## anonymous 5 years ago solve the following for x: 3e^(-0.15x+1)=6

1. anonymous

First isolate the $$e^{-.15x + 1}$$

2. anonymous

So divide 3 out?

3. anonymous

Yes

4. anonymous

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

5. anonymous

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

6. anonymous

Remember that $ln\ e^a = a$

7. anonymous

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

8. anonymous

I don't think so. $3e^{(-0.15x+1)}=6$ $\implies e^{-.15x + 1} = 2$ $\implies ln\ e^{-.15x + 1} = ln 2$ $\implies -.15x + 1 = ln 2$ $\implies -.15x = (ln2)-1$ $\implies x = \frac{(ln2) -1}{-.15}$

9. 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?

10. anonymous

Yes

11. anonymous

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

12. 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.

13. anonymous

0=-3, 2?

14. anonymous

k = -3 and k = 2.

15. anonymous

whoops that's what I meant!

16. anonymous

Since k = $$2^x$$ that means that $$2^x = -3$$ and $$2^x = 2$$ are solutions.

17. anonymous

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

18. anonymous

from which equation?

19. anonymous

From those last two equations.

20. anonymous

$2^x = -3$ and $2^x = 2$

21. 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

22. anonymous

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

23. 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.