A log question (probably easy)

[meee]

find the solution(s) to 2e^2x = e^x

its on a practise exam I am doing. i did with my calculator, but how could this be done without a calculator?

 

[Hurkyl]

You certainly seem to be aware that logarithms ought to be useful -- so use them.

 

[meee]

ohh yeah like this?

x=loge(2e^2x)
x = loge2 + loge(e^2x)
x = loge2 + 2xloge(e )
x = loge2 +2x
-x = loge2
x= -loge2

 

[Hurkyl]

That looks right.

Now, there's another way to solve this problem too. It comes up often enough that you should know about it (or at least will know about it). e^(2x) is just (e^x)^2 -- so your equation is a quadratic equation in e^x... and you know how to solve quadratic equations.