# can't solve exponential

• Jun 9th 2008, 07:12 AM
i_zz_y_ill
can't solve exponential
I am trying to draw x=3e^4t+12e^-t where t is x axis and x is y axis

I don't understand how from visualising this i can tell it is a growing function and doesntcross the t axis. I drew for instance y = -2e^4t +12e^-t since I knew it would cross the t axis as I could put y=0 and solve it.

Putting x=o however apparantly cannot be solved. Where am I going wrong,

0 = 3e^4t + 12e^-t
-e^4t = 4e^-t
e^4t = -4e^-t
lne^4t = -ln4 + lne^-t
4t = -ln4 -t
t = -(ln4)/5
• Jun 9th 2008, 07:21 AM
Isomorphism
Quote:

Originally Posted by i_zz_y_ill
e^4t = -4e^-t
lne^4t = -ln4 + lne^-t

Wrong!

$e^{4t} = -4e^{-t} \Rightarrow \ln e^{4t} = \ln(-4) + \ln{e^{-t}} \Rightarrow 4t = \ln(-4) - t \Rightarrow t = \frac{\ln(-4)}{5}$

But actually there is no real number called $\frac{\ln(-4)}{5}$ :)

What really should have been done was since $e^x > 0, \forall x \in \mathbb{R}$

$3e^{4t} > 0 , 12e^{-t} > 0 \Rightarrow x = 3e^{4t}+ 12e^{-t} > 0$

Thus $x \neq 0$