A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
solve the following for x: 3e^(0.15x+1)=6
anonymous
 5 years ago
solve the following for x: 3e^(0.15x+1)=6

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First isolate the \(e^{.15x + 1}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0leaves you with e^(0.15x+1)=(6/3) and I'm still stuck...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 6/3 = 2. Then you take the natural log of both sides

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Remember that \[ln\ e^a = a\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0obviously correct me if I'm wrong but would that leave me with 0.15x+1/x2/x=0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think so. \[3e^{(0.15x+1)}=6\] \[\implies e^{.15x + 1} = 2\] \[\implies ln\ e^{.15x + 1} = ln 2\] \[\implies .15x + 1 = ln 2\] \[\implies .15x = (ln2)1\] \[\implies x = \frac{(ln2) 1}{.15}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh. Hmm what if you have a polynomial. The example I have is 2^(2x)+2^(x)6=0. Are the steps much different?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For that you'd want to start by solving the quadratic.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let \(k = 2^x \implies 2^{2x} = (2^x)^2 = k^2\) \[\implies 2^{2x} + 2^x 6 = k^2 + k 6 = 0\] Use the quadratic formula (or factoring) to solve for k.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whoops that's what I meant!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since k = \(2^x\) that means that \(2^x = 3\) and \(2^x = 2\) are solutions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So to find x we again take the log of both sides.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From those last two equations.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2^x = 3\] and \[2^x = 2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry for the extremely late reply wouldn't that lead you with ln2^x=3ln and ln2^x=2ln? but to get x alone I'm lost

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, well when you take the log of something you can bring out (and down) the exponent. \[ln\ 2^x= x(ln\ 2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you have \[x(ln\ 2) = ln\ 2\] and \[x(ln\ 2) = ln\ 3\] But (ln 3) is not defined, so the first solution is the only valid one. And when you divide both sides by (ln 2) you have x = 1.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.