Posted by MV on Monday, December 5, 2011 at 10:48pm.
Use properties of logartithms to simplify and solve. Thanks.
Algebra2 - MV, Monday, December 5, 2011 at 11:00pm
I believe it may be x=2,x=13
Algebra2 - Reiny, Monday, December 5, 2011 at 11:21pm
Did 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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