Logarithm question (1 Viewer)

[tam89]

Solve for y in terms of t:
ln (y-1) - ln2 = t + lnt

--> I had a different answer: y = 2e^t + 2t + 1

Ans should be: y = 2te^t +1

 

[shafqat]

Solve for y in terms of t:
ln (y-1) - ln2 = t + lnt

--> I had a different answer: y = 2e^t + 2t + 1

Ans should be: y = 2te^t +1

ln (y-1) - ln2 = t + lnt
antilogs:
y - 1 = e(ln2 + t+ lnt)
= e^ln2 . e^t . e^lnt
y - 1 = 2. e^t .t
So y = 2te^t +1

 

[Emma-Jayde]

:Edit: Damn, beaten to it!

ln (y-1) - ln2 = t + lnt

y - 1 = e^{(ln2 + t+ lnt)}

y - 1 = e^ln2 x e^t x e^lnt

y - 1 = 2 x e^t x t

So y = 2te^t +1

 

[tam89]

Oh...
thankyou