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christinaxxx

how do you solve Log5 (2-x) – 1 = log5 (2x-7)

  • one year ago
  • one year ago

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  1. jennychan12
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    is that log base 5? if so just divide out the bases so you get 2-x = 2x-7 solve for x

    • one year ago
  2. christinaxxx
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    it is base five, but what about the minus 1?

    • one year ago
  3. jennychan12
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    move the logbase5(2x-7) to the other side and move the -1 over. try from there?

    • one year ago
  4. christinaxxx
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    i did and i got x=-3 and i feel like that's wrong

    • one year ago
  5. jennychan12
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    really? i got x = 3.

    • one year ago
  6. christinaxxx
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    snap, lemme see

    • one year ago
  7. christinaxxx
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    this is what i did. -1=logbase5 (2x+7/2-x) (2x+7)/(2-x)= 5^-1 (2x+7)/(2-x)=1/5 then i cross multiplied and got 5(2x+7)=1(2-x) 10x+35=2-x 11x+35=2 11x=-33 x=-3 righttttt?

    • one year ago
  8. jennychan12
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    Log5 (2-x) – 1 = log5 (2x-7) i meant like move the -1 to the right and the logbase5(2x-7) to the left so you get (2-x)/(2x-7) = 5 i got 3, but neither 3 nor -3 work as answers. and your 2x+7 should be a -7...unless u wrote the question wrong...

    • one year ago
  9. christinaxxx
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    nope, my question is from a worksheet. no error in my copying it down. but the negative one would be the exponent of the base 5, wouldn't it? so you'd have to raise five to the negative one?

    • one year ago
  10. jennychan12
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    i dunno why we're getting diff answers cuz u could do it multiple ways

    • one year ago
  11. christinaxxx
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    struggles.

    • one year ago
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