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| Question 608917:  ln(lne(exponent -x) =ln3
 ln6x-ln(x+1)=ln4
 lnx+3ln2=ln2/x
 Answer by lwsshak3(11628)
      (Show Source): 
You can put this solution on YOUR website! ln(lne(exponent -x) =ln3 lne^-x=-x (log of base raised to a power=power)
 ln(-x)=ln(3)
 -x=3
 x=3
 ..
 ln6x-ln(x+1)=ln4
 ln6x-ln(x+1)-ln4=0
 ln6x-(ln(x+1)+ln4)=0
 place under single log
 ln[6x/((x+1)*4)]=0
 convert to exponential form:
 e^0=6x/((x+1)*4)=1
 6x=4x+4
 2x=4
 x=2
 ..
 lnx+3ln2=ln2/x
 I will assume last term is meant to be ln(2/x) instead of (ln2)/x which I am not able to solve.
 lnx+3ln2=ln(2/x)
 lnx+3ln2=ln2-lnx
 2lnx+2ln2=0
 place under single log
 ln[x^2*2^2]=0
 ln[4x^2]=0
 convert to exponential form
 e^0=4x^2=1
 x^2=1/4
 x=-1/2 (reject, x>0)
 or
 x=1/2
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