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ilanshan · Jun 25, 2007, 04:34 AM

Hi friends

I need help with this question

3^x = 10/3 -3^-x (the sign of ^ mean Squared)

I need to find the rate of X (the answers are +1, -1)

I hope you will be able to help me and show me the way to solve this problem

thank you

Ilan :)

Capuchin · Jun 25, 2007, 04:52 AM

do you mean $3^x = \frac{10}{3}-3^{-x}$ ?

By rate do you mean the differential? With respect to what? At what point?

Otherwise I am not sure of what you mean by rate

galactus · Jun 25, 2007, 05:01 AM

What do you mean by 'rate of X'. It appears you have to solve for x.

$3^{x}=\frac{10}{3}-3^{-x}$

$3^{x}+\frac{1}{3^{x}}=\frac{10}{3}$

Multiply by $3^{x}$

$3^{x}\cdot{3^{x}}+3^{x}\cdot\frac{1}{3^{x}}=3^{x}* \frac{10}{3}$

$3^{2x}+1-3^{x}\cdot\frac{10}{3}=0$

Now, let $u=3^{x}$ and you get:

$u^{2}-\frac{10}{3}u+1=0$

Now solve the quadratic and remember what u equals.

ilanshan · Jun 25, 2007, 11:50 AM

Originally Posted by Capuchin

do you mean $3^x = \frac{10}{3}-3^{-x}$ ?

By rate do you mean the differential? With respect to what? At what point?

Otherwise I am not sure of what you mean by rate

Can you please see the answer that the other guy gave me. And maybe you will have another way to solve this problem
:)

Capuchin · Jun 25, 2007, 11:58 AM

I still don't know what you mean by rate.

ilanshan · Jun 26, 2007, 09:36 AM

Originally Posted by Capuchin

I still don't know what you mean by rate.

I want to know how to find the value of X

Thank you :)

Capuchin · Jun 26, 2007, 09:55 AM

Oh well then we're fine :)