rewrite the equation as a quadratic function where u = e^x then solve for x?

Question

e^-2x - 3e^-x = -2

i dont understand the directions.
i dont understand the problem.
IM LOST IN AP CALC ALREADY and this is only the summer assignment.
im going to get a [*]on the test at the end of the yr
oh well can someone help please?
oh and in case you were confused, e is the natural log e thing, not just an algebraic term

Answer 1

Hi,



u=e^x

plug in e^-2x-3e^-x+2=0

1-3e^x+2e^x=0

2u^2-3u+1=0

2u^2-2u-u+1=0

2u(u-1)-1(u-1)=0

(u-1)(2u-1)=0

u=1,u=1/2

u=e^x

1=e^x

so x=natural log1 to the base e=0, clearly x=0



1/2=e^x

x=log(1/2) to the base e=to get this answer calculate with calculator

Answer 2

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



This is the same as



1/[e^(2x)] - 3/e^x = -2



In order to eliminate all fractions, multiply both sides by e^(2x) (e^x).



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



And we can simplify the right hand side, because whenever we multiply same base exponentials, we can add the exponents.



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

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



Move everything to the left hand side,



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



And let's re-express e^(3x) as (e^x)^3 and e^(2x) as (e^x)^2.



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



Let's factor (e^x),



(e^x) [ 2 (e^x)^2 - 3e^x + 1 ] = 0



Since we know e^x cannot equal zero (The range of e^x is strictly all real numbers greater than 0), we can divide both sides of the equation by e^x, to get



[ 2 (e^x)^2 - 3e^x + 1 ] = 0



From here, we can use substitution

Let u = e^x. Then we get



2u^2 - 3u + 1 = 0



Which we solve as an ordinary quadratic.



(2u - 1)(u - 1) = 0



Therefore,



2u - 1 = 0

2u = 1

u = 1/2, but back-substitute u = e^x, we get

e^x = 1/2, therefore

x = ln(1/2)



u - 1 = 0

u = 1, which means

e^x = 1

x = ln(1)

x = 0



So our answer is x = { ln(1/2), 0 }

Answer 3

e^-2x - 3e^-x = -2

Multiply through by e^2x, to get it so that you only have positive exponents in e^x

now (remember you add powers when you multiply)

1 - 3e^x = -2e^2x

2e^2x - 3e^x +1 = 0

The trick here is to realise that this is

2(e^x)^2 - 3e^x +1 = 0

So you can treat this as a quadratic in e^x. So let u = e^x

2u^2 - 3u +1 = 0

(2u-1)(u-1) = 0

Therefore either u = 1 or u = 1/2

So e^x = 1 or e^x = 1/2

so x = ln1 = 0

or x = ln1/2



Hope that helps.

Answer 4

if u put u = e^x, the eqn becomes u^(-2) - 3u^(-1) = - 2

now multiply everything by u^2

1 - 3u = - 2u^2

so 2u^2 - 3u + 1 = 0

solve for u by factorizing

(2u - 1)(u -1) = 0

so e^x = u = 1/2 or = 1

so (take logs on both sides)

x = ln(1/2) = - ln(2) OR x = ln(1) = 0



(some books write log for ln, depends!)

Answer 5

e^-2x - 3*e^-x = -2

lets work by separated term to be more clear



1.- e^-2x => 1/ e^2x => 1 / (e^x * e^x)



2.- 3*e^-x => 3 / e^x





1 / (e^x * e^x) - 3 / e^x = - 2



(1 - 3*e^x) / e^x = - 2





multipling both sides by e^x :



1 - 3*e^x = - 2*e^x

1 = e^x



x = 0