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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    How can you solve this algebraically ln(x2-x) =ln 6

  2. anonymous
    • one year ago
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    well both arguments should be the same, just equal both and solve the equation

  3. anonymous
    • one year ago
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    x^2-x = 6 i know the answer is 3 but don't know how to express it algebraically

  4. Astrophysics
    • one year ago
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    Hint: \[\huge e^{lnx} = x\]

  5. anonymous
    • one year ago
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    Well I'm using the one to one property of exponents but don't know how i can algebraically express how to solve it??

  6. anonymous
    • one year ago
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    x^2 -x =6

  7. Astrophysics
    • one year ago
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    That's it now just factor

  8. Astrophysics
    • one year ago
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    No, remember how you factor, what two numbers add up to -1 and multiply to give -6, those are your two factors.

  9. anonymous
    • one year ago
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    (x-3) (x+2)

  10. anonymous
    • one year ago
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    x=3 x=-2

  11. Astrophysics
    • one year ago
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    There we go, so our roots are x = 3, and x = -2 :)

  12. anonymous
    • one year ago
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    Thanks! but we omit -2 because the natural logs domain is from (0,oo) right?

  13. Astrophysics
    • one year ago
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    Yes :)

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