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 :)

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

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.

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
:)

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

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

I want to know how to find the value of X

Thank you :)

Oh well then we're fine :)