Asked by kristie
how would you solve for y in this problem:
ln(y-1)-ln2 = x +lnx
ln(y-1)-ln2 = x +lnx Solve for y...
ln[(y-1)/2]]=x + lnx
take the antilog of each side
(y-1)/2= e^(x+lnx)
solve for y.
ln(y-1)-ln2 = x +lnx
ln(y-1)-ln2 = x +lnx Solve for y...
ln[(y-1)/2]]=x + lnx
take the antilog of each side
(y-1)/2= e^(x+lnx)
solve for y.