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anonymous

  • 5 years ago

solve5^x+1/5^1-x=1/25^x

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  1. anonymous
    • 5 years ago
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    First combine your like terms. What do you have after that?

  2. anonymous
    • 5 years ago
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    5^x+1/5^1-x = 5^x+1-1+x = 5^2x 1/25^x = 1/5^2x 5^2x = 1/5^2x 5^2x times 5^2x = 1 5^4x = 5^0 4x = 0 x=0

  3. anonymous
    • 5 years ago
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    I don't think so cause if I plug in 0 to the original equation I get: \[5^0 + 1/5^1 - 0 = 1/25^0\]\[\implies 1 + 1/5 = 1\]\[\implies 1 = 4/5\]\[\implies 5 = 4\] So yeah.

  4. anonymous
    • 5 years ago
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    did you figure this out? if not i can send you an answer.

  5. anonymous
    • 5 years ago
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    no and that would be great!

  6. anonymous
    • 5 years ago
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    ok let me go slow because it take a while for me to type this in. i will do it step by step. there may be a snap way to do this but i don't see it.

  7. anonymous
    • 5 years ago
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    \[5^x+\frac{1}{5}^{1-x}=\frac{1}{25^x}\]

  8. anonymous
    • 5 years ago
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    add the terms on the left: \[\frac{5^{x+1-x}+1}{5^{1-x}}=\frac{5^1+1}{5^{1-x}}=\frac{6}{5^{1-x}}\]\]

  9. watchmath
    • 5 years ago
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    Go ahead satellite :D

  10. anonymous
    • 5 years ago
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    i am a slow typer.

  11. anonymous
    • 5 years ago
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    so we have \[\frac{6}{x^{1-x}}=\frac{1}{25^x}\]

  12. anonymous
    • 5 years ago
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    the right hand side is \[\frac{1}{5^{2x}}\]

  13. anonymous
    • 5 years ago
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    \[\frac{6}{5^{1-x}}= \frac{1}{5^{2x}}\] \[6\times 5^{2x}=5^{1-x}\]

  14. anonymous
    • 5 years ago
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    take the log of both sides to get the variable out of the exponent: \[ln(6\times 5^{2x})=ln(5^{1-x})\] \[ln(6) + 2xln(5) = (1-x)ln(5)\]

  15. anonymous
    • 5 years ago
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    now it is algebra from here on in remembering that ln(6) and ln(5) are constants.

  16. anonymous
    • 5 years ago
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    \[ln(6) +2xln(5)=ln(5)-xln(5)\] \[2xln(5)+xln(5)=ln(5)-ln(6)\] \[3ln(5)x=ln(5)-ln(x)\] \[x=\frac{ln(5)-ln(6)}{3ln(5)}\]

  17. watchmath
    • 5 years ago
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    Agree! :) \(5^{x}+5^{x-1}=5^{-2x}\) \((1+\frac{1}{5})5^x=5^{-2x}\) \(\frac{6}{5}5^x=5^{-2x}\) Multiply by \(5^{2x}\) we have \(\frac{6}{5}5^{3x}=1\) Then \(125^x=\frac{5}{6}\) \(x=\ln(5/6)/\ln(125)\)

  18. anonymous
    • 5 years ago
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    if you choose you can rewrite this as \[\frac{ln(\frac{5}{6})}{3ln(5)}\]

  19. anonymous
    • 5 years ago
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    watchman much snappier as usual. how it is going?

  20. anonymous
    • 5 years ago
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    Awesome! Thank you!

  21. anonymous
    • 5 years ago
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    Hrm. I woulda just: \[5^x + \frac{1}{5^{1-x}} = \frac{1}{25^x}\]\[\implies 5^{x+1} + 5^x = \frac{1}{5^x}\] \[\implies 5^x(5+1) = \frac{1}{5^x} \implies 5^{2x} = \frac{1}{6}\]\[ \implies x(ln\ 25) = (ln\ 1) - (ln\ 6)\]\[\implies x = \frac{(ln\ 1) - (ln\ 6)}{(ln\ 25)}\]

  22. anonymous
    • 5 years ago
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    let me know if you do not understand any step

  23. anonymous
    • 5 years ago
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    Not sure if that's the same as what you did.

  24. anonymous
    • 5 years ago
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    It's not. Hrm.. now to see what went wrong.

  25. anonymous
    • 5 years ago
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    must not be the same because we got different answers. fairly certain of mine

  26. watchmath
    • 5 years ago
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    The \(5^{x+1}\) should be \(5^{x-1}\) Polpak.

  27. anonymous
    • 5 years ago
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    \[\frac{1}{5}^{1-x}\neq 5^{x+1}\]

  28. anonymous
    • 5 years ago
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    I multiplied the whole thing by 5. \[5^x + 5^{-1+x} = 5^{-2x}\] \[\implies 5^{x+1} + 5^x = 5^{-2x + 1}\]

  29. anonymous
    • 5 years ago
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    It was the right side that I screwed up.

  30. anonymous
    • 5 years ago
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    I canceled half the 5's from the \(1/25^x \) when I should have only cancelled one of them.

  31. anonymous
    • 5 years ago
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    So.. \[5^{x+1} + 5^x = 5^{1-2x}\]\[\implies 5^x(5+1) = 5^{1-2x}\]\[\implies x(ln\ 5) + (ln\ 6) = (1-2x)(ln\ 5)\]\[\implies x = \frac{(ln\ 5) - (ln\ 6)}{3(ln\ 5)}\]

  32. anonymous
    • 5 years ago
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    yeah, that's better.

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