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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

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

2. christinaxxx

it is base five, but what about the minus 1?

3. jennychan12

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

4. christinaxxx

i did and i got x=-3 and i feel like that's wrong

5. jennychan12

really? i got x = 3.

6. christinaxxx

snap, lemme see

7. christinaxxx

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

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

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

i dunno why we're getting diff answers cuz u could do it multiple ways

11. christinaxxx

struggles.

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