Solve e^ax = C(e^bx) ; a does not equal b

Question

anonymous · Answer

this is 
$$e^{ax}=C\times e^{bx}$$ right?

myininaya · Answer

$$C=e^{\ln(C)}  => C \cdot e^{bx}=e^{\ln(C)+bx}$$

myininaya · Answer

so we have
ax=ln(C)+bx

myininaya · Answer

assuming C>0

anonymous · Answer

take the log and get 
$$ax=ln(Ce^{bx})=\ln(C)+\ln(e^{bx})=\ln(C)+bx$$
$$ax-bx=\ln(C)$$
$$(a-b)x=\ln(C)$$
$$x=\frac{\ln(C)}{a-b}$$

myininaya · Answer

zarkon will you look at satellite and my debate?

anonymous · Answer

ok this time you really do win because you way is slicker.  same thing though

myininaya · Answer

http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e9e3faa0b8b81e688064d40

myininaya · Answer

lol

anonymous · Answer

which means i hope that when you teach finding derivatives of exponentials like 
$$x^{\sin(x)}$$ you rewrite them first as 
$$e^{\sin(x)\ln(x)}$$ and use the chain rule

myininaya · Answer

i like letting y=that mess and taking natural log of both sides

anonymous · Answer

inconsistent then aren't you?

myininaya · Answer

lol yes yoda

myininaya · Answer

you should have left then out
it would have sounded more like yoda

anonymous · Answer

i look like yoda too! well, maybe jabba the hut...