Thread: Natural Log

1. Natural Log

Hi All, please see if my steps are alright below:

Solve for x

$e^{2x} - 7e^x + 12$

$e^{2x} - 7e^x = -12$

$e^{2x} - (e^x)^{7} = -12$

$ln(\frac{e^{2x}}{e^(7x)} = e^{-12}$

$ln(e^{-5x} = e^{-12}$

$-5x = -12 \rightarrow x = \frac{12}{5}$

2. Re: Natural Log

Originally Posted by BobBali
Hi All, please see if my steps are alright below:

Solve for x

$e^{2x} - 7e^x + 12$

$e^{2x} - 7e^x = -12$

$e^{2x} - (e^x)^{7} = -12$

$ln(\frac{e^{2x}}{e^(7x)} = e^{-12}$

$ln(e^{-5x} = e^{-12}$

$-5x = -12 \rightarrow x = \frac{12}{5}$
1. I assume that you mean:

$e^{2x} - 7e^x + 12=0$

If so

2. Use the substitution $y = e^x$

Your equation becomes

$y^2-7y+12=0$

3. Use the quadratic formula to solve for y. Afterwards use the equation of #2 to determine x.

3. Re: Natural Log

Originally Posted by earboth
1. I assume that you mean:

$e^{2x} - 7e^x + 12=0$

If so

2. Use the substitution $y = e^x$

Your equation becomes

$y^2-7y+12=0$

3. Use the quadratic formula to solve for y. Afterwards use the equation of #2 to determine x.
Quadratic formula? Surely it's easier if you note that \displaystyle \begin{align*} y^2 - 7y + 12 &= (y - 3)(y - 4) \end{align*}...

4. Re: Natural Log

Originally Posted by Prove It
Quadratic formula? Surely it's easier if you note that \displaystyle \begin{align*} y^2 - 7y + 12 &= (y - 3)(y - 4) \end{align*}...
I get y=4 or 3

Solving for x using e^x =4. Gives ln4= 1.39 and 1.1

While my way x= 12/5 = 2.4

Does this mean i$m wrong??! 5. Re: Natural Log Originally Posted by BobBali Does this mean i$m wrong??!
You have a severe error in the OP.

It is: $7e^x\ne (e^x)^7.$

And that is only one of several.