precalculus

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solve the logarithmic equation . express solution in exact form

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

• precalculus -

1 = 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 -

log5(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 -

I 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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