Solve exponential equations

[Aladdin]

$\displaystyle e^3x+1 -5e^2x+1 +4e^x+1=0$

How would I type this correctly ......

Give me hints please : :

Let e^x+1 =X for example ?>

 

[pka]

Is the problem suppose to be $\displaystyle e^{3x+1}-5e^{2x+1}+4e^{x+1}$?

[tex]e^{3x+1}[/tex]

gives $\displaystyle e^{3x+1}$

 

[Aladdin]

Yes thanks pka ..

So .. ?

 

[mmm4444bot]

Hi Aladdin:

e^(3x + 1) = e^(3x) * e^1

This allows you to factor out the common factors of e from each term.

Next, e^(3x) = (e^x)^3.

This allows you to write a cubic equation, after substituting z = e^x.

z^3 - 5z^2 + 4z = 0

Do you need more hints?

Cheers,

~ Mark

 

[Aladdin]

Re:

Hi Aladdin:

e^(3x + 1) = e^(3x) * e^1

This allows you to factor out the common factors of e from each term.

Next, e^(3x) = (e^x)^3.

This allows you to write a cubic equation, after substituting z = e^x.

z^3 - 5z^2 + 4z = 0

Do you need more hints?

Cheers,

~ Mark

Thanks Mark , long time Mate .

Enough but let's wait for pka's reply .

 

[pka]

Re: Re:

Enough but let's wait for pka's reply .

Just finish what Mark has started.

 

[BigGlenntheHeavy]

$\displaystyle We \ have \ e^{3x+1}-5e^{2x+1}+4e^{x+1} \ = \ 0.$

$\displaystyle Using \ my \ trusty \ TI-89 \ ( F2-2), \ this \ factors \ to \ e^{x+1}(e^{x}-4)(e^{x}-1) \ = \ 0$

$\displaystyle Hence \ x \ = \ ln(4) \ or \ x \ = \ 0, \ QED.$

 

[Aladdin]

Re: Re:

Enough but let's wait for pka's reply .

Just finish what Mark has started.

Ok-

$\displaystyle (e^1)(z^3-5z^2+4z)=0$

e^1 can't be equal zero ... Impossible ?

z=1 or z=4

e^x=1 or e^x=4

x=0 or x=ln(4)

Correct .. ?

 

[Deleted member 4993]

Re: Re:

Enough but let's wait for pka's reply .

Just finish what Mark has started.

Ok-

$\displaystyle (e^1)(z^3-5z^2+4z)=0$

e^1 can't be equal zero ... Impossible ?

z=1 or z=4 or z = 0

e^x=1 or e^x=4

x=0 or x=ln(4)

Correct .. ?

 

[Aladdin]

yes yes Mr khan , you're right -- I thought it was quadratic , I didn't realize the cubic root.

 

[Aladdin]

Another Equation :

$\displaystyle e^{x}-4e^{-x}+3=0$

$\displaystyle e^{x}-4(\frac{1}{e^{x}})+3=0$

Next : ? : Would I let 1\e^x=Z

 

[Deleted member 4993]

Another Equation :

$\displaystyle e^{x}-4e^{-x}+3=0$

$\displaystyle e^{x}-4(\frac{1}{e^{x}})+3=0$

$\displaystyle e^{2x} - 4 + 3e^x \, = \, 0$

Next : ? : let e^x = Z

 

[Aladdin]

Another Equation :

$\displaystyle e^{x}-4e^{-x}+3=0$

$\displaystyle e^{x}-4(\frac{1}{e^{x}})+3=0$

$\displaystyle e^{2x} - 4 + 3e^x \, = \, 0$

Next : ? : let e^x = Z

Oh yes , you multiplied by e^x -- Now let e^x=Z -- And continue normal .

Thanks Mr khan.