solve the following for x: 3e^(-0.15x+1)=6

- anonymous

solve the following for x: 3e^(-0.15x+1)=6

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

First isolate the \(e^{-.15x + 1}\)

- anonymous

So divide 3 out?

- anonymous

Yes

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

leaves you with e^(-0.15x+1)=(6/3) and I'm still stuck...

- anonymous

Well 6/3 = 2. Then you take the natural log of both sides

- anonymous

Remember that \[ln\ e^a = a\]

- anonymous

obviously correct me if I'm wrong but would that leave me with -0.15x+1/x-2/x=0?

- anonymous

I 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

Oh. Hmm what if you have a polynomial. The example I have is 2^(2x)+2^(x)-6=0. Are the steps much different?

- anonymous

Yes

- anonymous

For that you'd want to start by solving the quadratic.

- anonymous

Let \(k = 2^x \implies 2^{2x} = (2^x)^2 = k^2\) \[\implies 2^{2x} + 2^x -6 = k^2 + k -6 = 0\] Use the quadratic formula (or factoring) to solve for k.

- anonymous

0=-3, 2?

- anonymous

k = -3 and k = 2.

- anonymous

whoops that's what I meant!

- anonymous

Since k = \(2^x\) that means that \(2^x = -3\) and \(2^x = 2\) are solutions.

- anonymous

So to find x we again take the log of both sides.

- anonymous

from which equation?

- anonymous

From those last two equations.

- anonymous

\[2^x = -3\] and \[2^x = 2\]

- anonymous

Sorry 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

Ah, well when you take the log of something you can bring out (and down) the exponent. \[ln\ 2^x= x(ln\ 2)\]

- anonymous

So 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.