Question

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

Answer 1

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

Answer 2

So divide 3 out?

Answer 3

Yes

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

Yes

Answer 11

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

Answer 12

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.

Answer 13

0=-3, 2?

Answer 14

k = -3 and k = 2.

Answer 15

whoops that's what I meant!

Answer 16

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

Answer 17

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

Answer 18

from which equation?

Answer 19

From those last two equations.

Answer 20

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

Answer 21

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

Answer 22

Ah, well when you take the log of something you can bring out (and down) the exponent.

\[ln\ 2^x= x(ln\ 2)\]

Answer 23

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.