Equation

[zeralda21]

Would like some help with an equation.

 

f(x)=e^2x

 

g(x)=lnx

 

The equation is: f(g(x))=g(f(x^3)). f(g(x))=e^2lnx and g(f(x^3))=ln(e^(2x^3)

 

e^2lnx=ln(e^(2x^3)

 

I´m stuck now. I don´t know how to continue to simplify further than this. Would like a hint or a tip so I can proceed.

Edited April 3, 2011 by zeralda21

[insane_alien]

i'd probably move the e^2 into the Ln(x) and then get rid of the ln's and go from there.

 

whether its the right approach or not, well, i suppose you'll find out.

[CaptainBlood]

Try to do e^(e^2lnx) = e^(ln(e^(2x^3)) then you should get e^2 = 2x^3 and solve for x

[zeralda21]

Thanks. I´ll try to work from here. I´ll be back with the result.

[` melody.]

e^2lnx = e^{lnx^2}

And given that e and ln are opposite functions, e^{lnx^2} cancel out to just = x²

So you probably would simplify it down to x² = 2 x³ and then solve for x.

I'm not 100% sure though. Hope it helps.

[Hal.]

More to follow ....

 

 

Edited April 5, 2011 by hal_2011

[zeralda21]

e^2lnx = e^{lnx^2}

And given that e and ln are opposite functions, e^{lnx^2} cancel out to just = x²

So you probably would simplify it down to x² = 2 x³ and then solve for x.

I'm not 100% sure though. Hope it helps.

 

Yes, you are right, I couldn´t figure out e^2lnx at first. x^2=2x^3 is the simplified equation. x^2-2x^3=0

x=0 or 1/2. But since ln is not defined for x=0, the answer is x=1/2.

 

Anyway, thanks!

[Hal.]

A solution has been found , the original poster is happy and I need the practice ! I didn't finish with a quadratic so ....... interesting ! Let me know if there is a mistake .

Edited April 6, 2011 by hal_2011

[zeralda21]

A solution has been found , the original poster is happy and I need the practice ! I didn't finish with a quadratic so ....... interesting ! Let me know if there is a mistake .

 

This is really good, different altough a correct answer. Thanks.