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Melissa_Something

  • one year ago

I got the answer, I just dont know how. LOG help :(

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  1. anonymous
    • one year ago
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    here to help m8

  2. Melissa_Something
    • one year ago
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    \[e^x-5=(1/e^4)^x+2\]

  3. Melissa_Something
    • one year ago
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    the x+2 is in the exponent with the x :/ Lol thanks

  4. anonymous
    • one year ago
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    u should see if tiger algebra will give the answer

  5. Melissa_Something
    • one year ago
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    My teacher said something about how a power and power cancel, and her next step was this \[e=e^-4(x+2)\]

  6. anonymous
    • one year ago
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    just sayin retipe it here if u can and i will solve

  7. anonymous
    • one year ago
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    retype

  8. Melissa_Something
    • one year ago
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    Whats tiger algebra?

  9. anonymous
    • one year ago
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    this is the website tiger-algebra.com

  10. anonymous
    • one year ago
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    type in your browser

  11. Melissa_Something
    • one year ago
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    I have the answer, just dont know how

  12. anonymous
    • one year ago
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    and post ypur answer

  13. anonymous
    • one year ago
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    your

  14. anonymous
    • one year ago
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    didi u do it

  15. anonymous
    • one year ago
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    that website will show u the steps too m8

  16. Melissa_Something
    • one year ago
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    It cant do what I needed :/

  17. anonymous
    • one year ago
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    damn what do u need m8

  18. anonymous
    • one year ago
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    sorry im part russian german and african american so im good in some things most i cant really do

  19. Nnesha
    • one year ago
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    is this ur question \[\huge\rm e^{x-5}=(\frac{1}{e^4})^{x+2}\] @Melissa_Something

  20. Nnesha
    • one year ago
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    i don't know if the left side is e^{x-5} or e^(x) - 5

  21. Nnesha
    • one year ago
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    but right side you can move e^4 to the numerator just like we did on the previous post

  22. Nnesha
    • one year ago
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    is this ur question \[\huge\rm e^{x-5}=(\frac{1}{e^4})^{x+2}\] `OR` \[\huge\rm e^x-5 =(\frac{1}{e^4})^{(x+2)}\]

  23. Nnesha
    • one year ago
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    here is an example \[\huge\rm \frac{1}{x^{-m}}= x^{m}\] when we flip the fraction sign of the exponent would change so \[\frac{1}{e^{4}}=?\]

  24. Nnesha
    • one year ago
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    and yes `e` and `ln` cancel each other out \[\large\rm e^{\ln x} = x\] \[\large\rm \cancel{e^{\ln} x} = x\]

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