Posted by MV on .
Use properties of logartithms to simplify and solve. Thanks.
ln(x3)ln(11x23)+in(x1)=0

Algebra2 
MV,
I believe it may be x=2,x=13

Algebra2 
Reiny,
Did you test your answers?
Obviously x= 2 cannot be right since it would make
the second term undefined,
ln(2223) = ln(1), which is undefined
using the rules of logs, we get
ln[(x3)(x1)/(11x  23) ] = 0
(x3)(x1)/(11x  23) = e^0 = 1
x^2  4x + 3 = 11x  23
x^2  15x+ 26 = 0
(x2)(x13) = 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