log functions help?

Question

Solve the equation for x by first making an appropriate substitution, such as u = ex.
5e^2x - 16e^x + 3 = 0

Answer 1

u=e^x

5u^2-16u+3=0
(5u-1)(u-3)=0
u=1/5 or 3
x=ln(1/5) or ln(3)

Answer 2

Let u = e^x

5e^2x - 16e^x + 3 = 0 becomes 5u^2 - 16u + 3 = 0.

Now factor the equation:

(5u -1)(u - 3) = 0

5u - 1 = 0 or u - 3 = 0

u = 1/5 or u = 3.

We've solved for u, so now we can back substitute using the term that we originally used in the substitution: u = e^x, so:

e^x = 1/5
ln(e^x) = ln(1/5)
x = ln(1/5)

Since 1/5 = 5^-1, ln(5^-1) = -ln 5.

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

Answer 3

I think you made a typing error.

Did you mean to say u = e^x?

If so, replace e^x with u in the equation and factor.

Your question 5e^2x - 16e^x + 3 = 0 becomes

5u^2 -16u + 3 = 0.

Can you solve for u now?

Guido

Answer 4

5e^2x - 16e^x + 3 = 0

let u = e^x
5*u^2 -16u +3

Now solve for u.
Once you have u
find ln(u) to get x

Answer 5

Well, if we let u = e^x, we get 5u^2 - 16u + 3 = 0, or (5u-1)(u-3) = 0.
So u = 0.2 or 3, so x = ln 0.2 or ln 3.