Solve the equation: [(3exp y).{8 exp (y/y+2)}] = 6?

Answer 1

Taking Log of both members of the eq. to the base 10, we get :
y log 3 + (y/y+2) log 2 = log 6-----------------------------1

Note: All the log taken are have the base 10. Since I was facing writing 10 in the base, so I didn't write anything.

rearranging eq. 1, we have
y^2 log 3 + y( 3 log 2 + 2 log 3 - log 6) - 2 log 6 = 0-----------------2

we now solve this quadratic equation....

in the original equation, manipulation can be done as :

3 exp y * 2 exp(3y/y+2) = 3 exp 1* 2 exp 1
which implies y = 1 or (3y/y+2) = 1.
Hence y=1 is a solution...
For this reason, by Viete's theorem the second root of quadratic equation is (- 2 log 6 )/ log 3.

Answer 2

[(3exp y).{8 exp (y/y+2)}] = 6
24 exp(y^2/(y+2)) = 6
exp(y^2/(y+2)) = 1/4
y^2/(y+2) = ln(1/4) = -ln4 = q
y2 = (y+2)q
y2 -qy -2q = 0
y = [q + or - root(q2 + 8q)]/2 with q = -ln4
y = [-ln4 + root{(ln4)^2 - 8ln4}]/2

Answer 3

yes take logs and solve the quadratic
so y=-3.6205 or -0.7658

Answer 4

3e^y*8e^(y/y+2)=6
3e^y*8e^(1+2)=6
24e^ye^3=6
e^(y+3)=1/4
y+3=-ln 4
y=3-ln 4