anonymous
  • anonymous
solve the following for x: 3e^(-0.15x+1)=6
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
First isolate the \(e^{-.15x + 1}\)
anonymous
  • anonymous
So divide 3 out?
anonymous
  • anonymous
Yes

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anonymous
  • anonymous
leaves you with e^(-0.15x+1)=(6/3) and I'm still stuck...
anonymous
  • anonymous
Well 6/3 = 2. Then you take the natural log of both sides
anonymous
  • anonymous
Remember that \[ln\ e^a = a\]
anonymous
  • anonymous
obviously correct me if I'm wrong but would that leave me with -0.15x+1/x-2/x=0?
anonymous
  • 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
  • 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
  • anonymous
Yes
anonymous
  • anonymous
For that you'd want to start by solving the quadratic.
anonymous
  • 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
  • anonymous
0=-3, 2?
anonymous
  • anonymous
k = -3 and k = 2.
anonymous
  • anonymous
whoops that's what I meant!
anonymous
  • anonymous
Since k = \(2^x\) that means that \(2^x = -3\) and \(2^x = 2\) are solutions.
anonymous
  • anonymous
So to find x we again take the log of both sides.
anonymous
  • anonymous
from which equation?
anonymous
  • anonymous
From those last two equations.
anonymous
  • anonymous
\[2^x = -3\] and \[2^x = 2\]
anonymous
  • 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
  • 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
  • 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.

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