# Homework Help: Isolating a variable in an exponent

1. Feb 24, 2013

### thatguythere

1. The problem statement, all variables and given/known data
So I had to find out
N=Noe-ux
Where No=1.5,e is the elementary charge, u=-0.068 and x=0.07
I came to the answer 1.221

Now I need to find x if u =-0.036, N=1.221 and No remains 1.5

2. Relevant equations

3. The attempt at a solution
I reduce the equation to the following

ln(N/No)=-ux

The problem is that when I plug my numbers from the original problem into this I end up getting the following

ln(1.221/1.5)=-(-0.068)x
-0.2058=0.068x
-3.02= x

This is clearly incorrect since I know x should be 0.07.
Any help would be greatly appreciated.
Thank you.

Last edited: Feb 25, 2013
2. Feb 25, 2013

### Mentallic

I get something different.

I'll assume u = -0.036 is a typo because it looks as though you want to use u = -0.068 again.

Just check your answer for N again, and see if that changes anything.

3. Feb 25, 2013

### thatguythere

My order of operations was wrong then?

So I get 1.273

Now -0.036 was NOT a typo, however I am ignoring it, as I am using the original numbers to try to figure out how to isolate x. However, when I plug this new number in, I still end up with x = -2.413 instead of 0.07 which it should be.
Clearly I am doing something very wrong and have no idea what.

4. Feb 25, 2013

### thatguythere

Let me illustrate step by step.

N=Noe-ux
N=1.5*1.6x10-19(0.068)(0.07)
N=1.5*0.0960.07
N=1.5*0.849
N=1.274

Now I try to reverse it to figure out how to isolate x, this is where I do something very wrong I believe.

N=Noe-ux
1.274=1.5*1.6x10-19(0.068)(x)
0.849=1.6x10-19(0.068)(x)
ln0.849=0.068x
-0.164=0.068x
-2.41=x

But we know x=0.07. I'm lost.

5. Feb 25, 2013

### Mentallic

Sorry, I mistook e as being the natural exponential 2.718...
We're supposed to be using $e=1.602\cdot10^{-19}$, right?

Using the figures:
$$N_0 = 1.5$$
$$e=1.602\cdot 10^{-19}$$
$$u=-0.068$$
$$x=0.07$$

You should get $N\approx 1.22$

6. Feb 25, 2013

### Mentallic

How did you go from the first to the second step?

7. Feb 25, 2013

### thatguythere

I got it.

8. Feb 25, 2013

### thatguythere

Haha Ok so I was correct in the first place when I posted 1.221. Then I was trying to figure out what I did wrong which would explain your confusion over my next process.

9. Feb 25, 2013

### Mentallic

Yes, it's not correct.

$$(n\cdot a^b)^{cd} = n^{cd}\cdot a^{bcd}$$

But you've done something else.
Of course, calculating them in this way is the long and inefficient way, so you'd instead enter it as so:

(1.6E-19)^(0.068*0.07)

EDIT: most of this is redundant, you already know it

Last edited: Feb 25, 2013
10. Feb 25, 2013

### Mentallic

Oh yeah sorry, that's my fault!

As for the second part of the problem where you're trying to find x, you've done essentially what confused me as well.

ln(N/No)=-ux

Remember that ln = loge is the natural logarithm, or in other words, it reverses the natural exponential operator.

If $a=e^{b}$ then $\ln(a)=b$ but we don't have the natural exponential (e=2.718...) here, we have another base altogether, so what we'd need to do to reverse the exponent operation is to take the log with base $1.6\cdot 10^{-19}$

But your calculator probably doesn't have the option of choosing any base for the logarithm, so you instead use the formula

$$\log_ab=\frac{\log_cb}{\log_ca}$$

So we have that

$$\log_{1.6\cdot 10^{-19}}(1.221/1.5) = \frac{\ln (1.221/1.5)}{\ln(1.6\cdot 10^{-19})}$$

11. Feb 25, 2013

### thatguythere

Let me try then.

12. Feb 25, 2013

### thatguythere

Are you saying that
$\frac{ln(1.221/1.5)}{ln(1.6E-19)}$=-ux?

13. Feb 25, 2013

### Mentallic

Yep!

14. Feb 25, 2013

### thatguythere

Yeah, I got that. So using -0.036, I come up with 0.132. Excellent.

15. Feb 25, 2013

### thatguythere

Thank you very much.

16. Feb 25, 2013

### Mentallic

You're welcome! And sorry about the confusion at the start.