5e^{2x} = 6 + 29e^x I would ususally multiply each side by a log value to get those exponents by themselfs, but that 6 is throwing me off. How do I get rid of it? Or do i just have to multiply it by log too?

Maribel Mcintyre

Maribel Mcintyre

Answered question

2022-10-20

5 e 2 x = 6 + 29 e x
I would ususally multiply each side by a log value to get those exponents by themselfs, but that 6 is throwing me off. How do I get rid of it? Or do i just have to multiply it by log too?

Answer & Explanation

Adalyn Pitts

Adalyn Pitts

Beginner2022-10-21Added 15 answers

Hint: Let y = e x and solve the quadratic equation you obtain by substitution.

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?