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christinaxxx

  • 2 years ago

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

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  1. jennychan12
    • 2 years ago
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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

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

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

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

  5. jennychan12
    • 2 years ago
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    really? i got x = 3.

  6. christinaxxx
    • 2 years ago
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    snap, lemme see

  7. christinaxxx
    • 2 years ago
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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?

  8. jennychan12
    • 2 years ago
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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...

  9. christinaxxx
    • 2 years ago
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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?

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

  11. christinaxxx
    • 2 years ago
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    struggles.

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