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December 17, 2014

December 17, 2014

Posted by **mz deedee** on Friday, November 11, 2011 at 12:00pm.

log5(x-9)+log5(x+4)=1+log5(x-5)

- precalculus -
**Steve**, Friday, November 11, 2011 at 12:30pm1 = log5(5), so we have

log5(x-9)+log5(x+4)=log5(5)+log5(x-5)

log5[(x-9)(x+4)] = log5[5(x-5)]

raise 5 to the powers, and we have

(x-9)(x+4) = 5(x-5)

x^2 - 5x - 36 = 5x - 25

x^2 - 10x - 11 = 0

(x-11)(x+1) = 0

Solutions are 11,-1

However, -1 does not fit the original equation: log of negatives are undefined.

- precalculus -
**Reiny**, Friday, November 11, 2011 at 12:40pmlog5(x-9)+log5(x+4)=1+log5(x-5)

log5(x-9)+log5(x+4)=log5(5)+log5(x-5)

log5[(x-9)(x+4)] = log5[5(x-5)}

(x-9)(x+4) = 5(x-5)

x^2 - 5x - 36 = 5x - 25

x^2 - 10x - 9 = 0

x = (10 ± √136)/2 = appr. 10.83 or -.83

but for each of the above to defined, x > 9

so x = (10 + √136)/2 = 5 + √34

check my arithmetic

- Steve had right equation, precalculus -
**Reiny**, Friday, November 11, 2011 at 12:43pmI have an error in my equation...

x^2 - 10x - 9 = 0 should be

x^2 - 10x - 11 - 0 , just like Steve had

then (x-11)(x+1) = 0

x = 11 or x = -1

so x = 11

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