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February 1, 2015

February 1, 2015

Posted by **MV** on Monday, December 5, 2011 at 10:48pm.

ln(x-3)-ln(11x-23)+in(x-1)=0

- Algebra2 -
**MV**, Monday, December 5, 2011 at 11:00pmI believe it may be x=2,x=13

- Algebra2 -
**Reiny**, Monday, December 5, 2011 at 11:21pmDid you test your answers?

Obviously x= 2 cannot be right since it would make

the second term undefined,

ln(22-23) = ln(-1), which is undefined

using the rules of logs, we get

ln[(x-3)(x-1)/(11x - 23) ] = 0

(x-3)(x-1)/(11x - 23) = e^0 = 1

x^2 - 4x + 3 = 11x - 23

x^2 - 15x+ 26 = 0

(x-2)(x-13) = 0

so x=2 or x=13

check for x=13

LS = ln10 - ln120 + ln12

= ln(10(12)/120) = ln 1 = 0 = RS

x = 13

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