Solving Logarithmic Functions

Mathematics
Tutor: None Selected Time limit: 1 Day

e^(2x+3) - 3e^(x) + 2 = 0

May 15th, 2015

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May 15th, 2015

e^(2x) - 3e^(x) + 2 = 0

put y = e^x

y^2-3y+2 =0

(y-3) (y-2) =0

y =3 or y =2

e^x = 3 or e^x =2

x = ln 3 or ln2

May 15th, 2015

I am sorry please ignore the above

e^(2x) - 3e^(x) + 2 = 0

put y = e^x

y^2-3y+2 =0

(y-1) (y-2) =0

y =1 or y =2

e^x = 1 or e^x =2

x = ln 1 or ln2

x =0 or ln2


May 15th, 2015

Hey I just noted that there is some problem with the question. Please ignore my first two comments.

e^(2x+3) - 3e^(x) +2 =0 is not solvable.

Writing the power term in brackets

it has to be "-2"

e^(2x+3) - 3e^(x)  - 2 =0  which can be simplified to 

e^(2x)*e^(3) - 3 e^(x) - 2 = 0

e^(3) = 20 and e^(x) = y

it will become 20 y^(2)  - 3y - 2 =0

Solving for y, you will get y = 0.4 as e^(x) cannot be a negative or an imaginary term

and x = -0.91629073187

Please confirm if the last term is -2 in the question.



May 15th, 2015

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