anonymous
  • anonymous
how do you solve Log5 (2-x) – 1 = log5 (2x-7)
Mathematics
katieb
  • katieb
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jennychan12
  • jennychan12
is that log base 5? if so just divide out the bases so you get 2-x = 2x-7 solve for x
anonymous
  • anonymous
it is base five, but what about the minus 1?
jennychan12
  • jennychan12
move the logbase5(2x-7) to the other side and move the -1 over. try from there?

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anonymous
  • anonymous
i did and i got x=-3 and i feel like that's wrong
jennychan12
  • jennychan12
really? i got x = 3.
anonymous
  • anonymous
snap, lemme see
anonymous
  • anonymous
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?
jennychan12
  • jennychan12
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...
anonymous
  • anonymous
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?
jennychan12
  • jennychan12
i dunno why we're getting diff answers cuz u could do it multiple ways
anonymous
  • anonymous
struggles.

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