e^2x − 5^ex − 36 = 0

Question

anonymous · Answer

do u guys know how to solve this?

zepdrix · Answer

This is the problem...?
$$\Large\rm e^{2x}-5^{ex}-36=0$$

zepdrix · Answer

Did you mean to write,$$\Large\rm e^{2x}-5e^x-36=0$$

anonymous · Answer

yes that is the problem

zepdrix · Answer

Rules of exponents let's us rewrite the first term,$$\Large\rm \left(e^{x}\right)^2-5e^x-36=0$$Then we can make a substitution to help us get through this,$$\Large\rm u=e^x$$Giving us,$$\Large\rm u^2-5u-36=0$$

zepdrix · Answer

It looks like it's factor-able from there.
Understand how to proceed?

anonymous · Answer

not quite

anonymous · Answer

its not factorable because 5 doesn't go into 36?

zepdrix · Answer

So we need factors of -36, they have to add to -5.

-6*6=-36
but -6+6 = 0.
Hmm those didn't work.

9*-4=-36
but 9+-4=5.
Hmm that didn't give us -5 but maybe we're on the right track with 9 and 4?

anonymous · Answer

i didn't get it because it needs to be a decimal number

zepdrix · Answer

That's because we're not done yet.
This is only half of the problem.
You need to be able to factor in order to get past this part though.