An exponential problem?

t=e^x

t+t^2-t^3=0

t(1+t-t^2)=0

t=0

and

1+t-t^2=0

t^2-t-1=0

t=(1-sqrt(5))/2 and t=(1+sqrt(5))/2

When t=0:

e^x=0 no solutions

When t=(1-sqrt(5))/2 :

e^x=(1-sqrt(5))/2 no solutions because (1-sqrt(5))/2 is negative

When t=(1+sqrt(5))/2 :

e^x=(1+sqrt(5))/2

x=ln((1+sqrt(5))/2)

x=0.48121...

ln is a logarithm with base e. Notation might be different in you country :)

...presentation:

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

so,

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

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

use "u-substitution"

u = e^x

u[u^2 - u - 1] = 0

u = 0

so, e^x = 0

there is no solution to this factor...extraneous

or

u^2 - u - 1 = 0

use the quadratic formula:

u = [1 ± √5]/2 [this is the golden ratio (φ) for the positive value]

when u = negative value, there is no solution...

so,

e^x = φ

x = ln φ

check

qed

Let y = e^x

Then, y + y^2 = y^3

y^2 - y - 1 = 0

This is a standard quadratic equation for which you have to find the positive solutions (note e^x can not be negative or zero)

y =(1 + sqrt(5))/2

x = ln(y) = ln((1 + sqrt(5))/2)) = 0.48121.....

e^x + e^2x = e^3x

e^x(1+e^x)=e^2x.e^x

e^x never be zero.

so 1+e^x=e^2x

ie,e^2x-e^x-1=0 which is a quadratic equation in e^x.

so e^x =[1+sqrt(1+4)]/2 or e^x =[1-sqrt(1+4)]/2.But the latter is inadmissible because e^x cant take negative values.so e^x =[1+sqrt(1+4)]/2

ie,e^x=1.618

ie,x=log 1.618=0.208978517=0.209

let e^x = y [x = ln(y)]

e^nx = [e^x]^n

so y + y^2 = y^3

y^3 - y^2 - y = 0

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

y=0 (impossible, see later) or y^2 - y - 1 = 0

(y - 0.5)^2 - 1 = 0.25

(y - 0.5)^2 = 5/4

y - 0.5 = rt(5)/2

y = 0.5 +/- rt(5)/2

If y = 0, x = ln(0) which is undefined

If y = 0.5 +/- rt(5)/2, x = ln[0.5 +/- rt(5)/2]

e^x + e^2x = e^3x

e^x(e^0 + e^x) = e^x * (e^x)^2

e^x = the Golden Mean (1.618033989...)

so x = ln(1.618033989...) = 0.481211825 or 0.481 to 3 sig.fig.