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solve: log8(w-6)=2-log8(w+15) note: the 8 is to be little at the bottom right corner of logAsked by Britt
solve: log8(w-6)=2-log8(w+15)
note: the 8 is to be little at the bottom right corner of log
note: the 8 is to be little at the bottom right corner of log
Answers
Answered by
Steve
since 64 = 8^2, then using base-8 logs, we have
log(w-6) = log(64)-log(w+15)
(w-6) = 64/(w+15)
(w-6)(w+15) = 64
w^2 + 9w - 154 = 0
w = 1/2 (-9±√697)
scrap the negative value (why?)
log(w-6) = log(64)-log(w+15)
(w-6) = 64/(w+15)
(w-6)(w+15) = 64
w^2 + 9w - 154 = 0
w = 1/2 (-9±√697)
scrap the negative value (why?)
Answered by
Britt
because the answer has got to be a real number which is not a negative number
Answered by
Britt
how do you get 1/2 as a solution?
Answered by
Britt
So technically there would be no solution right?
Answered by
Steve
So, technically, the solution is
x = 1/2 (-9+√697) = 8.70
where does the 1/2 come from? Think back ... back ... back to Algebra I and the quadratic formula.
x = 1/2 (-9+√697) = 8.70
where does the 1/2 come from? Think back ... back ... back to Algebra I and the quadratic formula.
Answered by
Britt
the quadratic formula is -b+-radical b^2-4ac/a
Answered by
Britt
SO HOW DO YOU GET 1/2 FORM THAT?
Answered by
Reiny
Britt, when Steve and I went to school, the quadratic formula used to be
(-b ± √(b^2 -4ac)/(2a) or
(1/2) (-b ± √(b^2 -4ac)/a , thus the 1/2
I was not aware of any recent changes to the formula.
(-b ± √(b^2 -4ac)/(2a) or
(1/2) (-b ± √(b^2 -4ac)/a , thus the 1/2
I was not aware of any recent changes to the formula.
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