How do I find a solution to e^{2x}=8e^{x}+20?

Alex Cervantes

Alex Cervantes

Answered question

2022-02-01

How do I find a solution to e2x=8ex+20?

Answer & Explanation

Larissa Hogan

Larissa Hogan

Beginner2022-02-02Added 10 answers

Step 1
e2x8ex20=0
By properties of exponents, we have:
(ex)28ex20=0
We let t=ex
t28t20=0
(t10)(t+2)=0
t=10 and 2
ex=10 and ex=2
Take the natural logarithm of both sides.
ln(ex=ln10 and lnex=ln2
Use the rule logan=nloga:
xlne=ln10 and xlne=ln2
ln and e are opposites, so equal 1
x=ln10 and x=0, since y=lnx has a domain of x>0

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?