 7, 2004
  6. Nov 7, 2004 #5
    4^x + 4^(x+1) = 40
     
  7. Nov 7, 2004 #6

    Pyrrhus

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    Halfway done:

    [tex] 4^x(1 + 4^1) = 40 [/tex]
     
  8. Nov 7, 2004 #7
    How did you do that?
     
  9. Nov 7, 2004 #8

    Pyrrhus

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    [tex] a^{c+d} = a^c a^d [/tex]
     
  10. Nov 7, 2004 #9
    OK but you only solved part of it, right?
     
  11. Nov 7, 2004 #10

    Pyrrhus

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    [itex] \frac{1}{16} [/itex] left to work out

    [tex] 4^x = 8 [/tex]

    [tex] 4^x = 2^3 [/tex]

    [tex] 2^{2x} = 2^3 [/tex]
     
  12. Nov 7, 2004 #11
    How come it's = to 8 all in a sudden? The question says it's = to 40.
     
  13. Nov 7, 2004 #12

    BobG

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    Look at post #6 again. What's 1 +4?

    That gives you
    [tex]4^x*5=40[/tex]

    Divide both sides by 5.
     
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