Posted by
**kristie** on
.

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.

Sunday

March 26, 2017
Posted by
**kristie** on
.

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.

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